$mathbb{E}(X_{Y+1}X_{2}^{2}X_{2}|x_{1})$ with $Xsim N(0,1)$ and $Ysim Pois(1)$ both independent












0














Let ${X_{i},iinmathbb{N}}$ be a sequence of independent standard normal random variables. Furthermore, $Y$ is a Poisson distributed random variable with parameter $lambda=1$, i.e., $mathbb{P}(Y=n)=frac{e^{-1}}{n!}, n=0,1,2,...$, and Y is independent of the sequence ${X_{i},iinmathbb{N}}$. Compute the following (conditional) expectation
$$
mathbb{E}(X_{Y+1}X_{1}^{2}X_{2}|X_{1})
$$






probability normal-distribution poisson-distribution expected-value







share|cite|improve this question
























  • Can your show your attempt?
    – Kavi Rama Murthy
    Nov 16 '18 at 11:55










  • Knowing, $X_{1}$ it is possible to take out $X_{1}^{2}$. Using the property of conditional expectation, we can condition on Y, which gives $mathbb{E}(X_{Y+1}X_{1}^{2}X_{2}|X_{1})=X_{1}^{2}mathbb{E}(mathbb{E}(X_{2}X_{Y+1}|Y))$. Now, we are able to use the distribution of $Y$ and the fact that $Y$ is independent of sequence ${X_{i},iinmathbb{N}}$, since for $Yneq1$ with probability $1-e^{-1}$ we have $mathbb{E}(X_{2}X_{Y+1})=0$. We may conclude that $mathbb{E}(X_{Y+1}X_{1}^{2}X_{2}|X_{1})=X_{1}^{2}(e^{-1}(1)+(1-e^{-1})(0))$, this was my attempt.
    – R.Sluij
    Nov 16 '18 at 12:05








  • 1




    Congratulations! You have answered the question yourselves.
    – Kavi Rama Murthy
    Nov 16 '18 at 12:12






  • 1




    The $lambda$ in Poisson distribution is supposed to be positive. It should be $lambda=1$.
    – StubbornAtom
    Nov 16 '18 at 12:14












  • Thank you both!
    – R.Sluij
    Nov 21 '18 at 12:46
















0














Let ${X_{i},iinmathbb{N}}$ be a sequence of independent standard normal random variables. Furthermore, $Y$ is a Poisson distributed random variable with parameter $lambda=1$, i.e., $mathbb{P}(Y=n)=frac{e^{-1}}{n!}, n=0,1,2,...$, and Y is independent of the sequence ${X_{i},iinmathbb{N}}$. Compute the following (conditional) expectation
$$
mathbb{E}(X_{Y+1}X_{1}^{2}X_{2}|X_{1})
$$






probability normal-distribution poisson-distribution expected-value







share|cite|improve this question
























  • Can your show your attempt?
    – Kavi Rama Murthy
    Nov 16 '18 at 11:55










  • Knowing, $X_{1}$ it is possible to take out $X_{1}^{2}$. Using the property of conditional expectation, we can condition on Y, which gives $mathbb{E}(X_{Y+1}X_{1}^{2}X_{2}|X_{1})=X_{1}^{2}mathbb{E}(mathbb{E}(X_{2}X_{Y+1}|Y))$. Now, we are able to use the distribution of $Y$ and the fact that $Y$ is independent of sequence ${X_{i},iinmathbb{N}}$, since for $Yneq1$ with probability $1-e^{-1}$ we have $mathbb{E}(X_{2}X_{Y+1})=0$. We may conclude that $mathbb{E}(X_{Y+1}X_{1}^{2}X_{2}|X_{1})=X_{1}^{2}(e^{-1}(1)+(1-e^{-1})(0))$, this was my attempt.
    – R.Sluij
    Nov 16 '18 at 12:05








  • 1




    Congratulations! You have answered the question yourselves.
    – Kavi Rama Murthy
    Nov 16 '18 at 12:12






  • 1




    The $lambda$ in Poisson distribution is supposed to be positive. It should be $lambda=1$.
    – StubbornAtom
    Nov 16 '18 at 12:14












  • Thank you both!
    – R.Sluij
    Nov 21 '18 at 12:46














0












0








0







Let ${X_{i},iinmathbb{N}}$ be a sequence of independent standard normal random variables. Furthermore, $Y$ is a Poisson distributed random variable with parameter $lambda=1$, i.e., $mathbb{P}(Y=n)=frac{e^{-1}}{n!}, n=0,1,2,...$, and Y is independent of the sequence ${X_{i},iinmathbb{N}}$. Compute the following (conditional) expectation
$$
mathbb{E}(X_{Y+1}X_{1}^{2}X_{2}|X_{1})
$$






probability normal-distribution poisson-distribution expected-value







share|cite|improve this question















Let ${X_{i},iinmathbb{N}}$ be a sequence of independent standard normal random variables. Furthermore, $Y$ is a Poisson distributed random variable with parameter $lambda=1$, i.e., $mathbb{P}(Y=n)=frac{e^{-1}}{n!}, n=0,1,2,...$, and Y is independent of the sequence ${X_{i},iinmathbb{N}}$. Compute the following (conditional) expectation
$$
mathbb{E}(X_{Y+1}X_{1}^{2}X_{2}|X_{1})
$$






probability normal-distribution poisson-distribution expected-value




probability normal-distribution poisson-distribution expected-value






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 21 '18 at 12:45

























asked Nov 16 '18 at 11:42









R.Sluij

236




236












  • Can your show your attempt?
    – Kavi Rama Murthy
    Nov 16 '18 at 11:55










  • Knowing, $X_{1}$ it is possible to take out $X_{1}^{2}$. Using the property of conditional expectation, we can condition on Y, which gives $mathbb{E}(X_{Y+1}X_{1}^{2}X_{2}|X_{1})=X_{1}^{2}mathbb{E}(mathbb{E}(X_{2}X_{Y+1}|Y))$. Now, we are able to use the distribution of $Y$ and the fact that $Y$ is independent of sequence ${X_{i},iinmathbb{N}}$, since for $Yneq1$ with probability $1-e^{-1}$ we have $mathbb{E}(X_{2}X_{Y+1})=0$. We may conclude that $mathbb{E}(X_{Y+1}X_{1}^{2}X_{2}|X_{1})=X_{1}^{2}(e^{-1}(1)+(1-e^{-1})(0))$, this was my attempt.
    – R.Sluij
    Nov 16 '18 at 12:05








  • 1




    Congratulations! You have answered the question yourselves.
    – Kavi Rama Murthy
    Nov 16 '18 at 12:12






  • 1




    The $lambda$ in Poisson distribution is supposed to be positive. It should be $lambda=1$.
    – StubbornAtom
    Nov 16 '18 at 12:14












  • Thank you both!
    – R.Sluij
    Nov 21 '18 at 12:46


















  • Can your show your attempt?
    – Kavi Rama Murthy
    Nov 16 '18 at 11:55










  • Knowing, $X_{1}$ it is possible to take out $X_{1}^{2}$. Using the property of conditional expectation, we can condition on Y, which gives $mathbb{E}(X_{Y+1}X_{1}^{2}X_{2}|X_{1})=X_{1}^{2}mathbb{E}(mathbb{E}(X_{2}X_{Y+1}|Y))$. Now, we are able to use the distribution of $Y$ and the fact that $Y$ is independent of sequence ${X_{i},iinmathbb{N}}$, since for $Yneq1$ with probability $1-e^{-1}$ we have $mathbb{E}(X_{2}X_{Y+1})=0$. We may conclude that $mathbb{E}(X_{Y+1}X_{1}^{2}X_{2}|X_{1})=X_{1}^{2}(e^{-1}(1)+(1-e^{-1})(0))$, this was my attempt.
    – R.Sluij
    Nov 16 '18 at 12:05








  • 1




    Congratulations! You have answered the question yourselves.
    – Kavi Rama Murthy
    Nov 16 '18 at 12:12






  • 1




    The $lambda$ in Poisson distribution is supposed to be positive. It should be $lambda=1$.
    – StubbornAtom
    Nov 16 '18 at 12:14












  • Thank you both!
    – R.Sluij
    Nov 21 '18 at 12:46
















Can your show your attempt?
– Kavi Rama Murthy
Nov 16 '18 at 11:55




Can your show your attempt?
– Kavi Rama Murthy
Nov 16 '18 at 11:55












Knowing, $X_{1}$ it is possible to take out $X_{1}^{2}$. Using the property of conditional expectation, we can condition on Y, which gives $mathbb{E}(X_{Y+1}X_{1}^{2}X_{2}|X_{1})=X_{1}^{2}mathbb{E}(mathbb{E}(X_{2}X_{Y+1}|Y))$. Now, we are able to use the distribution of $Y$ and the fact that $Y$ is independent of sequence ${X_{i},iinmathbb{N}}$, since for $Yneq1$ with probability $1-e^{-1}$ we have $mathbb{E}(X_{2}X_{Y+1})=0$. We may conclude that $mathbb{E}(X_{Y+1}X_{1}^{2}X_{2}|X_{1})=X_{1}^{2}(e^{-1}(1)+(1-e^{-1})(0))$, this was my attempt.
– R.Sluij
Nov 16 '18 at 12:05






Knowing, $X_{1}$ it is possible to take out $X_{1}^{2}$. Using the property of conditional expectation, we can condition on Y, which gives $mathbb{E}(X_{Y+1}X_{1}^{2}X_{2}|X_{1})=X_{1}^{2}mathbb{E}(mathbb{E}(X_{2}X_{Y+1}|Y))$. Now, we are able to use the distribution of $Y$ and the fact that $Y$ is independent of sequence ${X_{i},iinmathbb{N}}$, since for $Yneq1$ with probability $1-e^{-1}$ we have $mathbb{E}(X_{2}X_{Y+1})=0$. We may conclude that $mathbb{E}(X_{Y+1}X_{1}^{2}X_{2}|X_{1})=X_{1}^{2}(e^{-1}(1)+(1-e^{-1})(0))$, this was my attempt.
– R.Sluij
Nov 16 '18 at 12:05






1




1




Congratulations! You have answered the question yourselves.
– Kavi Rama Murthy
Nov 16 '18 at 12:12




Congratulations! You have answered the question yourselves.
– Kavi Rama Murthy
Nov 16 '18 at 12:12




1




1




The $lambda$ in Poisson distribution is supposed to be positive. It should be $lambda=1$.
– StubbornAtom
Nov 16 '18 at 12:14






The $lambda$ in Poisson distribution is supposed to be positive. It should be $lambda=1$.
– StubbornAtom
Nov 16 '18 at 12:14














Thank you both!
– R.Sluij
Nov 21 '18 at 12:46




Thank you both!
– R.Sluij
Nov 21 '18 at 12:46










1 Answer
1






active

oldest

votes


















0














I would suggest the following:



$$mathbb{E}left[ X_{Y+1} X_1^2 X_2 | X_1 right] = mathbb{P}(Y=1) mathbb{E} left[ X_2X_1^2 X_2 | X_1 right] + mathbb{P} (Ygeq 2) mathbb{E}left[X_3X_1^2X_2 | X_1right]\ = mathbb{P}(Y=1) mathbb{E}[X_2^2] X_1^2 + mathbb{P}(Ygeq 2) mathbb{E}[X_2]^2 cdot X_1^2
$$



And now you simply use the distribution of $Y$ and $X_i$...






share|cite|improve this answer





















  • Thanks, however that gives the same value
    – R.Sluij
    Nov 22 '18 at 19:54











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

votes






active

oldest

votes









0














I would suggest the following:



$$mathbb{E}left[ X_{Y+1} X_1^2 X_2 | X_1 right] = mathbb{P}(Y=1) mathbb{E} left[ X_2X_1^2 X_2 | X_1 right] + mathbb{P} (Ygeq 2) mathbb{E}left[X_3X_1^2X_2 | X_1right]\ = mathbb{P}(Y=1) mathbb{E}[X_2^2] X_1^2 + mathbb{P}(Ygeq 2) mathbb{E}[X_2]^2 cdot X_1^2
$$



And now you simply use the distribution of $Y$ and $X_i$...






share|cite|improve this answer





















  • Thanks, however that gives the same value
    – R.Sluij
    Nov 22 '18 at 19:54
















0














I would suggest the following:



$$mathbb{E}left[ X_{Y+1} X_1^2 X_2 | X_1 right] = mathbb{P}(Y=1) mathbb{E} left[ X_2X_1^2 X_2 | X_1 right] + mathbb{P} (Ygeq 2) mathbb{E}left[X_3X_1^2X_2 | X_1right]\ = mathbb{P}(Y=1) mathbb{E}[X_2^2] X_1^2 + mathbb{P}(Ygeq 2) mathbb{E}[X_2]^2 cdot X_1^2
$$



And now you simply use the distribution of $Y$ and $X_i$...






share|cite|improve this answer





















  • Thanks, however that gives the same value
    – R.Sluij
    Nov 22 '18 at 19:54














0












0








0






I would suggest the following:



$$mathbb{E}left[ X_{Y+1} X_1^2 X_2 | X_1 right] = mathbb{P}(Y=1) mathbb{E} left[ X_2X_1^2 X_2 | X_1 right] + mathbb{P} (Ygeq 2) mathbb{E}left[X_3X_1^2X_2 | X_1right]\ = mathbb{P}(Y=1) mathbb{E}[X_2^2] X_1^2 + mathbb{P}(Ygeq 2) mathbb{E}[X_2]^2 cdot X_1^2
$$



And now you simply use the distribution of $Y$ and $X_i$...






share|cite|improve this answer












I would suggest the following:



$$mathbb{E}left[ X_{Y+1} X_1^2 X_2 | X_1 right] = mathbb{P}(Y=1) mathbb{E} left[ X_2X_1^2 X_2 | X_1 right] + mathbb{P} (Ygeq 2) mathbb{E}left[X_3X_1^2X_2 | X_1right]\ = mathbb{P}(Y=1) mathbb{E}[X_2^2] X_1^2 + mathbb{P}(Ygeq 2) mathbb{E}[X_2]^2 cdot X_1^2
$$



And now you simply use the distribution of $Y$ and $X_i$...







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 21 '18 at 13:04









cptflint

208




208












  • Thanks, however that gives the same value
    – R.Sluij
    Nov 22 '18 at 19:54


















  • Thanks, however that gives the same value
    – R.Sluij
    Nov 22 '18 at 19:54
















Thanks, however that gives the same value
– R.Sluij
Nov 22 '18 at 19:54




Thanks, however that gives the same value
– R.Sluij
Nov 22 '18 at 19:54


















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