We can conclude that the results from the signaling coexpression network are very different from all other network modules. The cancer dependency adn the homology don't appear to be that different, although this is might be because they have similar module densities and very different module densities compared to the other methods. The two protein-protein interaction networks are similar to each other, which is expected. Adding a group of random module predictions would maybe help. It is not clear that the top two methods for each network type are very different compared to the overall variance of all methods for that network type. Interestingly, it seems that the top 5 are not really that similar compared to the consensus (supplementary figure 2). In fact, I would like to see the scoring of the consensus without the top 5. 3/5 of the top 5 methods seem to be similar to each other.

What restriction prevents a model from maximizing the number of clusters by splitting clusters that should contain 6 elements to two clusters that contain 3 elements?

Assume X, Y, Z, W are singleton sets containing the nodes X, Y, Z, W. Note that X is not d-separated from Y conditional on Z because Z is a collider between X and Y. However, X is d-separated conditional on the empty set. Then, the path from X to Y contains a collider Z not in the empty set. Then, X is d-separated from Y dependent on the empty set.

X and Y are independent of W conditional on Z. Any path from X or Y to W contains Z. Z is not a collider.

Suppose there is no such partition. Then, (1) \(\nexists X_i \)such that \(X_i \cup C_i \subset H_j\). However, (1) contradicts property 2 of the separator. Note that vertices can belong to both H_1 and H_2, but that doesn't mean that they have to belong to A_j.

Candidate Search: For each \(X_i\), we consider whether \(pa(X_i) \in Pa^G(X_i) \ \forall pa(X_i) \in Pa^H(X_i)\). \(|Pa^H(X_i)| \leq k\). Then, this calculation is \(O(2^k)\) for each \(X_i\).

Scoring: initially, every node is checked with k potential parents at most. In subsequent iterations, we consider whether each of those k parents is still optimal or not.

Note that P(A)P(B) > P(A,B) can be true for some A and B. if x and y are independent, then P(x,y) = 0. if not, there can be some negative terms, but the sum is always positive.

Note that the causal Markov Assumption implies statistical independence for non-immediate ancestors. It says nothing about the causality itself.

actually, under the causal Markov assumption, if X is not the immediate ancestor, it is not causal.

If Y -> A and X -> A, then by theorem 2.1, this structure is the same for all members of the equivalence class. This means that Y -> A and X ->A. By definition, a Markov relation indicates that two variables are related in some joint process.

G is overloaded. The G in the left is an arbitrary graph. The G on the right is variable.

strictly speaking, this is an image set.

the probability of \(F(t + \Delta t) \cap R(t) = F(t + \Delta t) - F(t)\)

Again, they defined 5.149 as the lognormal parameter, which is different from the normal.

this is wrong. probability is 1 - 0.1314.

this was defined as the lognormal parameter mu, not the normal parameter mu.

the meaning of mu and sigma is overloaded!

beta reduces the maximum (spreads) the data faster than alpha (by the square of beta). Both alpha and beta translate the distribution at the same rate. This agrees with my graphical interpretation from the previous comment.

alpha translates the pulse across the x axis but also decreases the maximum value. \(\Beta\) slows down the exponential decay but also decreases the maximum.

expression is missing a sigma!

What if one of these is zero? I'm assuming that in that case \(\sigma_{XY} = 0\)

How does this look on a Venn Diagram?

probability of even is 2 times odd. There is only one event in B and it is even, so 2w = p(B|A).

how do the results of the other models compare when there are no priors.

They were statistically unlikely in the rewired networks, so why did they hypothesize that they are suppressed in evolution?

Here, k should be <= 4.

Why does the p > 0 case have P(k) decrease exponentially with k?

if n1 + ... + nk = n, then theorem 2.5 is equivalent to theorem 2.5.

Only the first r factors are relevant. 25 students can take the first award, 24 the second, ...

The theorem relies only on \(\lnot\), \(\land\), and \(\lor\).

My "proof" needs revision.

Let C be the set of all wffs generated from all sentence symbols with the five sentence operations. Define \(C_n\) such that \(C_n \cup C_s = C\) and \(C_s \cap C_n = \emptyset\). \(C_s \neq C\) because \(C_s\) was not generated from all sentence symbols and sentence operators, so \(C_n \neq \emptyset\).

We must show that every element of \(C_n\) is not special. Suppose there were some special expression \(\alpha \in C_n\). Then, \(\alpha \notin C_s\). Apply the parsing algorithm to that \(\alpha\). The following conditions must apply (If not, \(\alpha\) is not special): 1. At each vertex, the connective symbol must be \(\lnot\) or \(\to\) 2. All minimal vertices must contain \(A_i \in {A_2, A_3, A_5}\).

If these conditions are met, then \(\alpha\) also belongs to \(C_s\) because \(C_s\) is closed under \(\epsilon_\lnot\) and \(\epsilon_\to\) and includes \(A_i \in {A_2, A_3, A_5}\).

Assumptions in bold

[1-2] v is acceptable --> the domain of \(\bar{h}\) is inductive --> \(\bar{h}\) is a function --> \(\bar{h}\) is an acceptable function

[3] g and f are free (given) + \(\bar{h}\) is an acceptable function --> v is acceptable

I'm assuming they used the same argument that v must equal h for some x and y.

Why is K nonempty?

Why must there be an acceptable v? This is because \(\bar{h}\) is defined as the union of all acceptable functions. Then, if we assume that \(\bar{h}(x)\) is defined, then v(x) is also defined.

Suppose there is a function h(x) and a function j(x) that satisfy all the requirements. Then, h(x) = j(x) for x in B. Additionally, h(f(x,y)) = j(f(x,y)) and h(g(x)) = j(g(x)). Because C is generated from B by f and g, h(x) = g(x) for all x in C.

h(1) = h(g(0)) = 0 + 2

h(1) = h(f(g(0),g(0)) = f(h(0),h(0)) = f(0,0) = 0

Basically, C is the smallest inductive set, so if S is smaller or equal to C, then S = C.

Based on context, I will define a "proper segment" as a segment that is a wff.

This is only possible because the operations undo each other.

Let \(\Sigma\) be the set of all inductive sets of U. By definition, \(B \subseteq S \ \forall S \in \Sigma\). Then, the intersection must contain B. If B = S for some S \(\in \Sigma\), then B is the smallest member of \(\Sigma\). Then, B = C*. If B is smaller than all S, then B is not closed under f and g because otherwise B would be inductive and therefore belong to \(\Sigma\), which is a contradiction. But this means that all S must have a subset D such that \(D \cap B = \emptyset\) and that satisfies closure under f and g for all elements of B. Then, D also belongs to the intersection and \(B \cup D\ \in \Sigma\). Since no element of B or D can be removed without losing the inductive property, \(B \cup D\) must be the smallest member of \(\Sigma\). Then, \(B \cup D =\) C*.

They proved earlier that applying a finite sequence of operations \(<\epsilon_1, ..., \epsilon_n >\) gives a unique expression \(\alpha\). However, here they are treating f and g abstractly.

Because \(\alpha_0\) and \(\beta_0\) are proper initial segments in step 2 and 5, these two cases have an excess of left parentheses. Furthermore, in cases 3 and 6, since \(\alpha\) and \(\beta\) are not proper initial segments, by Lemma 13A, \(\alpha\) and \(\beta\) must have the same number of left and right parentheses. Then, in cases 3 and 6, \((\alpha\) and \((\beta\) also have an excess of left parentheses.

the third value could be "unprovable"

equivalent to the axiom of choice (see Zorn's Lemma).

if they were not disjoint, then it would not be well defined because \(\kappa + \kappa = card(A \cup A)\), but \(A \cup A = A\) and \(\kappa + \kappa = 2\kappa\).

A is still smaller than \(\mathbb{N}\), and A still maps into \(\mathbb{N}\). However, we can also map A onto \(\mathbb{N}\)? Note that we can always make an f' that is one to one and onto for any A that has an f that is one to one and into \(\mathbb{N}\). Therefore, the definition only requires for there to exist an f; the existence of f' is irrelevant.

the extra dimension is the output variable.

If there were a member of A that were a finite sequence of members, m is not necessarily = n.

The empty set contains only itself. Therefore the power set of the empty set contains only the empty set.

The power set of the power set of the empty set contains itself and the empty set. Note that {\(\emptyset\)} is distinct from \(\emptyset\) because {\(\emptyset\)} is the set that contains the empty set and \(\emptyset\) is simply the empty set.

Note that this definition does not include closure, so a group under this operation is not necessarily a group (see example 5).

we are saying it is a group relative to the operation. It is not a group however.

Define a binary operator under substraction B. Then B(3, 2) \(\neq\) B(3,2) 3 - 2 \(\neq\) 2 - 3

the negative rationals are also a group?

if -a, then -a1/-a = 1. 1-a = -a*1 = -a.

Suppose there is a \(b \in \mathbb{Z}\) such that 5b = 1. Then, 5 | 1. But if a | 1, a = 1 for \(a \in \mathbb{Z}\) > 0.

Let a, b, c \(\in \mathbb{Z}\). Then (a + b) + c = a + (b + c). a - a = 0. a + 0 = 0 + a = a.

You cannot use induction to prove the principle of induction, or can you?

Assume S is nonempty. It must have a smallest member a \(\in\) S. Then a + 1 (\in\) S. Assume it holds for k > a. Show it holds for k + 1. Since k (\in\) S, then k + 1 is in S. If the theorem is true, then this is a proof. It is a proof, but it is a useless proof because we do not know if it is a proof unless we know whether the theorem is true or not.

proof?

How, if the molecule is 3 dimensional? It does not have "2 faces"? I suppose we can split it with a plane in half, where 1 of the hydrogens and the nitrogen are on the plane, and the other 2 hydrogens are above and below. Then those two hydrogens can be projected onto the plane, so that it can be visualized as a triangle. However, this does not exclude other forms of symmetry.

-a -> (-a)^2 = a^2 and a -> a^2. For the onto part, there is no integer which squared gives 2. There is also no integer which squared gives a negative integer.

Let a be an integer. a -> |a| = a and -a -> |-a| = a.

Each negative integer is mapped to the nonnegative counterpart, and each positive is mapped to itself. Thus, the entire range of the the nonnegative integers is assigned to an element of the domain twice except 0.

why is n positive? It does not seem necessary in the proof.

1 = 23s + 25t for \(s,t \in \mathbb{Z}\) can be solved using Euclid's algorithm. Is there another method?

Each element in A can only have one element in B assigned to it. If an equivalence class corresponds to a single element of A, the entire class must be assigned only one element in B. We should be able to define the function \(\phi\) if each element of the equivalence class [a/b] were an element of A, but then the domain would not be the rational numbers. It would be a larger set \({\frac{a}{b} | a,b \in \mathbb{N}, b \neq 0}\).

This theorem doesn't say that you can't use two negative integers to give a positive integer. The product is unique only when you consider other prime combinations. However, it is not unique if you include the negative numbers. Of course, a combination of negative integers that gives a positive product will be equivalent to the same combination of positive integers times negative one.

Is this the correct logic?

if r' - r > 0, then $$ 0 \leq r' \le b $$ is not true. Since b divides r' - r, then if r' = r, then q' = q.

Look at individual neuron. Now they are using electrophysiology response isntead of light to qualify response. Response at the single cell level but activating inhibitory cells in bulk. Calcium is fairly slow temporal dynamics. b. Probing individual putative pyramidal responses c. Confirming it is pyramidal using the spike rate d. responses of single pyramidal cell over time. The inhibition pattern is clearly pattern between SOM and PV population. e. The inhibition of PV cells is greater at the peak response, but it is uniform in the SOM population. The reduction is uniform, there is an increase in orientation selectivity in the SOM

why was this restriction added.

How is this object instantiated inside the class that defines it?

what does this mean?

above, they included the % when they defined the format specifier. But here, they did not add the - to the "beginning of the format specifier (before the %).

Note that these classes are special because: 1. instead of storing variables, they store constants 2. there are no static subroutines 3. the constants are stored into variables of type Season. Therefore, the static constants behave like objects. We can conclude that classes are not limited to storing variables; they can also store constants. The definition of an enum must integrate the constants with subroutines to create objects.

The corners are not counted and position 1 is in the edge but position 39 is not. Thus, a total of 19 spatial positions are preserved with a kernel of 3.

Why does it hold for this interval? Is it the error in approximation because the function may require more than 3 Taylor terms? If so, why did they not use Big O notation as above?

Here, they use the term in vivo. However, did they incubate outside of the body? Figure 3l shows what happens in the body, so they prob did this in vitro. Check methods.

The assumption is that the lack of disruptions in in vitro morphology of cells and EVs means there is no or minimal disruption in vivo.

Treated EVs have higher fluorescence. Fluorescence was used to quantify the capture on beads. Capture efficiency could be high enough that a big fluorescence difference corresponds to a smaller capture difference.

Different cell types display different levels of EVs.

• What is the mechanism of Ac4ManAz?

As explained below, System.out.println is the subroutine. This means a command is a subroutine with an inputted data (in this case "Hello World!").

should be p(u,v)

The k's cancel out and the denominator is equal to 1; the numerator is equal to \(\beta\).

The nullity is unknown but we know it is > 0. This assumes that the dimension of the null space is equivalent to the rank r. However, if q = 0, so that \(t - \lambda_1\) is the zero map. But if \(t - \lambda_1\) is the 0 map, then the rank is the empty set, but the the nullity > 0. This is a contradiction. This proves that it cannot be the zero map.

The exact number of y's is not essential for the proof because they all go to \(\overrightarrow{0}\) (see end of proof).

These are cases where lambda is not equal to 0.

The original function was applied to the domain \(\mathbb{C}^n\). Now the transformation (with the same name) is applied to the domain \(W\).

They prove that if the eigenvalues are distinct, the eigenvectors are linearly independent. Does this statement hold in the other direction? That is, if the eigenvectors are linearly independent, are the eigenvalues distinct?

By theorem 3.18 (introduced later), for any distinct set of eigenvalues, there is a linearly independent set of eigenvectors.

Need to prove that every t-string basis gives a matrix similar to a matrix that is all zeros expept for blocks of subdiagonal ones. The examples show this but a formal proof has not been shown.

Note that $$B_{R}^\frown B_N$$ is not a basis for V. (1) The dimension may be equivalent but range(h) + null(h) is not necessarily equal to V unless h is onto and h:V-->V. (2) In addition, the null space and the range space may overlap more than just the zero vector, in which case null(h) + range(h) is not a direct sum and \(B_R\) and \(B_N\) are not linearly independent. See page 435, where the diagram and proof clearly show this.

The sum of the two subspaces is by definition the span of the union of the range space and null space, which is less than V because we are assuming nilpotency > 1 for the inductive step. However, the sum rank(t) + nullity(t) = dim(V). This is because for every vector missing from the domain, there is a redundant vector that is in both the null and range spaces (see diagram above). However, the sum of the null and range spaces does not give V. Only the dimension is equivalent.

The problem with this concatenation is that C hat includes the basis for the intersection of the range and null spaces. I think he meant the concatenation B conc Z. In other words, the null space and range space are not independent. The equation below writes this relationship correctly because they subtract the dimension of the intersection.

We know the left-hand disjoint string basis form is the correct one (see map action graph above), so \(\beta_2\) and \(\beta_4\) are elements of the range. Any two vectors that satisfy the map action graph will do.

This is because the powers between 0 and n are not necessarily one to one so more than one vector may be sent to the zero vector and overlap with the range space (see example below).

They do not have to shrink. There is the case where k = 1.

Note that they prove that if \(t(\beta_i) = \lambda_i\beta_i \), then it is diagonalizable. But they do not prove that if it is diagonalizable, then \(t(\beta_i) = \lambda_i\beta_i \). It is trivial to prove this because it is true by the definition of diagonalizability of transformations.

if \(d(x) = x - \lambda\), then \(d(\lambda) = 0 \) and \( \ p(\lambda) = r(\lambda).\) Because we know that r(x) must be constant (based on division theorem for polynomials), then \(r(x) = c_0\). Thus, \(r(\lambda) = r(x)\).

Theorem 4.3 says that "a matrix is invertible if and only if it is nonsingular".

See the definition for a nonsingular linear map in Definition 2.7 and also the requirements for an inverse function in the appendix.

zero is considered "even".

This proof assumes that a function with the properties in Definition 2.1 exists. However, they have yet not proven that a function with these properties exists.

check proof later

Note that if D has less than n elements, it is not a basis for the enclosing set, but it is for a subset. The theorem proves that D cannot have more than n elements but not that it cannot have less. Thus, spanning the set is not required for D to be linearly independent.

He finally defines what a "spanning set" is.

They used polynomial division to get this result. However, multiplying by \(\frac{ 1/x}{1/x}\) is simpler. $$\lim_{x\to\infty} \frac{x}{x+1} = \lim_{x\to\infty} \frac{1}{1+ \frac{1}{x}} = 1$$

1 | a. Also, a | 1 if and only if a = 1. Therefore, pk = 1. However, by definition, every prime is divisible by one, so 1 is not a prime number.

I have not seen a theorem that shows that if ab | cd then a | cd (where they are all primes). However, they assume that r < s, so it is feasible that r = 1 with this assumption. Why can they assume this without "loss of generality"?

Refer to theorem 1.2.3. We know \(m_1\ | \ m_1\) (a). Thus, since \(m_1 \ | \ m_1\) and \(m_2 \ | \ n_1\), \(m_1m_2 \ | \ m_1n_1\) (f).

They prove using only the natural numbers and the multiplicative property. I think it could also be proven using the rational numbers and the multiplicative property by multiplying 1/c on both sides of ac = bc. However, they did not do this probably because they want to only use the natural numbers, and it is not necessary. The contrapositive statement, if a = b, then ac = bc is clear using the multiplicative property of equality.

Chain rule on the left side. We do not know y explicitly, so we factor out y'. However, we can easily solve for y' by implicit differentiation to check our solution.

What do they mean by sensible? I think they had commented earlier.

Let $$r = \begin{pmatrix} r_1 \ r_2 \end{pmatrix}$$. According to Lemma 2.3, when multiplying a linear combination of linear combinations by a factor, in this case r,

$$r_1s_1 + r_2s_2$$

the variables can be rearranged into a singular linear combination by regrouping the shared variables. In this case, the variables are the shared vectors.

$$\begin{pmatrix} 2 \ 1 \ 0 \end{pmatrix}(r_1y_1 + r_2y_2) + \begin{pmatrix} -1 \ 0 \ 1 \end{pmatrix}(r_1z_1 + r_2z_2) = \begin{pmatrix} 2 \ 1 \ 0 \end{pmatrix}r(y_1 + y_2) + \begin{pmatrix} -1 \ 0 \ 1 \end{pmatrix}r(z_1 + z_2) $$

Because \(y,z \in \mathbb{R}\), \(y_1 + y_2, z_1 + z_2 \in \mathbb{R}\). Also, \(r(z_1 + z_2), r(y_1 + y_2) \in \mathbb{R}\). Therefore, the elements of the set of vectors above are equivalent to the elements of S; i.e., the two sets are equal. QED

See example 2.8. Notice the wording in the next sentence:

we cannot ... be confident the result is an element of T.

Many of the vectors in T will satisfy the conditions, but not all.

Stated in other words, they are proving that any system's solution set differs from the corresponding homogenous system's solution set only by the particular solution.

It is a bit confusing, because in the Lemma, \(\overrightarrow{h}\) is assumed to be a set of 1. I am not using \(\overrightarrow{s}\) and \(\overrightarrow{h}\) in the same way as the proof. \(\overrightarrow{s}\) and \(\overrightarrow{h}\) represent sets with \(i\) solutions.

Part I. From Appendix A-5, if two sets have the same members, they are equivalent. For set inclusion the first way (\(\overrightarrow{s}_i \in \overrightarrow{p} + \overrightarrow{h}\)). In this case, for any solution in \(\overrightarrow{s}\), \(\overrightarrow{s}_i - \overrightarrow{p}\) gives \(\overrightarrow{0}\), just as any solution of the associated homogenous system \(\overrightarrow{h_i}\) gives \(\overrightarrow{0}\) (i.e. the particular solution of any homogenous system is \(\overrightarrow{0}\)).

Part II. Likewise, for set inclusion the second way (\( \overrightarrow{p} + \overrightarrow{h}_i \in \overrightarrow{s}\)): any solution of the solution set of the associated homogenous system plus the particular solution of the system is sufficient to solve any system. By definition \(\overrightarrow{h}_i\) always gives \(\overrightarrow{0}\), but adding \(\overrightarrow{p}\) will give the complete solution set because in the first part of the proof they showed that \(\overrightarrow{s}_i - \overrightarrow{p} = \overrightarrow{h}_i.\).

Thus, we can use the solution set of an associated homogenous system to get the solution set of any system (if we have the particular solution). The other lemma proved that any homogenous system could be expressed in the pattern: particular + unrestricted combination. Combining that lemma with this one shows that any system can be expressed in that pattern because you only need to add the particular solution to get an equivalent solution set. However, they do not show this proof.

They prove that if the first condition on addition is satisfied for the subset, that all other conditions on addition are satisfied.

For if P, then Q statements, the statement is true only if Q is true, independent of P. For equivalent statements, both if P then Q and if Q then P need to be true. (3) --> (1) is the hardest, so the "intermediate" (2) statement was made to help prove (3) --> (1). In addition, (2) is more practical when trying to prove that S is a subspace of V.

The following were treated as self-evident: If (3) is true, then (2) is true because a pair of vectors in S is a subset of any vectors in S (3). If (1) is true, then (3) must be true by the definition of a vector space.

Vector spaces are "safe" spaces in which vector addition and multiplication can be performed. Although the name is "vector space", it encompasses all collections of linear combinations, not just vector structures or linear systems.

(2) They are proving that \(\overrightarrow{w}\) = \(\overrightarrow{v}\cdot-1\). Thus, we can use \(\overrightarrow{-v} = \overrightarrow{w}\) without worrying about confusing the term with \({\overrightarrow{v}\cdot-1}\).

\(a_n = (n-1)^2 + 1\) or \(a_{n-1} + 2n - 3, a_0 = 2\)

Was this an exercise?

Here they are using a homogenous system. They proved in the ealier chapter that an associated homogenous system differs from a system only by the particular solution.

By theorem here, he means the Lemma.

He has not stated the theorem yet. It's possible the author wrote the proofs in one coherent argument, but then split the theorem into Lemmas and did not bother to patch them up properly.

Proof by contradiction. Assume there is a row in R that is a linear combination of the others and see if the Lemma 2.5 holds.

(From the previous section) Linear surfaces are defined by a set of vectors. This means that there exists a vector that connects any two points in the surface. This definition unifies the properties of the vector with the properties of the linear surface, which allows us to explain the properties of the surface with the theorem. Thus, if the theorem does not hold in higher dimensions, the lines in higher dimensional spaces would not be flat.

The theorem says (in other words): take any two vectors (\(\overrightarrow{u}\) and \(\overrightarrow{v}\)) connecting two points in the plane they define, and no matter how much they approximate linearity, \(|\overrightarrow{u} + \overrightarrow{v}|\) will always have less length than the sum of their lengths. If they are multiples of each other, they are parallel and thus they do not form a plane.

In other words, any set of linear combinations of a variable can be regrouped into a single linear combination of the variable. Furthermore, any scalar multiplied to each element of the set of linear combinations becomes part of the new coefficients of the new linear combination of the variable.

This is important because as they showed in the examples above, Gaussian row operations essentially create sets of linear combinations of the variable within a row, but Lemma 2.3 shows that the variables can be regrouped so that set of linear combinations collapses to singular linear combination of the variable.

The Lemma is a bit confusing because they are using "linear combination" in a general and a specific sense. What they mean is:

A linear combination of linear combinations of x is a linear combination of x.

They use "linear combination" in the general sense because they create a linear combination of the scalar "d" with the set of linear combinations of x. Then they regroup to create a singular linear combination of x.

They are proving the theorem by lack of contradiction to inequality theorems. See Equivalent statements in appendix A-2. This seems to be a proof by "lack of contradiction". Assume that P is true, and see if Q holds because P is true if and only if Q is true.

I think the author may have initially used \(p_j \) and \(s_j\) instead of \(p_n \) and \(s_n\).

According to the definition of a partition, overlapping parts are equal. So all of the \(S_x\) parts (there are infinite) that share at least one \(y\) will collapse into one part. The result is two subsets that compose \(\Omega.\) From what I can tell, in the context of partitions, \(S = \Omega.\)

Earlier (A-4), they defined \(\hat{n}\) using \(n = 2^{k_n} \cdot \hat{n}\), where \(k_n\) refers to the number of 2's in \(n\). This definition is extended to \(\hat{d}\): \(d = 2^{k_d} \cdot \hat{d}\). However, in this context, \(k\) refers to the number of 2's that satisfy the following condition: \(\hat{n}d = n\hat{d}\).

This must be a typo because -1 \(\neq\) 1.