$mathbb{E}(X_{Y+1}X_{2}^{2}X_{2}|x_{1})$ with $Xsim N(0,1)$ and $Ysim Pois(1)$ both independent
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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
asked Nov 16 at 11:42
R.Sluij
236
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up vote
0
down vote
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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
asked Nov 16 at 11:42
R.Sluij
236
Can your show your attempt?
– Kavi Rama Murthy
Nov 16 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 at 12:05
1
Congratulations! You have answered the question yourselves.
– Kavi Rama Murthy
Nov 16 at 12:12
1
The $lambda$ in Poisson distribution is supposed to be positive. It should be $lambda=1$.
– StubbornAtom
Nov 16 at 12:14
Thank you both!
– R.Sluij
Nov 21 at 12:46
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
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
asked Nov 16 at 11:42
R.Sluij
236
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
asked Nov 16 at 11:42
R.Sluij
236
asked Nov 16 at 11:42
R.Sluij
236
edited Nov 21 at 12:45
asked Nov 16 at 11:42
R.Sluij
236
asked Nov 16 at 11:42
R.Sluij
236
asked Nov 16 at 11:42
R.Sluij
236
236
Can your show your attempt?
– Kavi Rama Murthy
Nov 16 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 at 12:05
1
Congratulations! You have answered the question yourselves.
– Kavi Rama Murthy
Nov 16 at 12:12
1
The $lambda$ in Poisson distribution is supposed to be positive. It should be $lambda=1$.
– StubbornAtom
Nov 16 at 12:14
Thank you both!
– R.Sluij
Nov 21 at 12:46
add a comment |
Can your show your attempt?
– Kavi Rama Murthy
Nov 16 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 at 12:05
1
Congratulations! You have answered the question yourselves.
– Kavi Rama Murthy
Nov 16 at 12:12
1
The $lambda$ in Poisson distribution is supposed to be positive. It should be $lambda=1$.
– StubbornAtom
Nov 16 at 12:14
Thank you both!
– R.Sluij
Nov 21 at 12:46
Can your show your attempt?
– Kavi Rama Murthy
Nov 16 at 11:55
Can your show your attempt?
– Kavi Rama Murthy
Nov 16 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 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 at 12:05
1
1
Congratulations! You have answered the question yourselves.
– Kavi Rama Murthy
Nov 16 at 12:12
Congratulations! You have answered the question yourselves.
– Kavi Rama Murthy
Nov 16 at 12:12
1
1
The $lambda$ in Poisson distribution is supposed to be positive. It should be $lambda=1$.
– StubbornAtom
Nov 16 at 12:14
The $lambda$ in Poisson distribution is supposed to be positive. It should be $lambda=1$.
– StubbornAtom
Nov 16 at 12:14
Thank you both!
– R.Sluij
Nov 21 at 12:46
Thank you both!
– R.Sluij
Nov 21 at 12:46
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
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 $X_i$...
answered Nov 21 at 13:04
cptflint
115
Thanks, however that gives the same value
– R.Sluij
Nov 22 at 19:54
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
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 $X_i$...
answered Nov 21 at 13:04
cptflint
115
Thanks, however that gives the same value
– R.Sluij
Nov 22 at 19:54
add a comment |
up vote
0
down vote
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 $X_i$...
answered Nov 21 at 13:04
cptflint
115
Thanks, however that gives the same value
– R.Sluij
Nov 22 at 19:54
add a comment |
up vote
0
down vote
up vote
0
down vote
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 $X_i$...
answered Nov 21 at 13:04
cptflint
115
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 $X_i$...
answered Nov 21 at 13:04
cptflint
115
answered Nov 21 at 13:04
cptflint
115
answered Nov 21 at 13:04
cptflint
115
answered Nov 21 at 13:04
cptflint
115
115
Thanks, however that gives the same value
– R.Sluij
Nov 22 at 19:54
add a comment |
Thanks, however that gives the same value
– R.Sluij
Nov 22 at 19:54
Thanks, however that gives the same value
– R.Sluij
Nov 22 at 19:54
Thanks, however that gives the same value
– R.Sluij
Nov 22 at 19:54
add a comment |
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Can your show your attempt?
– Kavi Rama Murthy
Nov 16 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 at 12:05
1
Congratulations! You have answered the question yourselves.
– Kavi Rama Murthy
Nov 16 at 12:12
1
The $lambda$ in Poisson distribution is supposed to be positive. It should be $lambda=1$.
– StubbornAtom
Nov 16 at 12:14
Thank you both!
– R.Sluij
Nov 21 at 12:46