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Martin Sleziak
Martin Sleziak
1:08 PM
Maybe we could check also some macros on Cross Validated.
This answer contains the macro \E defined in the question: Derivation of normalizing transform for GLMs. stats.stackexchange.com/posts/253110/revisions
In principle, just by knowing $f(x,y)$, you implicitly know the distributions of $X$ and $Y$ including their expectations and also $\E(X \mid Y)$. So, all of the latter "extra" information is superfluous. Note that the transformation $(X,Y) \mapsto (X,X+Y)$ is one-to-one and so transformation-of-variables will give a density for $(X,X+Y)$ (which is easy to write in terms of $f$). From this, you can then easily recover a simple formula for $\E(X \mid X+Y)$. But, saying anything more explicit within the very general assumptions you've made doesn't seem likely. — cardinal Jan 2 '12 at 19:41
great explanation. Why do you say $\E_\theta \dot{\ell}(\theta) =0$? it's a function of $\theta$ - isn't it 0 only when evaluated at the true parameter $\theta_0$? — ihadanny Aug 13 '16 at 6:33
The question in comment : NO, $\E (O_i | \lambda_i)=\lambda_i E_i$, but $\E O_i = E_i$, since $\E \lambda_i=1$. — kjetil b halvorsen Dec 17 '16 at 15:30
The sign changed as $\E(X) = -\E(Y)$ (fact 2). I posted the answer as I thought readers from What is cov(X,Y), where X=min(U,V) and Y=max(U,V) for independent uniform(0,1) variables U and V? could be interested. — Franck Dernoncourt Jan 9 '17 at 1:21
No, the return values are $\E(Z_+)$ and $\Cov(Z_+)$; what I printed was the $L_\infty$ distance between Monte Carlo estimators of those quantities and the computed value. You could maybe invert these expressions to get a moment-matching estimator for $\mu$ and $\Sigma$ – Rosenbaum actually does that in his section 3 in the truncated case – but that's not what I wanted here. — Dougal Feb 1 '18 at 21:37
No. By exchangeability, all the expectations summed over are equal, so the sum is that common expectation times $n!$. So the expectation is $\mu \times \frac{n+1}{2}$, which you can see equals $\E X_j \times \E R_j$. It will follow that the correlation is zero. — kjetil b halvorsen Nov 9 '18 at 13:30
This answer contains macro \Cov defined in the question: Covariance of Wiener Process. stats.stackexchange.com/posts/230759/revisions
This one was already mentioned, since it contains \E:
No, the return values are $\E(Z_+)$ and $\Cov(Z_+)$; what I printed was the $L_\infty$ distance between Monte Carlo estimators of those quantities and the computed value. You could maybe invert these expressions to get a moment-matching estimator for $\mu$ and $\Sigma$ – Rosenbaum actually does that in his section 3 in the truncated case – but that's not what I wanted here. — Dougal Feb 1 '18 at 21:37
Here is another example:
@StellaLee To answer your comment-question in a comment (if it goes any further, you should ask this as a separate question): Note that $$\DeclareMathOperator{\E}{\mathbb E}\E[(X Y)_{ij}] = \E\left[\sum_k X_{ik} Y_{kj}\right] = \sum_k \E[X_{ik} Y_{kj}] = \sum_k \E[X_{ik}] \E[Y_{kj}] + \Cov(X_{ik}, Y_{kj}).$$ If you just want $\E[X Y]$, then doing it elementwise like this suffices, though if you want something like $\E[\lVert X Y \rVert]$ then it gets trickier. — Dougal Mar 23 '16 at 18:25
This answer was already mentioned, since it contains both \Cov and \Cor: Prove using correlation to do t-test is equivalent to the standard t-test formula in linear regression?
I did not find broken comments.
 

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