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visits member for 2 years, 7 months
seen Mar 11 at 19:31

Dec
16
answered Proving an optimization problem
Dec
16
comment Proving an optimization problem
I didn't know this property of entropy. Could you tell me where you got it from?
Dec
15
comment Simple probability question related to IT
In case there are maximally 5000 users, a Binomial distribution might be more appropriate. However, since the request rate is probably low and the number of users is high, the Poisson distribution will probably do. But the problem remains: You need to know or assume the rate of requests.
Dec
14
comment Simple probability question related to IT
Yes, but as I understand it you don't want to many other users sending requests during that time.
Dec
14
answered Simple probability question related to IT
Dec
7
accepted Conditional Independence and Mutual information
Oct
12
comment Conditional Independence and Mutual information
That may very well be. I tried being somewhat rigorous by using $X, Y, Z$ for random variables, and $x,y,z$ for actual values of the random variables. However, since I never learned measure theory properly (any good book to recommend?) the notation might not be reasonable. Did I understand you correctly now, that $p(x,y|y)=p(x|z)p(y|z)$ holds only almost surely in the continuous setting? Thank you for spending all the time to solve my issue.
Oct
11
comment Conditional Independence and Mutual information
I was not expressing myself very clearly. What confuses me is: Say the expectation is taken over a function $f$ of a uniform random variable. $f(u)=1$ for $u\in[0,.5)\cup(.5,1]$ and $f(.5) = e$. Then $E[\log f(U)] = \log E[f(u)]$ but not $f \equiv 1$ everywhere. This might hold with probability one, but not for every $u$. Does that mean that $p(x,y|z)=p(x|z)p(y|z)$ also holds only with probability one?
Oct
11
comment Conditional Independence and Mutual information
Thanks for answering (and pointing to the fact, that I still had an error in the definition of conditional mutual information which is corrected now). If I understand you correctly, you argue that we only have equality if we integrate over $\log c$. However, what if $p(x|z)p(y|z)/p(x,y|z)\not=c$ at finitely many points $z$? The integral would still be the same, but we would not have $p(x|z)p(y|z)=p(x,y|z)$ at each $z$. Or am I missing something?
Oct
11
revised Conditional Independence and Mutual information
added 2 characters in body
Oct
11
comment Conditional Independence and Mutual information
I just realized that I had a mistake in the definition. I am not sure whether I understand you comment. Do you want to say that the second definition has to be satisfied for each $z$ (instead of the mean) in order to be equivalent to the first?
Oct
11
revised Conditional Independence and Mutual information
added 31 characters in body
Oct
10
asked Conditional Independence and Mutual information
Sep
1
awarded  Yearling
Aug
23
comment Minimize Variance of a Linear Function
@AnonSubmitter85 You are probably right. In that case yours is the right answer, of course.
Aug
23
comment Minimize Variance of a Linear Function
@AnonSubmitter85 What I don't understand is: If $x$ are vectors, then $y = Ax+c$ is a vector as well. Does he mean $\mbox{var}(y)=\frac{1}{n}\sum_i (y_i - \overline y)^2$ in which case it would work for a single vector, or does he mean $\mbox{cov}(y)$ in which case he'd have to specify what minimizing variance meant. In the former case, one could of course additionally minimize the variance over the vector entries w.r.t. different samples $x_i$.
Aug
23
comment Minimize Variance of a Linear Function
Hi Tom, can you be a bit more specific? Do you mean the variance over the vector entries? Is $A$ square or can it have any shape? What have you tried already?
May
6
accepted Efficient (approximate) projection onto the special orthogonal group
May
6
comment Efficient (approximate) projection onto the special orthogonal group
Yes, of course! I totally missed that. Thanks a lot.
May
6
comment Efficient (approximate) projection onto the special orthogonal group
I hadn't thought about that. Do you know how I would efficiently transform $Z$ into the block-diagonal form (i.e. are there standard algorithms that do that)?