 Last 7 days

arxiv.org arxiv.org

[35,
citation 35 and 6 are the same citation (one in the conference, and one in arxiv). I think they should be merged.

width
square root of width?


arxiv.org arxiv.org

n≤d
\(d\) is the input dimension. But also the number of parameters, because we are looking at a linear model here.

We assume the feature matrix can be decomposed into the formX=X+ZwhereXis lowrank(i.e. rank(X)=r<<n) with singular value decompositionX=UVTwithU∈Rn×r,∈Rr×r,V∈Rd×r,andZ∈Rn×dis a matrix with i.i.d.N(0;2x~n)entries.
Noise model. Information space is the component of the inputs that live in a lowdim space (low rank component), and the nuisance space is the component that corresponds to i.i.d. noise, which will w.h.p. be of maximum rank

 Jul 2019

arxiv.org arxiv.org

r large, we obtain a low limit training accuracy and do not observe overfitting, asurprising fact since this amounts to solving an overparameterized linear system. This behavioris due to a poorly conditioned linearized model, see Appendix C.
Wait, so it seems that in all the experiments with CNNs you just found that the lazy training didn't converge to a global minimum of training error. So it doesn't mean they aren't generalizing well!
Is your Jacobian degenerate for the first set of experiments (with squared loss), because if not, then your theorem implies that they should converge to a global minimum right?

hat manages to interpolate the observations with just a smalldisplacement in parameter space (in both cases, near zero training loss was achieved).
zero training loss is achieved both in the lazy and nonlazy regime, but the nonlazy solution generalizes much better

Cover illustration.
I suppose that in both the lazy and nonlazy regime, it has reached a global minimum of training loss?

Theorem 2.5(Underparameterized lazy training).Assume thatFis separable,Ris stronglyconvex,h(w0) = 0andrankDh(w)is constant on a neighborhood ofw0. Then there exists0>0such that for all0the gradient flow(4)converges at a geometric rate (asymptoticallyindependent of) to a local minimum ofF.
Convergence to local minimum, removing assumption about nondegeneracy of Jacobian

In terms of convergence results, this paper's main new result is the convergence of gradient flow, and showing that it stays close to the tangent (linearized) gradient flow.
And saying this for general parametrized models. The assumption of nondegenerate Jacobian is related to overparametrization, as nondegeneracy is more likely when one is overparametrized.

he gradient flow needs to be integrated with a stepsize of order1=Lip(rF) = 1=Lip(h)2
size of step size for gradient flow to be a good approximation

s!1,supt0kw(t)w0k=O(1=)
How come it can find a minimum arbitrarily close to the initialization?
Ah I see by the nondegenerate Jacobian assumption, you can find a local change that will fit \(y^*\), and \(\alpha\) large is just needed to reach the overall size/scale of \(y^*\) with the local change

kh(w0)y?kis bounded
How realistic is this?

squareintegrablefunctions with respect tox
why do we need them to be squareintegrable?

are bound to reach the lazy regime as the sizes of all layers grow unbounded
and the learning rate tending to zero..

r
Remember this nabla is w.r.t. to its argument not parameters \(w\)


arxiv.org arxiv.org

Theorem 2.Assume thatis a Lipschitz, twice differentiable nonlinearity function, with boundedsecond derivative. For anyTsuch that the integralRT0kdtkpindtstays stochastically bounded, asn1;:::;nL1!1, we have, uniformly fort2[0;T]
Under the NTK parametrization, which they use, this limit implies that the learning rate (for GD on the standardparametrization) is \(O(1/\sqrt{n}\) (where \(n\) is layer size). So the parameters move less and less for a fixed \(T\), in this limit, which is, intuitively, why the NTK stays constant for this period of time until \(T\)
The interesting thing is that the function \(f\) can change, as all the parameters "conspire" for it to change. Therefore it can potentially fit a function, and find a global minimum, while the parameters have almost not moved at all.
I think the intuition for this "conspiracy" is that the total change in \(f\) is given by a sum over all the parameters' individual gradients. The number of parameters grows like \(n^2\). gradient w.r.t. last hidden layer activations is \(O(1/\sqrt{n})\), w.r.t to second to last hidden layer activations is \(O(\sqrt{n}(1/\sqrt{n})^2) = O(1/\sqrt{n})\), where the \(\sqrt{n}\) comes from variance of summing over all the activations in last hidden layer. This means that the gradient w.r.t. to a weight, in NTK parameterization, is \(O((1/\sqrt{n})^2)=O(1/n)\) In GD, each weight changes by an ammount of the same order as the gradient (assumin \(O(1)\) learning rate, which we assume for NTKparametrization learning rate), so each weight contributes to change \(f\) by \(O(1/n^2)\). Therefore the total contribution from all the weights is \(O(1)\). Note that the contributions all have the same sign as they are essentially the gradient w.r.t. that weight, squared, so they add linearly, (and not growing like \(\sqrt{n}\) if they were all randomly signed)

~(`)1pn`W(`)
i guess the first product here is elementwise, although it's not explicitly said

The connection weightsW(L)ijvary at rate1pnL, inducing a change of the same rate to the whole sum
From chain rule

N(0;1)
huh? no \(1/\sqrt{n}\) ?

ANNrealization functionF(L):RP! F, mapping parameterstofunctionsfin a spaceF.
:O we studied the same object in our paper! But we called it the parameterfunction map!

This shows a direct connection to kernel methodsand motivates the use of early stopping to reduce overfitting in the training of ANNs
But early stopping doesn't seem to often help in ANN training


arxiv.org arxiv.org

Consider the class of linear functions overX=Rd, with squared parametrization as follows
Seems quite artificial, but ok

duplicating units and negating their signs, the Jacobian of the modelis degenerate at initialization, or in their notationmin= 0
is this if the weights are tied only? Do they assume they are tied?

The data are generated by a5sparse predictoraccording toy(n)N(h;x(n)i;0:01)withd= 1000andN= 100.
perhaps large initialization is like a small L2 norm bias, and small initialization like an L1 norm bias. So the kernel regime is bad for learning sparse networks (I think Lee also says this in his talk)

training with gradient descent has the effect of finding the minimum RKHS norm solution.
they showed that for GD and logistic regression, but what about SGD, and square loss? I think for square loss you need either early stopping or regularization to get min norm solution?


arxiv.org arxiv.org

which forinstance can be seen from the explicit width dependence of the gradients in the NTK parameterization
Yeah but the NTK parametrization makes the gradients much smaller. For normal parametrization, gradient of individual weights is not infinitesimal right?


arxiv.org arxiv.org

Can we moreprovide theoretical justications for this gap?
are all our base belong to us?


arxiv.org arxiv.org

distance
the kernel distance?


arxiv.org arxiv.org

the case of a regression loss, the obtained modelbehaves similarly to a minimum norm kernel least squares solution
Only its expected value, see page 7 in Jacot2018, if I understood correctly

Stateoftheart neural networks are heavily overparameterized, making the optimization algorithma crucial ingredient
The fact that most naive learning algorithms work well, makes me question the "crucial" qualifier..


www.wikiwand.com www.wikiwand.com

which is a nonlinear
typo on above equation? this appears to be the same as the Schrodinger equation


arxiv.org arxiv.org

raining just the top layer with anℓ2loss is equivalent to a kernel regression for the following kernel:kerx,x′=Eθ∼W[f(θ,x)·fθ,x′],
This is the expected value of the kernel, not the actual kernel, which would correspond to a random features kernel right?
Hmm I think I remember random features converging when their number grows to infinity, but the product \(f(\theta,x)f(\theta,x')\) doesn't stochastically converge when the width grows to infinity right? Only its expectation converges

aGaussian Process (GP)[Neal,1996].This model as well as analogous ones with multiple layers [Lee et al.,2018,Matthews et al.,2018]and convolutional filters [Novak et al.,2019,GarrigaAlonso et al.,2019] make up the GaussianProcess view of deep learning. These correspond to infinitely wide deep nets whose all parametersare chosen randomly (with careful scaling), and only the top(classification) layer is trained.
Maybe, but these kernels also correspond to those of a fully trained ideal Bayesian neural network, with prior over weights given by the iid initialization

 Jun 2019

www.marxists.org www.marxists.org

He is not like that on account of a cowardly heart or lungs or cerebrum, he has not become like that through his physiological organism; he is like that because he has made himself into a coward by actions.
philosology does affect you as well though...


arxiv.org arxiv.org

f= logpd.
If the optimum of (2) is given by this when function is unrestricted, if we consider a family with zero "approximation error" (so that the optimum is in the family), then the optimum on the family is the same as over all functions


arxiv.org arxiv.org

we can employ efficient offpolicy reinforcement learningalgorithms that are faster than current metaRL methods,which are typically onpolicy (Finn et al., 2017a; Duanet al., 2016).
Why are previous metaRL algorithms typically onpolicy?

 May 2019

dspace.mit.edu dspace.mit.edu

D∗[T∗μ,T(Zm0)]
I see this as one of the main innovations of the paper. This term is a discrepancy between the sample, and the true distribution \(\mu\). This would allow Z_m to be sampled from a different distribution for instance, allowing to get bounds that account for distributional drift, for instance.

V[f]
This basically offers a measure of the variance of the loss (in a nonstatistical sense) over the instance space, of the learned function.

hus, in classical bounds including datadependent ones,asHgets larger and more complex, the bounds tend to become more pessimistic for theactual instance ˆyA(Sm)(learned with the actual instanceSm), which is avoided in Theorem1.
Sure, but that is also avoided in some statistial learning approaches, like Structural Risk Minimization, PACBayes, and the luckiness framework, which you cite!


arxiv.org arxiv.org

the total computational costis similar to that of singlehead attention with full dimensionality
smaller?

Multihead attention allows the model to jointly attend to information from different representationsubspaces at different positions. With a single attention head, averaging inhibits this.
So if I understand correctly, with a single head, different parts of the d_modeldimensional query vector may "want" to attend to different parts of the key, but because the weight of the values is computed by summing over all elements in the dot product, it would just average these local weights. Sepparating into different heads, allows to attend to different value vectors for different "reasons".

 Apr 2019

Local file Local file

theprobability
log probability

say
~~

too weak
for Kmax

tighter
in relative terms

o generatex
given the map \(f\)

rst e
first give x, and then enumerate ... identifying all inputs p mapping to x, namely \(f^{1}(x)\)

lays a key rolein
is the main component in

,
:

function
of

NI= 2n.
for binary strings

derived
suggested

Since manyprocesses in science and engineering can be described asinputoutput maps that are not UTMs
Perhaps say "This suggests that, even though many maps are not UTMs, the principle that low K are high P should hold widely"
because it is not because they are not UTMs, but it is in spite of them not being UTMs, I would argue.

, a classic categorization of machinesby their computational power,
in parenthesis


www.cs.toronto.edu www.cs.toronto.edu

k(y,x,x′,y′)
should be \(k(y,x,y',x')\) right?


distill.pub distill.pub

If we add a periodic and a linear kernel, the global trend of the linear kernel is incorporated into the combined kernel.
Remember that kernel functions with one of its arguments evaluated are members of the reproducing kernel Hilbert space to which all the functions supported by a particular Gaussian process belong.
Therefore adding kernels, amounts to adding the functions on these two spaces. That is why the resulting functions work like this when combining kernels!


www.jmlr.org www.jmlr.org

concave in both arguments. Jensen’s inequality (f(x,y)concave⇒E f(x,y)≥f(Ex,Ey))
Actually it's convex

 Mar 2019

www.jmlr.org www.jmlr.org

A stochastic error rate,ˆQ(~w,μ)S=E~x,y∼S ̄F(μγ(~x,y)
Remember that the w sampled from the "posterior" isn't necessarely parallel to the original w, so that the stochastic classification rate isn't simply F(sign(margin)) but something more complicated; see the proof.

Since the PACBayes bound is (almost) a generalization of the Occam’s Razor bound, the tightnessresult for Occam’s Razor also applies to PACBayes bounds.
Oh, c'mon :PP You are just showing that PACBayes is tight as a statement for all Q and for a particular P. As in you are saying that if we only let it depend on the quantities it can depend (namely KL divergence between Q and P, delta, etc), then it can't be made tighter, because then it would break for the particular choice of D, hypothesis class, Q, and for any value of KL in that case in the Theorem 4.4 above.
> What I mean is this: that we say the bound is a function f(KL, delta, m, etc). Theorem 4.4 shows that there is a choice of learning problem and algorithm such that these arguments could be anything, and the bound is tight. Therefore, we can't lower this bound without it failing. It is tight in that sense. However, it may not be tight if we allow the bound to depend on other quantities!

The lower bound theorem implies that we can not improve an Occam’s Razor like statement.
Yeah, as in if it only depends on \(P(c)\) and the other quantities expressed there, and have it not depend on the algorithm, so it should be a general function that takes \(P(c)\), \(\delta\) etc, but the same function for any algorithm. Then yes. And this is what they mean here.

For all P(c), m, k,δthere exists a learning problem D andalgorithm such that
Depends what do you mean by For all \P(c)) are you fixing the hypothesis class or what? Because your proof assumes a particular type of hypothesis class... For P(c) having support over a hypothesis class where the union bound doesn't hold, then it is not tight any more..

The distributionDcan be drawn by first selectingYwith a single unbiased coin flip, and thenchoosing theith component of the vectorXindependently, Pr((X1, ...,Xn)Y) =Πni=1Pr(XiY). Theindividual components are chosen so Pr(Xi=YY) =Bin(m,k,δP(c)).The classifiers we consider just use one feature to make their classification:ci(x) =xi. The trueerror of these classifiers is given by:cD=Bin(m,k,δP(c))
Ok, so this has proven that the Occam bound is tight for this particular \(D\) for this particular hypothesis class, which is quite special, because it has the property that the union bound becomes tight. But that is a very special property of this hypothesis class (or more general, of this choice of support for \(P\) right??)

if any classifier has a toosmall train error, thenthe classifier with minimal train error must have a toosmall train error
this is because having "toosmall train error" here means (is equivalent to) having train error smaller than \(k\), so that the classifier with smallest train error also has it smaller than \(k\) and therefore it also has too small train error.

The differences between the agnostic and realizable case are fundamentally related to the decrease inthe variance of a binomial as the bias (i.e. true error) approaches 0.
If we observe a zero empirical error, then the probability of observing that decreases very quickly with increasing true error. So I wouldn't really say it's the decrease in variance of the binomial (one could imagine distributions where the variance doesn't decrease as you go to 0, but which still have the property that makes the realizable case error rate smaller in the same way as here)

cS
Remember they define training error as error count

PrS∼DmcD≤Bin(m,ˆcS,δ)≥1−δ.
Btw, this approach only works because \(c_D\) is a one dimensional random variable, so that {the set of \(k\) such that \(Bin(m,k,p)<\delta\)} equals {the set of \(k\) less than or equal to {the maximum \(k\) such that \(Bin(m,k,p)<\delta\)}}.
This happens because the different ("confidence interval") set of \(k\) defining \(Bin(m,k,p)\) (i.e. all \(k\) smaller than or equal to some \(k\)) are all nested. In more general situations with other confidence intervals defined (like for 2D cumulative distributions) this may not happen.

δ∈(0,1]
No,
Because Lemma 3.6 requires \(\frac{k}{m}<p\), then we need \(\delta\) to be small enough such that {{the \(c_D\) that results from solving the Chernoff bound = \(\delta\)} be larger than \(\frac{k}{m}\)}</p>

ε
\(c_D\)

The test set bound is, essentially, perfectly tight. For any classifier with a sufficiently large trueerror, the bound is violated exactly aδportion of the time
And any tighter bound would be violated a larger portion of the time, at least for some value of the true error (note that for fixed or constrained true error one can have tighter bounds).
The need for a sufficiently large true error is because for instance for true error zero, the bound is never violated.
But still the fact of it being perfectly tight is because of what I said above.

more
less

All of the results presented here fall in the realm of classical statistics. In particular, all randomizations are over draws of the data, and our results have the form of confidence intervals.
So not Bayesian statistics
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www.gaussianprocess.org www.gaussianprocess.org

he normalizing term of eq. (3.53),ZEP=q(yX), is the EP algorithm’s approximationto the normalizing termZfrom eq. (3.48) and eq. (3.49)
so the EP approximation to the marginal likelihood?

in the EP framework we approximate the likelihood by alocal likelihood approximation13in the form of an unnormalized Gaussian function inthe latent variablefi
how is the EP approximation good when the probit function as likelihood is shaped so differently from the unormalized Gaussian which is used to approximatie it!?


arxiv.org arxiv.org

n adversarial network is used to define a loss function which sidestepsthe need to explicitly evaluate or approximatepx(·)
Adversarial training as an alternative to maximum likelihood training!


papers.nips.cc papers.nips.cc

Dirichlet prior over these parameterss
You'd need to sum over all \(y\) for that?

q(zjx)q(yjx),
Ehm, in the line below you have \(q_\phi(zy,z)\) not \(q_\phi(zx)\)


arxiv.org arxiv.org

Inheriting from the properties of stochasticprocesses, we assume thatQis invariant to permutationsofOandT.
In principle, just like for consistency, they could choose not to enforce permutation invariance (e.g. using just an RNN), and rely on the model learning it from data.
However, the contributions they are making (here and in the Neural Processes paper) is ways of putting the structure that Bayesian inference on stochastic processes should satisfy explicitly on the model, so that it has to learn well.
We are putting prior knowledge about how to use prior knowledge

without imposing consistency
I think in the context of CNPs consistency would imply things like
\(\sum_{y_1} p(y_1)p(y_2y_1) =p_2\)
These things are not automatically guaranteed by the framework used here. The data should constrain the network to satisfy these approximately


arxiv.org arxiv.org

Since the decodergis nonlinear, we can use amortisedvariational inference to learn it.
So, the nice thing about NPs is that if you did the inference of \(z\) exactly it would be a stochastic process exactly (unlike CNPs that don't have a simple interpretation/approach that guarantees being an exact stochastic process). However, because of not doing the inference of \(z\) exactly, consistency of the resulting marginal distributions is not exactly guaranteed.

nontrivial ‘kernels’
or even stochastic processes that aren't GPs and may not be describable by kernels

Given that this is a neural approximationthe curves will sometimes only approach the observationspoints as opposed to go through them as it is the case forGPs.
?? For GPs with Gaussian likelihood the functions don't pass exactly through the observation points, just near, as in here?


arxiv.org arxiv.org

h1;:::;hk
I think he meant \(h^{j_1},...,h^{j_k}\)

(W0>Wx)i= 0whereW0is an iid copy ofW
So we sample the A transposes independently from the As I see. That is the gradient independence assumption.
Btw is this related to some stuff I saw Hinton spoke about showing that you could do backprop with weights different from the forward prop, and they couldn't understand why. Is this the explanation? Could this, as they suggested, be related to a way the brain could be doing backprop?

The sampling ofinput Gvars models the distribution of the first hidden layer across multiple inputs, sampling of the first layer parameters(see Appendix B.1 for an example), and/or sampling of bias vectors.
"upon sampling of the first layer parameters" or "sampling the first layer parameters", I guess he meant?
oh ok , he kinda means sampling of the last layer parameters, but when we do backprop, it's the first layer..

In general, the multiplefvigallow for multiple NN outputs.
Ah ok, the \(v^i\) are like the weights of the outputs in the loss function
NO. each \(v_i\) is a vector, the vector of weights which when multiplied by some \(\mathtt{g}\) or \(\mathtt{h}\) give a single realvalued output labelled \(i\). This we call a linear readout layer.
When we back propagate the loss we would multiply each of these vectors by the derivative of the loss w.r.t. to each of these outputs.
Can this be done in this formalism?

batchnorm, whose Jacobian has a singularity on a 1dimensional affine subspace(and in particular, at the origin).
so that the \(\mathtt{f^l}\) are not polynomially bounded

am
again should be \(a_{j_i}^l\)

m
don't need \(m\) here?

has enough power to express thebackpropagation offand compute the gradients with respect to hidden states.
Note that the Nonlin functions in the appended lines are allowed to depend on the previous \(\mathtt{g}\)s, which is necessary to compute the backpropagation of the gradient through the nonlinearity. The odness works for backprop because the Nonlin is just linear w.r.t. the \(v\)s

Theorem 4.3.
Generalized law of large numbers extended to a case where the i.i.d. r.v.s have a distribution that changes as we increase the number of samples (although the way it changes is constrained).
EDIT: see comment
The reason this is nontrivial is that the \(\phi(g_i^{\frak{c}t})\) have a distribution that changes with \(t\), although it approaches a limit. So it is different from the standard law of large numbers.
In words, here we are saying that the empirical averages as we increase \(t\) approach (almost surely) the expected value of \(Z\) distributed according to the limit distribution of \(\phi(g_i^{\frak{c}t})\)

theLat which the exponentiating effect ofWLkicks in increases withn
the \(L\) at which exponential effect kicks increases like \(\sqrt{n}\) it seems according to these arguments. Nice
seems related to things in here http://papers.nips.cc/paper/7339whichneuralnetarchitecturesgiverisetoexplodingandvanishinggradients

zN(c;Kc)
\(\mu^\frak{c}\) is the vector of means of the tuple of \(i\)th components of the vectors in the argument of either \(\mathtt{f}^a\) or \(\mathtt{f}^b\), and \(K^\frac{c}\) is the covariance matrix of this tuple/vector.
Note (also useful for the above definition) that we are defining means and covariances for any individual component of the vectors \(\mathtt{g}\). That is, we are describing the distribution of \(\mathtt{g}^{\frak{c}}_i=(\mathtt{g}^l_i)_{g^l\in\frak{c}}\) for any \(i\). Different tuples of components are independently distributed, as explained in a comment in the beggining of the Setup section above

c(gl)
This is defined for \(g^l \in \frak{c}\)

Kc(gl;gm)
This is defined for \(g^l, g^m \in \frak{c}\)

amji
should be \(a^l_{j_i}\)

gcinti N(cint;Kcint)for eachi;j
Note the subscript \(i\), so this is the distribution for the tuple of the \(i\)th components of all the Invecs in \(\frac{c}\).
We therefore allow the \(i\)th component of two different Invecs to be correlated (useful to model the distribution of the first hidden layer, as per the usual NNGP analysis). But we don't allow different components of Invecs, \(g_i^{lt}\) and \(g_j^{mt}\) for \(i\neq j\), to be correlated.
Thre is a typo, it should say for each \(i\).

sequence (int2N)
what does this mean? Ah \(t\) is like a "time", so it is the index of the sequence. \(lt\) just represents two indices (not their product!)
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 Feb 2019

arxiv.org arxiv.org

that such functions should be simple with respect to all the measures of complexity above,
Why?? Do you show that all functions that have the property of being insensitive to large changes in the inputs have high probability? If so, then say it, if not. Then not all such functions need be simple w.r.t. to all measures of complexity that are found to correlate with probability for a random NN

Schmidhuber, 1997, Dingleet al., 2018].
Dingle et al explore bias towards low complexity, but not the relation to generalization.

our result implies that the probability of this function is exponentially small inn.
Although I think it is exponentially small in \(n\), why is that implied by your result? All that you know from what we've been told up to this point in the paper, is that it's probability has to be smaller than sqrt(log(n)/n), or using symmetry, we can divide this by 1/n

ne,
for \(n>> 1\), the probability of all of the Hammingdistance1 neighbours giving the same result goes to 0. So the weight of the term corresponding to Hamming distance 1 goes to 1, in the average

wo
Approximately a geometric distribution with success probability p=2, for large \(n\), and the mean of a geometric distribution is 1/p = 2

 Jan 2019

arxiv.org arxiv.org

(3)
This is a Schmitt trigger :D, see here: https://youtu.be/Iu21laCEsVs?t=2m31s

Inspired by the method of Lagrange Multipliers
How is this inspired by the method of Lagrange Multipliers?


yann.lecun.com yann.lecun.com

0AB
This is not \(\delta u_k\), why is he using \(u_k\) ? I don't see what's the justification


www.jmlr.org www.jmlr.org

other nonquantitative uses of bounds (such as providing indirect motivations forlearning algorithms via constant fitting) do exist. We do not focus on those uses here
What do they mean by "learning algorithms via constant fitting"?


www.youtube.com www.youtube.comYouTube1

haha long boy

 Dec 2018

arxiv.org arxiv.org

Hyperparameter I guess

it does not distinguish between models which fit the training data equally well
Well, if it is regularized (so there is an effective prior), then this isn't true!


openreview.net openreview.netpdf7

on networks and problems at a practical scale
Yeah. But not on the original networks, but on the compressed ones!

L()
This is a 01 loss. It can be applied to top1 or top5 defining the loss appropriately!

The pruned model achieves a validationaccuracy of60 %.
this is far from SOTA. right?

A direct application of our naïve Occam bound yields nonvacuousbound on the test error of98:6%(with95%confidence)
Ok, that is nonvacuous, but just, right?

Our simple Occam bound requires only minimal assumptions, and can be directly applied to existingcompressed networks
But remember that the compression scheme technically shouldn't depend on the data, for the bound to be valid (as the PACBayes prior depends on the compression scheme)

We obtain a bound on the training error of46%
On the training error, or the test error?

We prunethe network using Dynamic Network Surgery (Guo et al., 2016), pruning all but1:5%of the networkweights.
wow, they are very weightcompressible indeed


emtiyaz.github.io emtiyaz.github.io

Meaneld underestimates the variance
Independence increases concentration

REq=i(w=i)[logp(D;w)]dw
You are normalizing the log not the distribution here!


www.cs.cmu.edu www.cs.cmu.edu

/
Exp missing

 Nov 2018

emtiyaz.github.io emtiyaz.github.io

DKL[qkp]
It is minus this!


openreview.net openreview.netpdf6

jSjc+jCjc
These are the sizes of the codes for encoding the vector of weight indices and discretized weight values, according to coding scheme \(c\)

264bytes, which is272bits
isn't \(2^{64}\ bytes \(2^{67}\) bits, as each byte is \(2^3\) bits?

Theorem 4.1
Bound on KL divergence for Universal prior, and point posterior

Ascis injective onHc, we have thatZ1.
By Kraft inequality, and \(m\) being a probability measure

The generalization bound can be evaluated by compressing a trained network, measuring theeffective compressed size, and substituting this value into the bound.
An important question is: are the bounds valid for the compressed network, or for the original one as well? (as isn't the case in the related work: https://arxiv.org/abs/1802.05296 )

empirical evidence suggests that they fail to improve performance in practice(Wilson et al., 2017)
Well. If this is the case it means that proving a generalization bound for one of these procedures "implies" a bound for the standard training procedure, as you are claiming that the standard one generalizes not (much) worse than these ones


www.youtube.com www.youtube.com

To understand phenomenon described by Saxe and in the video at 43:00, we can think of this: Low eigenvalues in XX^T correspond to directions with little variations in the input. However, by the random fluctuation eta, the output could have an O(1) variation, even for arbitrarily small input variation, which requires a large weight to fit, and produces large generalization error.
for $\alpha<1$ the probability of this directions in input space with low variation decreases, as we get less directions overall with points I think (directions with no points/variation are ignored for the algorithm which projects weight into input subspace, and these are the 0 eigenvalue parts of the MarchenkoPastur distr)


arxiv.org arxiv.org

weakinteractions between layers can cause the network to have high sharpness value.
why does interacting weakly cause high sharpnes??
I think the reason is that for layers which interact strongly random perturbations tend to cause smaller relative change on the output, than for layers that interact strongly. And this smaller relative change on outputs probably translates into a smaller absolute change in the Loss...
think about aligned eigenvectors and stuff

depends only linearly on depth and does not have any exponential dependence, unlike other notionsof generalization.
Aren't VC dim bounds for NNs linear in depth also?

forming a confusion set that includessamples with random labels.
This is what's done in Wu et al https://arxiv.org/abs/1706.10239 also

middle and right
For a fixed expected sharpness the KL (and thus the effective capacity/bound) increases less for true labels than for random lables. Also, in both cases, it increases monotonically

capacity is proportional toCMmarginndiamM(X).
This is the "curse of dimensionality"!
Probably can prove using covering numbers which bound RadComp (via Massart's lemma)

simply bounding the Lipschitz constantof the network is not enough to get a reasonable capacity control
but are the generalization error bounds for Lipschitz functions tight?

However, the covering number of the input domaincan be exponential in the input dimension and the capacity can still grow as
Unless inputs lie on a slow dimensional manifold I guess

The bounds based on`2path normand spectral norm can be derived directly from the those based on`1path norm and`2norm respectively
Hmm. how?

Instead, to meaningfully compare norms of the network, we should explicitly take into account thescaling of the outputs of the network. One way this can be done, when the training error is indeedzero, is to consider the “margin” of the predictions in addition to the norms of the parameters.
This has been used in several papers on generalization bounds for neural nets.
I think the idea is like that for margin bounds for SVMs. You can't bound the RadComp using the 01 loss as it's Lipschitz (related to above arguments). But you can bound the Hinge loss (which is related to having a margin!), and then use the fact that the Hinge loss upper bounds the 01 loss.
The intuition I guess is that the previous troublesome cases with normbased capacity can be solved by taking margin into account, as the extremes in both cases are reducing/increasing the margin (for weights going to zero/infinity).

proportional toQdi=1kWik21;1,wherekWik1;1is the maximum over hidden units in layeriof the`1norm of incoming weights tothe hidden unit [4].
This comes from the Rademacher complexity of the neural network. See here: http://www.stats.ox.ac.uk/~rebeschi/teaching/AFoL/18/material/lecture3.pdf#page=4 for a derivation

One can then ensure generalization of a learned hypothesishin terms of the capacity ofHM;M(h)
They just state this, but they should maybe cite some work on nonuniform learnability, SRM, MDL, so that people unfamiliar with it can see why this is!

anymeasure which is uniform across all functions representable by a given architecture, is not sufficientto explain the generalization ability of neural networks trained in practice. For linear models, normsand marginbased measures, and not the number of parameters, are commonly used for capacitycontrol [4,8,24].
These are all SRM based.
They depend (most often implicitly via the training data) on the datadistribution/target function, and the algorithm (also implicitely) If you get a good bound with SRM it's most likely because your algorithm is biased towards the right kind of solution (kind basically corresponds to the classes used in SRM). In other words, a bad algorithm (relative to target fun) will be very unlikely to produce a good bound. On the other hand, a good algorithm may produce a bad bound, if the prior is chosen badly. So that SRM bounds may not be tight in general


arxiv.org arxiv.org

typo \(\theta\) shouldn't be here

 Oct 2018

www.stats.ox.ac.uk www.stats.ox.ac.uk

www.stats.ox.ac.uk www.stats.ox.ac.uk

+ 1
I don't think the \(+1\) is actually necessary

Proposition 5.1
Remember: larger metric correspond to smaller rulers

the
there

5.2
5.3


www.stats.ox.ac.uk www.stats.ox.ac.uk

2
shouldn't there be a \(\lambda\) here too?
And in fact another \(\lambda\) from Rad(A_1)?

=
this should be \(\leq\),
thought I would have used \(\mathcal{L} \cup \mathcal{L}\) instead

0
appostrophe shouldn't be there

h
hull

=
missing parenthesis

l
$$j$$


www.stats.ox.ac.uk www.stats.ox.ac.uk

=
\(\leq\)


arxiv.org arxiv.org

(U) =RSP(Ujv)d(v).
? How does this define what \(\rho\) is?


arxiv.org arxiv.org

he largestlast width function seems to converge slightly faster than the largest last width function
typo

more
less


courses.cs.washington.edu courses.cs.washington.edu

distribution

 Aug 2018

arxiv.org arxiv.org

DdCid!dCjd!E=1NdCd!2+F(!)ij
I don't get this formula. What are C_i and C_j supposed to mean. Or rather, where is the randomness over which we average?


arxiv.org arxiv.org

is not surprising and there are suitable options, even without GANs
Well, what is nontrivial about this is generalizing (as usual), how do you generate novel samples, what are samples that "look like" the rest of the data. How do you define that?


openreview.net openreview.netpdf1

The alignment problem is how to make sure that the input image in A is mapped (via image generation) to an analog image in B, where this analogy is not defined by training pairs or in any other explicit way (see Sec. 2).
Well, you are attempting to define it as "the alignment given by the lowest complexity meme".
I think this doesn't address the question, which is more generally addressed to the whole field of crossdomain mapping, I think. Why aren't the new neural network approaches compared with previous approaches of unlabelled crossdomain mapping, which can be formalized as different forms of alignment between sets of points. I think the main difference here is that in those approaches you typically know the whole target domain, while here we don't quite know it, we just have a few samples from it, but I feel this is a surmontable problem

 Jul 2018

www.wikiwand.com www.wikiwand.com

The position, C {\displaystyle C} , of the camera expressed in world coordinates is C = − R − 1 T = − R T T {\displaystyle C=R^{1}T=R^{T}T} (since R {\displaystyle R} is a rotation matrix).
The minus is because we are translating the world in the opposite way that the camera was translated to get it right relative to the camera, while putting the camera at the origin. The R^1 is because we first rotate by R and then translate by T, so to get the original translation vector we need to undo the rotation to T

u 0 {\displaystyle u_{0)) and v 0 {\displaystyle v_{0)) represent the principal point, which would be ideally in the centre of the image.
It is multiplied by the z component of the point because we want to translate by a fixed amount in the uv plane. However, after being acted by the camera matrix, they are in homogeneous coordinates, scaled by the z component!


arxiv.org arxiv.org

dJd=dLd
Why??
I see, I think the derivative wrt \(\theta\) here is supposed to be while z satisfies the constraints in (32)...

restricted architectures which partition the hidden units.Our approach does not have these restrictions
But it has whatever restriction being an ODE imposses. How does the expressivity and learnability of the ODEnet compare with ResNets?

Minibatching
Can you not just compute the gradient for each input, and average them?

Second
typo?

tendtstart(z(t))dt
This second term comes from the fact that we want the probability of observations at times \(t_1, ..., t_N\), and at no other time between \(t_{start}\) and \(t_{end}\)

layers of only a single hidden unit
corresponding to weight matrices of rank 1

Instantaneous Change of Variables
This is just the differential form of the continuity equation

6 residual blocks, which are replaced by an ODESolve
What are the residual blocks precisely? Single layer + ReLU?

the number of evaluations of the hidden state dynamics required, a detail delegated to the ODE solverand dependent on the initial state or input.
Adaptive computation time

\(\theta(t)\)


www.youtube.com www.youtube.com

1:00:00 scale separation as a way to go beyond Markov (local interactions) assumption

~32:00 What about the domain of the function being effectively lower dimensional, rather than a strongly regularity assumption? That would also work, right? Could this be the case for images? (what's the dimensionality of the manifold of natural images?)
Nice. I like the idea of regularity <> low dimensional representation. I guess by that general definition, the above is a form of regularity..
He comments about this on 38:30
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pdfs.semanticscholar.org pdfs.semanticscholar.org

it should be +? This typo carries through to later parts


arxiv.org arxiv.org

to have identical norm at the input layer,
This works because the marginal variance is independent of the orientation of the input vector. This can be proven by noting that a random Gaussian matrix doesn't change its distribution when rotated by an orthonormal similarity matrix..


arxiv.org arxiv.org

(2)
Why is this the variational loss???

 Jun 2018

arxiv.org arxiv.org

Since GPs give exact marginal likelihood estimates, this kernel construction mayallow principled hyperparameter selection, or nonlinearity design, e.g. by gradient ascent on the loglikelihood w.r.t. the hyperparameters. Although this is not the focus of current work, we hope toreturn to this topic in followup work
Cool idea!

will be a sum of i.i.d. terms
Not true (for finite width nets) I think. the terms are conditionally independent, when conditioned on the value of the final hidden layer. If we consider the joint distribution over all the weights over all layers, then the terms won't be independent. However, in the limit of infinite widths, they do become independent. This is because their covariance is \(0\) and in the infinite width limit, they become Gaussian with that 0 covariance, and therefore independent.
Remember also that this is for a fixed input (and later analysis is for a finite collection of inputs)


www.princeton.edu www.princeton.edu

concentration
is this concentration phenomenon related to the asymptotic equipartition properties of large collections of independent (or weakly dependent) variables?

 May 2018

www.stats.ox.ac.uk www.stats.ox.ac.uk

−1
this means the functional inverse

φ
this should be the other phi?

 Apr 2018

www.lesswrong.com www.lesswrong.com

all the semantic consequences.
What are semantic consequences? Is there such a thing as syntactic consequences? Is the statement said here different for them? I don't think so right?

all possible subsets of a set?
subsets are defined as properties in Logic. Secondorder logic goes above firstorder logic by being able to quantify over properties (and thus over subsets)

 Jan 2018

static.googleusercontent.com static.googleusercontent.com

he matrixYl(Yl)Tis thelthlayer data covariance matrix. The distribution of its eigenvalues (or thesingular values ofYl) determine the extent to which the input signals become distorted or stretchedas they propagate through the network
I would say more that it determines the appearance of correlations between the elements of Y.
Also, this isn't quite the covariance matrix, unless the mean of Y is zero?


arxiv.org arxiv.org

While local optima may not be a problem withdeep neural networks in supervised learning where the correct answer is always given (Pascanu et al., 2014), the sameis not true in reinforcement learning problems with sparseor deceptive rewards.
Reason why evolutionary methods are useful in RL, but not so much in SL. What about UL?

random search
Is this what they mean https://en.wikipedia.org/wiki/Random_search ?

 Dec 2017

www.people.fas.harvard.edu www.people.fas.harvard.edu

irst, eigenvalueswhich are exactly zero (λi= 0) correspond to directions with no learning dynamics so thatthe parametersziwill remain atzi(0) indefinitely. These directions form afrozen subspacein which no learning occurs. Hence, if there are zero eigenvalues, weight initializations canhave a lasting impact on generalization performance even after arbitrarily long training.
No training on singular directions.
However, it seems like sloppy directions. Those with small eigenvalue in input correlation matrix causes overfitting. This is because in these direction there is small variability on the inputs, and yet there is variability on the output due to noise. To fit the small variability, we learn large weights, which are in fact too large, and don't generalize well..


arxiv.org arxiv.org

(1)
See here. This is the crossentropy between the likelihood for $\theta$, $q(X\theta)$, and $q(X\theta_0)$. It is the same as the disorderaveraged energy of state $\theta$, that is averaged over $X$. Note that because the energy is a mean over $x$, and each $x$ is independent, then for the disorderaveraged energy, we can just calculate the average for one $X$.
If we evaluate this at $\theta=\theta_0$, then we have the Shannon entropy of $q(X\theta_0$. $\theta_0$ is the disorderdaveraged ground state (it's not hard to see that it is the $\theta$ which minimizes $H(\theta;\theta_0), this is just a property of crossentropy).
The other one is the empirical version of this.

(3)
This is equal to
$$\int_\Theta d\theta \bar{\omega}(\theta)\prod_{X\in x^N}q(X\theta)$$
which is just the probability of observing the samples $x^N$.
But the interesting thing is that it can be written as a partition function where $\theta$ is the physical state vector as said after.

 Dec 2016

journal.frontiersin.org journal.frontiersin.org

nodeperturbationlike correlations
What is this?
