Poisson distribution, and embedded poisson distribution

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Assume that $X_j$ and $Y_j$ are both independent poisson distributed RV with the same rate $lambda>0$ for all $j=0,1,2,....$.

Now define $U_j$ such that $U_j(omega)=Y_X_j(omega)(omega)$ for all $omega in Omega$.

I now want to find $mathbbE[U_j]$.

My approach has been the following argument:

Since $X_jsim Pois(lambda)$ and IID, we know that for some $jin 0,1,2...$ that $X_jin 0,1,2...$, hence will we have that:

$mathbbE[U_j]=mathbbE[Y_X_j]=mathbbE[Y_k]=lambda$ for some $kin 0,1,2...$ since all $Y_j$ and $X_j$ are IID.

I just feel that this is a bit to easy....

poisson-distribution

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asked Aug 2 at 14:09

Jonathan Kiersch

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Assume that $X_j$ and $Y_j$ are both independent poisson distributed RV with the same rate $lambda>0$ for all $j=0,1,2,....$.

Now define $U_j$ such that $U_j(omega)=Y_X_j(omega)(omega)$ for all $omega in Omega$.

I now want to find $mathbbE[U_j]$.

My approach has been the following argument:

Since $X_jsim Pois(lambda)$ and IID, we know that for some $jin 0,1,2...$ that $X_jin 0,1,2...$, hence will we have that:

$mathbbE[U_j]=mathbbE[Y_X_j]=mathbbE[Y_k]=lambda$ for some $kin 0,1,2...$ since all $Y_j$ and $X_j$ are IID.

I just feel that this is a bit to easy....

poisson-distribution

share|cite|improve this question

asked Aug 2 at 14:09

Jonathan Kiersch

374

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

Assume that $X_j$ and $Y_j$ are both independent poisson distributed RV with the same rate $lambda>0$ for all $j=0,1,2,....$.

Now define $U_j$ such that $U_j(omega)=Y_X_j(omega)(omega)$ for all $omega in Omega$.

I now want to find $mathbbE[U_j]$.

My approach has been the following argument:

Since $X_jsim Pois(lambda)$ and IID, we know that for some $jin 0,1,2...$ that $X_jin 0,1,2...$, hence will we have that:

$mathbbE[U_j]=mathbbE[Y_X_j]=mathbbE[Y_k]=lambda$ for some $kin 0,1,2...$ since all $Y_j$ and $X_j$ are IID.

I just feel that this is a bit to easy....

poisson-distribution

share|cite|improve this question

asked Aug 2 at 14:09

Jonathan Kiersch

374

Assume that $X_j$ and $Y_j$ are both independent poisson distributed RV with the same rate $lambda>0$ for all $j=0,1,2,....$.

Now define $U_j$ such that $U_j(omega)=Y_X_j(omega)(omega)$ for all $omega in Omega$.

I now want to find $mathbbE[U_j]$.

My approach has been the following argument:

Since $X_jsim Pois(lambda)$ and IID, we know that for some $jin 0,1,2...$ that $X_jin 0,1,2...$, hence will we have that:

$mathbbE[U_j]=mathbbE[Y_X_j]=mathbbE[Y_k]=lambda$ for some $kin 0,1,2...$ since all $Y_j$ and $X_j$ are IID.

I just feel that this is a bit to easy....

poisson-distribution

share|cite|improve this question

asked Aug 2 at 14:09

Jonathan Kiersch

374

share|cite|improve this question

share|cite|improve this question

asked Aug 2 at 14:09

Jonathan Kiersch

374

asked Aug 2 at 14:09

Jonathan Kiersch

374

asked Aug 2 at 14:09

Jonathan Kiersch

374

374

add a comment | 

add a comment | 

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Careful, $X_j$ is a random variable that is nonconstant as $omega$ varies.

I think there is no way around it: You have to compute. This is done as follows:

$$
mathbb E[Y_X_j] = sum_k=0^n mathbb P(X_j = k) mathbb E[Y_k|X_j = k] oversettextindependence= lambda sum_k=0^infty fraclambda^k e^-lambdak! = lambda.
$$

(Apparently, your solution was correct anyway.)

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edited Aug 2 at 14:47

answered Aug 2 at 14:18

AlgebraicsAnonymous

66111

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
0
down vote



accepted

Careful, $X_j$ is a random variable that is nonconstant as $omega$ varies.

I think there is no way around it: You have to compute. This is done as follows:

$$
mathbb E[Y_X_j] = sum_k=0^n mathbb P(X_j = k) mathbb E[Y_k|X_j = k] oversettextindependence= lambda sum_k=0^infty fraclambda^k e^-lambdak! = lambda.
$$

(Apparently, your solution was correct anyway.)

share|cite|improve this answer

edited Aug 2 at 14:47

answered Aug 2 at 14:18

AlgebraicsAnonymous

66111

add a comment | 

up vote
0
down vote



accepted

Careful, $X_j$ is a random variable that is nonconstant as $omega$ varies.

I think there is no way around it: You have to compute. This is done as follows:

$$
mathbb E[Y_X_j] = sum_k=0^n mathbb P(X_j = k) mathbb E[Y_k|X_j = k] oversettextindependence= lambda sum_k=0^infty fraclambda^k e^-lambdak! = lambda.
$$

(Apparently, your solution was correct anyway.)

share|cite|improve this answer

edited Aug 2 at 14:47

answered Aug 2 at 14:18

AlgebraicsAnonymous

66111

add a comment | 

up vote
0
down vote



accepted

up vote
0
down vote



accepted

Careful, $X_j$ is a random variable that is nonconstant as $omega$ varies.

I think there is no way around it: You have to compute. This is done as follows:

$$
mathbb E[Y_X_j] = sum_k=0^n mathbb P(X_j = k) mathbb E[Y_k|X_j = k] oversettextindependence= lambda sum_k=0^infty fraclambda^k e^-lambdak! = lambda.
$$

(Apparently, your solution was correct anyway.)

share|cite|improve this answer

edited Aug 2 at 14:47

answered Aug 2 at 14:18

AlgebraicsAnonymous

66111

Careful, $X_j$ is a random variable that is nonconstant as $omega$ varies.

I think there is no way around it: You have to compute. This is done as follows:

$$
mathbb E[Y_X_j] = sum_k=0^n mathbb P(X_j = k) mathbb E[Y_k|X_j = k] oversettextindependence= lambda sum_k=0^infty fraclambda^k e^-lambdak! = lambda.
$$

(Apparently, your solution was correct anyway.)

share|cite|improve this answer

edited Aug 2 at 14:47

answered Aug 2 at 14:18

AlgebraicsAnonymous

66111

share|cite|improve this answer

share|cite|improve this answer

edited Aug 2 at 14:47

edited Aug 2 at 14:47

edited Aug 2 at 14:47

answered Aug 2 at 14:18

AlgebraicsAnonymous

66111

answered Aug 2 at 14:18

AlgebraicsAnonymous

66111

answered Aug 2 at 14:18

AlgebraicsAnonymous

66111

66111

add a comment | 

add a comment | 

 

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