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2d
comment $\mathrm E [X \mid X=x] = x$?
Not all the answers, not necessary, not a problem either.
2d
comment $\mathrm E [X \mid X=x] = x$?
To answer the question... When $P(X=x)=0$, $E(Y\mid X=x)$ is defined as $a(x)$ where the function $a$ is measurable and such that $E(Y\mid X)=a(X)$ almost surely. The "almost surely" makes that $a$ is only unique up to negligible sets for $P_X$, and this has unfortunate consequences. For example, assume that $X=Y$ is unform on $(0,1)$ and let $a(y)=y+(z-y)\mathbf 1_{y=x}$ for some $z\ne x$. Then $E(X\mid X)=a(X)$ almost surely (one can check the two conditions necessary for $a(X)$ to be (a version of) $E(X\mid X)$ hold) but, obviously, $E(X\mid X=x)=z\ne x$. To sum up, the result is wrong.
2d
comment $\mathrm E [X \mid X=x] = x$?
To be clear: the result that the question asks for and that this answer "proves", is wrong.
2d
comment $\mathrm E [X \mid X=x] = x$?
When $P(X=x)=0$, $E(Y\mid X=x)$ is defined as $a(x)$ where the function $a$ is measurable and such that $E(Y\mid X)=a(X)$ almost surely. The "almost surely" makes that $a$ is only unique up to negligible sets for $P_X$, and this has unfortunate consequences. For example, assume that $X=Y$ is unform on $(0,1)$ and let $a(y)=y+(z-y)\mathbf 1_{y=x}$ for some $z\ne x$. Then $E(X\mid X)=a(X)$ almost surely (one can check the two conditions necessary for $a(X)$ to be (a version of) $E(X\mid X)$ hold) but, obviously, $E(X\mid X=x)=z\ne x$. To sum up, the result is wrong.
2d
comment $\mathrm E [X \mid X=x] = x$?
When $P(X=x)=0$, $E(Y\mid X=x)$ is defined as $a(x)$ where the function $a$ is measurable and such that $E(Y\mid X)=a(X)$ almost surely. The "almost surely" makes that $a$ is only unique up to negligible sets for $P_X$, and this has unfortunate consequences. For example, assume that $X=Y$ is unform on $(0,1)$ and let $a(y)=y+(z-y)\mathbf 1_{y=x}$ for some $z\ne x$. Then $E(X\mid X)=a(X)$ almost surely (one can check the two conditions necessary for $a(X)$ to be (a version of) $E(X\mid X)$ hold) but, obviously, $E(X\mid X=x)=z\ne x$. To sum up, the result is wrong.
2d
comment When the sum of Markov chains is a Markov chain: “dumb” algorithm
"No", applying to which part of what I said exactly? Yes there are sporadic cases when X+Y is Markov, such as the highly symmetrical one in your comment. I do not think one can expect general criteria though. For a more general framework, the notion of lumpability might help, since your question amounts to determine when the function X+Y of the Markov chain (X,Y), is still Markov.
2d
comment Evaluating $\lim_{n\to\infty} e^{-n} \sum\limits_{k=0}^{n} \frac{n^k}{k!}$
Oh yes, a revolution, what else? I have absolutely no idea what you are talking about (and I wonder if you have). No big deal anyway.
Jul
31
comment Evaluating $\lim_{n\to\infty} e^{-n} \sum\limits_{k=0}^{n} \frac{n^k}{k!}$
In theory, sure. (In the specific case at hand, "deeper approach", coming after "profound answer", is rather hilarious, though.)
Jul
31
comment When the sum of Markov chains is a Markov chain: “dumb” algorithm
The problem as you stated it was to determine the sums of independent sums of independent increments which were Markov chains. The answer is: all of them. Now if you have a different problem in mind, please be specific. (Note that the "other" case you describe in your comment is actually a special case of independent sums of independent increments.)
Jul
31
comment Generalizing the pull-out property in conditional expectations
(P.S.: There should be a system on math.SE to signal non empty questions such as the present one.)
Jul
31
comment Generalizing the pull-out property in conditional expectations
The problem lies in the negligible sets necessary to define conditional expectations. Consider $X=Y$ uniform on $(0,1)$, then $E(Y\mid X)=X$ almost surely and, for every $x$, $E(xY\mid X)=xX\mathbf 1_{X\ne x}$ almost surely since $P(X=x)=0$. Assume that these (legal) choices are made for $E(Y\mid X)$ and for $E(xY\mid X)$ for every $x$. Then $X(\omega)E(Y\mid X)(\omega)=X^2(\omega)$ and $E(X(\omega)Y\mid X)(\omega)=0$ for every $\omega$ hence, in this case, the suggested identity holds for no $\omega$ at all.
Jul
31
comment When the sum of Markov chains is a Markov chain: “dumb” algorithm
Indeed the sum of two independent Markov chains may not be a Markov chain, but when each Markov chain is a sum of independent increments, then their sum is again Markov and again a sum of independent increments. If the increments of the two chains at time $n$ have distributions $p_n$ and $q_n$ respectively, then the increments of their sum have distribution $p_n\ast q_n$ at time $n$ (convolution). I have the feeling that this simple observation answers every part of your question but I might be mistaken.
Jul
31
comment Does $\lim_{x\to\infty}f(x)e^{-x^2/2}=0$ imply $\lim_{x\to\infty}f'(x)e^{-x^2/2}=0$?
Why still no accepted answer to this?
Jul
31
comment Probability distribution of derivative of function of random variable
You said you know the distribution of $g(X)$ for every $g$, try $g=f'$.
Jul
31
comment A nonlinear differntial equation $f'''-(f')^2+1=0,$
*Sorry, the two $5$ in the formula for $f'(x)$ above should read $(\sqrt3+\sqrt2)^2$. I think. Or something similar... :-)
Jul
31
comment Probability distribution of derivative of function of random variable
?? Apply what you can do to the function $f'$.
Jul
31
comment A nonlinear differntial equation $f'''-(f')^2+1=0,$
And, unrelatedly, this should be closed for lack of personal input.
Jul
31
comment A nonlinear differntial equation $f'''-(f')^2+1=0,$
@YiorgosS.Smyrlis Indeed the initial conditions allow to complete this, see my comment to main.
Jul
31
comment A nonlinear differntial equation $f'''-(f')^2+1=0,$
@ClaudeLeibovici See above.