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Code / solutions for Mathematics for Machine Learning (MML Book)
First of all, thanks for your solution sheet!
For 5.6, I believe d(g)/d(X) =ATBT. Because A is DxE, and B is FxD. ATBT is valid, while BTAT is not.
x is [1,2], thus magnitude of x is sqrt(5) instead of sqrt(6).
Hi @ilmoi, firstly, thanks for this incredible work. It as helped me so much. While solving Question 10 in Chapter 6. I came across an abnormality in your calculations for its solution.
I think that in places where I put an exclamation mark, there is supposed to be a power to -1 if we expand the bracket correctly. Furthermore, I maybe wrong, or missing the point somewhere. This is why, it would be really helpful for me if anybody could help me solve this issue.
Thanks.
Q: A = {(λ, λ + µ3, λ − µ3) | λ, µ ∈ R}
A: This has a basis of {(1,1,1)T,(0,1,−1)T}, so it is a subspace of R3.
Hi, could you explain this or give me some reference? Why we can say V has a basis of xxx, so it is a subspace of R3?
In my opinion there is one too many conditions regarding the search for the local maximum of the bivariate Guassian p(x, y). In fact it would not be enough to impose the Hessian determinant of p (x, y) to be > 0, and, using Sylvester's criterion, to impose the top left entry (of the Hessian) to be < 0 too to have the guarantee that both eigenvalues of 'Hessian are negative (ie local maximum condition).
Therefore the condition for which the second derivative of p with respect to y is < 0 is too many.
For more info:
https://math.stackexchange.com/questions/1985889/why-how-does-the-determinant-of-the-hessian-matrix-combined-with-the-2nd-deriva
For question 2.5 a, The value of the A_13(if we use 0 index) should be -5 not 5. The system should not have any solution
I think that in each part (1,2,3) of this exercise the conditional component is not considered.
For example in the first, you considered p(x(t+1)) = p(x(t+1) | y1, ... , yt), but the knowledge of y1, ... , yt influences the density of x(t+1) since x(t+1) depends on y(t) (y(t) influences x(t) which in turn influences x(t+1)) and so on...
I think that you should use the formula of conditional Gaussian to compute p(x(t+1) | y1, ... , yt) since both x(t+1) and y1, ... , yt are Gaussian multivariate densities.
Perhaps you can help me with my confusion about this problem. It appears in the r2-2r1 step, your solution accidentally performed -A24-2A14 to get a value of -3 for A24 instead of A24-2A14 to get a 7 for A24. I don't think the correct solution is not an empty solution set. There are four equations and four unknowns and the solution is [x1 x2 x3 x4]T = [26/15 -13/30 2/5 -1/10]T. Am I thinking about this the wrong way?
Rank of B (made from columns of U) is 4 and not 3.
Verified using online matrix reduction calculator.
EDIT: Made a grave error, it turns out that the matrix has actually rank 3.
I'm not 100% sure but I believe the value for y_2 is incorrect.
It should be: y_2 = x_1*z_2 - y_1
Note: Thanks for making this repo! it's a great help.
Hi, thank you for the solutions. This is super minor, but I thought I'd point it out anyway. I was wondering if for question 5 part c in Chapter 5 it should say "It's first column consists of" instead of "it's first row consists of" when talking about the Jacobian since we've already established the dimension would be n^2 * n and so it could only be every column that has n^2 entries.
The derivative df/dx is not an E x D matrix but a (D x 1)-dimensional vector.
z = Ax + b and therefore Ax is a (E x D) * (D x 1) product -> E x 1 (also after adding up vector b and take the cos() of each component).
We transpose it in a (1 x E)-dimensional vector indeed (this is the gradient of f respect to z vector).
Later, we multiply it with matrix A (dz/dx gradient) -> (1 x E) * (E x D) -> 1 x D and take the transpose -> D x 1.
Now the dimension of the gradient df/dx mathes with the dimension of the input vector x (D x 1).
For my understanding:
The Intersection of U1 and U2 has only one independent column. Let's call the combination of the matrices V.
Therefore av1+...av6 can be reduced by a singel linear combination a*v?
I understand that because dim(U1)=2 and dim(U2) we can set 0=a1v1+a2v2-a4v4-a5v5
Then the solution says v1,v2,v4 are linearly independent and we can set a5 to any number e.g. 9.
How does the augmented matrix look like to solve for v?
| a1
? | a2
| a4
| 9
Thanks for the solutions,
I believe that your matrices are have the variables defined in the wrong order.
You have them defined as
1 x y
0 1 z
0 0 0
Whereas the book has them defined as
1 x z
0 1 y
0 0 0
I believe the 2nd to last line should be....= {(3x1 + 2x2 + x3 The x3 you have as x1
Hi,
For problem #9 in chapter 5, I think there is currently a missing partial with respect to v when computing dq/dv. Essentially, I believe this should be
dq/dv = aq/az * az/av + aq/av * dv/dv = aq/az * az/av + aq/av
the solution currently does not include this last term aq/av. (Note that "a" was used throughout to denote partial derivative.)
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