Same stochastic momenta => identical distribution
I got a stochastic problem and hope some of you can help me!
It's about stochastic variables $X,Y in L^{infty}(P)$, which have the same (finite) stochstic moments $forall n in mathbb{N}: ~mathbb{E}X^n = mathbb{E}Y^n $
(a) First I need to show, that for an arbitrary integrable and continous function $f $ follows: $mathbb{E}f(X) = mathbb{E}f(Y)$
(b) Furthermore I should show, that the distributions $X$ and $Y$ even are identical.
I've already thought about the two problems.
I guess the integrability is only necessary for definig/calculation $mathbb{E}f(X)$, so this is not the propoerty, which is used in the proof. My approach for (a) was the Weierstraß - Theorem. The theorem claims, that $forall f in C([a,b])$ with $-infty < a < b < infty$ there holds: $forall epsilon >0 ~ exists n in mathbb{N}: exists p in P_n: ||f-p ||_{infty} < epsilon$ on $[a,b]$.
So for a continous $f$ I can approximate it by $p in P_n$: $f(z) approx sum_{k=0}^{n} a_k z^k$. So it follows $mathbb{E}f(X) = mathbb{E}left[ sum_{k=0}^{n} a_k z^k right] = sum_{k=0}^{n} a_k mathbb{E}X^n = sum_{k=0}^{n} a_k mathbb{E}Y^n = mathbb{E}f(Y)$.
At least that's the plan. The requirement for the Weierstraß-Theorem was a finite interval $[a,b]$, but according to the indication I cannot restrain to finite $a,b$. How can I generalize my result? (Or is there a general misstake in my approach?)
For (b) I dont't have an idea how to proof this. It might be possible, that I Need to apply a) but don't see how, due to I made a misstake at a)?
I am a little bit confused, due to I thought $X,Y$ having same stochastical momenta is equal to $X$ idential to $Y$. But obviously this is exactly what we need to proof.
I hope some of you can me!
Thanks a lot! :)
probability-theory measure-theory probability-distributions
asked Dec 10 '18 at 21:15
pcalc
27518
add a comment |
I got a stochastic problem and hope some of you can help me!
It's about stochastic variables $X,Y in L^{infty}(P)$, which have the same (finite) stochstic moments $forall n in mathbb{N}: ~mathbb{E}X^n = mathbb{E}Y^n $
(a) First I need to show, that for an arbitrary integrable and continous function $f $ follows: $mathbb{E}f(X) = mathbb{E}f(Y)$
(b) Furthermore I should show, that the distributions $X$ and $Y$ even are identical.
I've already thought about the two problems.
I guess the integrability is only necessary for definig/calculation $mathbb{E}f(X)$, so this is not the propoerty, which is used in the proof. My approach for (a) was the Weierstraß - Theorem. The theorem claims, that $forall f in C([a,b])$ with $-infty < a < b < infty$ there holds: $forall epsilon >0 ~ exists n in mathbb{N}: exists p in P_n: ||f-p ||_{infty} < epsilon$ on $[a,b]$.
So for a continous $f$ I can approximate it by $p in P_n$: $f(z) approx sum_{k=0}^{n} a_k z^k$. So it follows $mathbb{E}f(X) = mathbb{E}left[ sum_{k=0}^{n} a_k z^k right] = sum_{k=0}^{n} a_k mathbb{E}X^n = sum_{k=0}^{n} a_k mathbb{E}Y^n = mathbb{E}f(Y)$.
At least that's the plan. The requirement for the Weierstraß-Theorem was a finite interval $[a,b]$, but according to the indication I cannot restrain to finite $a,b$. How can I generalize my result? (Or is there a general misstake in my approach?)
For (b) I dont't have an idea how to proof this. It might be possible, that I Need to apply a) but don't see how, due to I made a misstake at a)?
I am a little bit confused, due to I thought $X,Y$ having same stochastical momenta is equal to $X$ idential to $Y$. But obviously this is exactly what we need to proof.
I hope some of you can me!
Thanks a lot! :)
probability-theory measure-theory probability-distributions
asked Dec 10 '18 at 21:15
pcalc
27518
Your random variables are given to be bounded a.s., so you can take the relevant $[a,b]$ in the Weierstrass theorem to be $[-M,M]$ where $M$ is the $L^infty$ norm of the variables. Now $E[p(X)]=E[p(Y)]$ for any polynomial $p$, and now you just need to improve that to $E[f(X)]=E[f(Y)]$ for any bounded continuous function $f$.
– Ian
Dec 10 '18 at 21:21
@lan Thanks for your quick respond. That solves my problem with a)!
– pcalc
Dec 10 '18 at 21:40
Moments don't always determine the distribution, but do so in many cases. Maybe relevant: mathoverflow.net/questions/3525/…
– jdods
Dec 11 '18 at 1:23
@jdods But here the variables are given to be bounded a.s., which is well more than enough.
– Ian
Dec 11 '18 at 15:43
add a comment |
I got a stochastic problem and hope some of you can help me!
It's about stochastic variables $X,Y in L^{infty}(P)$, which have the same (finite) stochstic moments $forall n in mathbb{N}: ~mathbb{E}X^n = mathbb{E}Y^n $
(a) First I need to show, that for an arbitrary integrable and continous function $f $ follows: $mathbb{E}f(X) = mathbb{E}f(Y)$
(b) Furthermore I should show, that the distributions $X$ and $Y$ even are identical.
I've already thought about the two problems.
I guess the integrability is only necessary for definig/calculation $mathbb{E}f(X)$, so this is not the propoerty, which is used in the proof. My approach for (a) was the Weierstraß - Theorem. The theorem claims, that $forall f in C([a,b])$ with $-infty < a < b < infty$ there holds: $forall epsilon >0 ~ exists n in mathbb{N}: exists p in P_n: ||f-p ||_{infty} < epsilon$ on $[a,b]$.
So for a continous $f$ I can approximate it by $p in P_n$: $f(z) approx sum_{k=0}^{n} a_k z^k$. So it follows $mathbb{E}f(X) = mathbb{E}left[ sum_{k=0}^{n} a_k z^k right] = sum_{k=0}^{n} a_k mathbb{E}X^n = sum_{k=0}^{n} a_k mathbb{E}Y^n = mathbb{E}f(Y)$.
At least that's the plan. The requirement for the Weierstraß-Theorem was a finite interval $[a,b]$, but according to the indication I cannot restrain to finite $a,b$. How can I generalize my result? (Or is there a general misstake in my approach?)
For (b) I dont't have an idea how to proof this. It might be possible, that I Need to apply a) but don't see how, due to I made a misstake at a)?
I am a little bit confused, due to I thought $X,Y$ having same stochastical momenta is equal to $X$ idential to $Y$. But obviously this is exactly what we need to proof.
I hope some of you can me!
Thanks a lot! :)
probability-theory measure-theory probability-distributions
asked Dec 10 '18 at 21:15
pcalc
27518
I got a stochastic problem and hope some of you can help me!
It's about stochastic variables $X,Y in L^{infty}(P)$, which have the same (finite) stochstic moments $forall n in mathbb{N}: ~mathbb{E}X^n = mathbb{E}Y^n $
(a) First I need to show, that for an arbitrary integrable and continous function $f $ follows: $mathbb{E}f(X) = mathbb{E}f(Y)$
(b) Furthermore I should show, that the distributions $X$ and $Y$ even are identical.
I've already thought about the two problems.
I guess the integrability is only necessary for definig/calculation $mathbb{E}f(X)$, so this is not the propoerty, which is used in the proof. My approach for (a) was the Weierstraß - Theorem. The theorem claims, that $forall f in C([a,b])$ with $-infty < a < b < infty$ there holds: $forall epsilon >0 ~ exists n in mathbb{N}: exists p in P_n: ||f-p ||_{infty} < epsilon$ on $[a,b]$.
So for a continous $f$ I can approximate it by $p in P_n$: $f(z) approx sum_{k=0}^{n} a_k z^k$. So it follows $mathbb{E}f(X) = mathbb{E}left[ sum_{k=0}^{n} a_k z^k right] = sum_{k=0}^{n} a_k mathbb{E}X^n = sum_{k=0}^{n} a_k mathbb{E}Y^n = mathbb{E}f(Y)$.
At least that's the plan. The requirement for the Weierstraß-Theorem was a finite interval $[a,b]$, but according to the indication I cannot restrain to finite $a,b$. How can I generalize my result? (Or is there a general misstake in my approach?)
For (b) I dont't have an idea how to proof this. It might be possible, that I Need to apply a) but don't see how, due to I made a misstake at a)?
I am a little bit confused, due to I thought $X,Y$ having same stochastical momenta is equal to $X$ idential to $Y$. But obviously this is exactly what we need to proof.
I hope some of you can me!
Thanks a lot! :)
probability-theory measure-theory probability-distributions
probability-theory measure-theory probability-distributions
asked Dec 10 '18 at 21:15
pcalc
27518
asked Dec 10 '18 at 21:15
pcalc
27518
asked Dec 10 '18 at 21:15
pcalc
27518
asked Dec 10 '18 at 21:15
pcalc
27518
asked Dec 10 '18 at 21:15
pcalc
27518
27518
Your random variables are given to be bounded a.s., so you can take the relevant $[a,b]$ in the Weierstrass theorem to be $[-M,M]$ where $M$ is the $L^infty$ norm of the variables. Now $E[p(X)]=E[p(Y)]$ for any polynomial $p$, and now you just need to improve that to $E[f(X)]=E[f(Y)]$ for any bounded continuous function $f$.
– Ian
Dec 10 '18 at 21:21
@lan Thanks for your quick respond. That solves my problem with a)!
– pcalc
Dec 10 '18 at 21:40
Moments don't always determine the distribution, but do so in many cases. Maybe relevant: mathoverflow.net/questions/3525/…
– jdods
Dec 11 '18 at 1:23
@jdods But here the variables are given to be bounded a.s., which is well more than enough.
– Ian
Dec 11 '18 at 15:43
add a comment |
Your random variables are given to be bounded a.s., so you can take the relevant $[a,b]$ in the Weierstrass theorem to be $[-M,M]$ where $M$ is the $L^infty$ norm of the variables. Now $E[p(X)]=E[p(Y)]$ for any polynomial $p$, and now you just need to improve that to $E[f(X)]=E[f(Y)]$ for any bounded continuous function $f$.
– Ian
Dec 10 '18 at 21:21
@lan Thanks for your quick respond. That solves my problem with a)!
– pcalc
Dec 10 '18 at 21:40
Moments don't always determine the distribution, but do so in many cases. Maybe relevant: mathoverflow.net/questions/3525/…
– jdods
Dec 11 '18 at 1:23
@jdods But here the variables are given to be bounded a.s., which is well more than enough.
– Ian
Dec 11 '18 at 15:43
Your random variables are given to be bounded a.s., so you can take the relevant $[a,b]$ in the Weierstrass theorem to be $[-M,M]$ where $M$ is the $L^infty$ norm of the variables. Now $E[p(X)]=E[p(Y)]$ for any polynomial $p$, and now you just need to improve that to $E[f(X)]=E[f(Y)]$ for any bounded continuous function $f$.
– Ian
Dec 10 '18 at 21:21
Your random variables are given to be bounded a.s., so you can take the relevant $[a,b]$ in the Weierstrass theorem to be $[-M,M]$ where $M$ is the $L^infty$ norm of the variables. Now $E[p(X)]=E[p(Y)]$ for any polynomial $p$, and now you just need to improve that to $E[f(X)]=E[f(Y)]$ for any bounded continuous function $f$.
– Ian
Dec 10 '18 at 21:21
@lan Thanks for your quick respond. That solves my problem with a)!
– pcalc
Dec 10 '18 at 21:40
@lan Thanks for your quick respond. That solves my problem with a)!
– pcalc
Dec 10 '18 at 21:40
Moments don't always determine the distribution, but do so in many cases. Maybe relevant: mathoverflow.net/questions/3525/…
– jdods
Dec 11 '18 at 1:23
Moments don't always determine the distribution, but do so in many cases. Maybe relevant: mathoverflow.net/questions/3525/…
– jdods
Dec 11 '18 at 1:23
@jdods But here the variables are given to be bounded a.s., which is well more than enough.
– Ian
Dec 11 '18 at 15:43
@jdods But here the variables are given to be bounded a.s., which is well more than enough.
– Ian
Dec 11 '18 at 15:43
add a comment |
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Your random variables are given to be bounded a.s., so you can take the relevant $[a,b]$ in the Weierstrass theorem to be $[-M,M]$ where $M$ is the $L^infty$ norm of the variables. Now $E[p(X)]=E[p(Y)]$ for any polynomial $p$, and now you just need to improve that to $E[f(X)]=E[f(Y)]$ for any bounded continuous function $f$.
– Ian
Dec 10 '18 at 21:21
@lan Thanks for your quick respond. That solves my problem with a)!
– pcalc
Dec 10 '18 at 21:40
Moments don't always determine the distribution, but do so in many cases. Maybe relevant: mathoverflow.net/questions/3525/…
– jdods
Dec 11 '18 at 1:23
@jdods But here the variables are given to be bounded a.s., which is well more than enough.
– Ian
Dec 11 '18 at 15:43