Timeline for Mean and Variance of a function of a multivariate normal
Current License: CC BY-SA 4.0
20 events
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| Mar 11, 2020 at 15:55 | history | edited | jld | CC BY-SA 4.0 |
improved latex
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| Oct 19, 2016 at 20:02 | history | edited | jld | CC BY-SA 3.0 |
added simpler form for E(W) in the general case.
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| Oct 19, 2016 at 17:07 | comment | added | jld | @Peter I see what you're saying -- if we have a satisfactory solution for $\int_c^\infty x e^{-x^2/2}dx$ then we can also handle $\int_0^\infty x e^{-(x-\mu)^2/2}dx$. My point was that this 'satisfactory solution' in general necessarily contains special functions such as $\Phi$ or $\text{erf}$ so it's just an integral in disguise. It's just a matter then of manipulating the integrals in my answer into whatever form you like best (also thanks for the non-paywall source). | |
| Oct 19, 2016 at 15:15 | comment | added | jld | Well, let's say we have $X \sim \mathcal N(\mu, 1)$ so $$E(\max(X, 0)) \propto \int_0^\infty x e^{-(x-\mu)^2/2} = e^{-\mu^2/2} + \mu \sqrt{2\pi} \left( 1 - \Phi(-\mu)\right).$$ If this integral has a closed form solution in terms of elementary functions of $\mu$ then it looks like we can turn that into a closed-form solution to $\Phi(\mu)$, which is provably impossible. So if this is all correct then the only possible 'closed forms' for $E(\max(X, 0))$ will have special functions in them and aren't very different from expressions with integrals. Does this seem right? | |
| Oct 19, 2016 at 13:49 | comment | added | Peter | It does use a zero-mean multivariate normal, but allows for arbitrary cutoffs, so the problem is equivalent to arbitrary-mean with zero-cutoffs. It's also dealing with a truncated-normal, rather than a rectified-normal (as is the case in my problem), but I think this is a pre-requesite for finding the solution for the rectified-normal case. (Non-paywall source) | |
| Oct 19, 2016 at 13:42 | comment | added | jld | @Peter If I'm reading these correctly it seems that they're answering slightly different things than your question. In the Lee paper that you link to, they're deriving the results for error terms which importantly have mean 0 which makes the math way easier (see the first two sections of my answer, for example). Lee cites Tallis (1961), where Tallis derives the MGF of the truncated normal. I found the first page of this paper (rest was paywalled) and it looks like it is again the mean 0 multivariate normal. I definitely could be wrong about this though | |
| Oct 19, 2016 at 7:45 | comment | added | Peter | There're a bunch of papers on it from the 70's: sciencedirect.com/science/article/pii/0165176579901113 | |
| Oct 18, 2016 at 20:08 | comment | added | jld | @Peter Thanks. I'd be very interested in seeing a simpler solution if it exists. | |
| Oct 17, 2016 at 14:40 | comment | added | Peter | Thanks @Chaconne, will get back to this and mark to either mark your response complete or post a simplified solution later. | |
| Oct 13, 2016 at 19:49 | history | edited | jld | CC BY-SA 3.0 |
replaced "F << m + delta" with an explanation of what this means
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| Oct 13, 2016 at 17:19 | history | edited | jld | CC BY-SA 3.0 |
fixed typo in update 1
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| Oct 13, 2016 at 17:07 | comment | added | jld | @Peter I believe that I've completed extended this to the multivariate case at this point. | |
| Oct 13, 2016 at 17:06 | history | edited | jld | CC BY-SA 3.0 |
added 1550 characters in body
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| Oct 13, 2016 at 16:15 | comment | added | jld | @Peter I've added a bit on $X \sim \mathcal N(\mu, \sigma)$. Also note that this does address your general case for $E( \bf Y)$ since the expectation of a random vector is the vector of expectations of each element, and we've got $E(Y_i)$ for each $i$. | |
| Oct 13, 2016 at 16:07 | history | edited | jld | CC BY-SA 3.0 |
added general normal case
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| Oct 13, 2016 at 15:55 | history | edited | jld | CC BY-SA 3.0 |
edited body
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| Oct 13, 2016 at 15:37 | comment | added | Peter | Thanks for your response. However, I have trouble seeing how this could be extended to answer the original problem. The main difficulty would be in solving the integral when (a) $\mu$ is not zero, and (b) $\Sigma$ is not diagonal. | |
| Oct 13, 2016 at 15:13 | comment | added | jld | True, but at least this is still a principled approach. | |
| Oct 13, 2016 at 15:10 | comment | added | Josh Magarick | Unfortunately it won't be nearly as nice when $\mu \neq 0$. You lose symmetry which makes it quite a bit sloppier. | |
| Oct 13, 2016 at 15:05 | history | answered | jld | CC BY-SA 3.0 |