PRML-Solution-Manual
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zhengqigao
My Own Solution Manual of PRML
Hi, Zhengqi, as usual, I learnt a lot from your solution, like the one to exercise 4.26 which uses a different approach, simple and elegant compared to taking differentiation and...
Hi, Zhengqi, I found an issue about Problem 4.6 solution. At the bottom of page 93, the left side of the equation we need to prove is $\sum\limits_{n=1}^N({\bf x}_n{\bf x}_n^T)-N{\bf...
for Gam function h(x) function should only depend on x but in the solution it also depends on eta1. Alternative solution, taking x^eta1 as exp(eta1*ln(x)) we get eta = same...
At the second term of the lower bound $\mathbb{E}[\ln p(\boldsymbol{w}|\alpha)]_{\boldsymbol{w},\alpha}$ , the last term should be $$ \mathbb{E}[\boldsymbol{w}^{\text{T}} \boldsymbol{w}]_{\boldsymbol{w}} $$ 
The first term in the solution will not simply so easily, because we are doing summation(xn-mu(N))^2 and not summation(xn-mu(N-1))^2. Using sequential expansion for mu(N) = mu(N-1) + (xn - mu(N-1))/N...
Hello Zhengqi, first thanks for the publication of the solution manual! Starting after the paragraph > For E[σ2 ML ], we need to take advantage of (1.56) and what has...
Exc. 1.1
Hi! This is more of a question (albeit perhaps a stupid one), but I do not understand how the derivative of y(x_n,w) in exercise 1.1 is simply (x_n)^i. I see...
 About problem 6.1, is this really a linear combination? I do not really understand what is happening here. It seems you are using variables as the denominator of the...
I believe that in the denominator of the final expression of the gradient the two sumatories simplify to N.
Thank you for sharing your solutions! I think there's a minor issue in your solution to Exercise 10.26. Shouldn't the result for $-\mathbb{E}[ln q^*(\beta)]$ be negated? Essentially, it is analogous...
