Timeline for Log-convexity of conditional variances
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 25, 2017 at 20:07 | vote | accept | Xiaosheng Mu | ||
| Jan 25, 2017 at 14:37 | comment | added | Xiaosheng Mu | Got it. I forgot to use symmetry in that derivation. That's a wonderful proof. | |
| Jan 25, 2017 at 5:15 | comment | added | Suvrit | Recall that $\text{tr}(MHMH) = \langle \text{vec}(MHM), \text{vec}(H) \rangle = h^T(M\otimes M)h$ where $h=\text{vec}(H)$. Hence, I stated the inequality in terms of the the Kronecker product alone. | |
| Jan 24, 2017 at 21:17 | comment | added | Xiaosheng Mu | If I'm not mistaken, $$d^2/dt^2 F(I+A(D+tH)) = -Tr[ ((A^{-1}+D+tH)^{-1} H)^2].$$ Then we need to show $$Tr[ ((A^{-1}+D)^{-1} H)^2] \leq Tr[ ((B^{-1}+D)^{-1} H)^2].$$ How would you show this for general $H$? | |
| Jan 24, 2017 at 18:35 | comment | added | Suvrit | Try doing this: $d^2/dt^2 F(D+tH)$, where $F(D)$ is the difference of log-dets up there; you'll obtain something along the lines I mentioned; have a look at mathoverflow.net/questions/214908/… for instance...; in particular because of how we identify the derivative of the inverse of a matrix -- I just differentiated the first derivative wrt to the matrix $D$ to obtain the hessian actually... | |
| Jan 24, 2017 at 17:16 | comment | added | Xiaosheng Mu | It does seem true that if we restrict attention to diagonal entries of D, then the corresponding (smaller) Hessian matrix is the Hadamard square of $A^{−1}+D$, which is a principal minor of the Kronecker product. So your argument at least resolves my original question :) | |
| Jan 24, 2017 at 16:57 | comment | added | Xiaosheng Mu | That's a very neat idea. I was able to follow your argument until you identify the Hessian matrix with the Kronecker product. What I got is that $$\frac{\partial^2 \det(A+ID)}{\partial D_{ij} \partial D_{i'j'}} = (A^{-1}+D)^{-1}\mid_{j'i} \cdot (A^{-1}+D)^{-1}\mid_{ji'},$$ which doesn't seem to correspond to the Kronecker product. Am I missing something obvious? | |
| Jan 24, 2017 at 7:45 | comment | added | Suvrit | To avoid the "\epsilon" argument above, alternatively, we could instead work with $D$ instead of $D^{-1}$ and establish log-concavity in $D$, from which log-convexity in $D$ should follows; I have not verified these details though. | |
| Jan 24, 2017 at 7:44 | history | answered | Suvrit | CC BY-SA 3.0 |