Is there an extension of Wald's Equation to the expectation of a product of random variables?
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So Wald's Equation states that for a real-values, independent and identically distributed sequence of random variables $(X_{n})_{ninmathbb{N}}$ and a nonnegative integer $N$, which is independent of the sequence, we have that:
$mathbb{E}[X_{1}+dots+X_{N}]=mathbb{E}[N]mathbb{E}[X_{1}]$
Under the assumption that both have finite expectation.
My question is if there exists some extension or similar identity that can be applied to the same type of problem except then with multiplication. I'm not sure if this simply follows from independence in most cases though. For example, given the same setting as earlier, do we have that:
$mathbb{E}[prod^{N}_{i=1}X_{i}]=prod^{N}_{i=1}mathbb{E}[X_{i}]$
Or is the outcome something comparable to the Wald's Equation?
Any help is appreciated!
probability probability-theory expected-value
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up vote
1
down vote
favorite
So Wald's Equation states that for a real-values, independent and identically distributed sequence of random variables $(X_{n})_{ninmathbb{N}}$ and a nonnegative integer $N$, which is independent of the sequence, we have that:
$mathbb{E}[X_{1}+dots+X_{N}]=mathbb{E}[N]mathbb{E}[X_{1}]$
Under the assumption that both have finite expectation.
My question is if there exists some extension or similar identity that can be applied to the same type of problem except then with multiplication. I'm not sure if this simply follows from independence in most cases though. For example, given the same setting as earlier, do we have that:
$mathbb{E}[prod^{N}_{i=1}X_{i}]=prod^{N}_{i=1}mathbb{E}[X_{i}]$
Or is the outcome something comparable to the Wald's Equation?
Any help is appreciated!
probability probability-theory expected-value
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
So Wald's Equation states that for a real-values, independent and identically distributed sequence of random variables $(X_{n})_{ninmathbb{N}}$ and a nonnegative integer $N$, which is independent of the sequence, we have that:
$mathbb{E}[X_{1}+dots+X_{N}]=mathbb{E}[N]mathbb{E}[X_{1}]$
Under the assumption that both have finite expectation.
My question is if there exists some extension or similar identity that can be applied to the same type of problem except then with multiplication. I'm not sure if this simply follows from independence in most cases though. For example, given the same setting as earlier, do we have that:
$mathbb{E}[prod^{N}_{i=1}X_{i}]=prod^{N}_{i=1}mathbb{E}[X_{i}]$
Or is the outcome something comparable to the Wald's Equation?
Any help is appreciated!
probability probability-theory expected-value
So Wald's Equation states that for a real-values, independent and identically distributed sequence of random variables $(X_{n})_{ninmathbb{N}}$ and a nonnegative integer $N$, which is independent of the sequence, we have that:
$mathbb{E}[X_{1}+dots+X_{N}]=mathbb{E}[N]mathbb{E}[X_{1}]$
Under the assumption that both have finite expectation.
My question is if there exists some extension or similar identity that can be applied to the same type of problem except then with multiplication. I'm not sure if this simply follows from independence in most cases though. For example, given the same setting as earlier, do we have that:
$mathbb{E}[prod^{N}_{i=1}X_{i}]=prod^{N}_{i=1}mathbb{E}[X_{i}]$
Or is the outcome something comparable to the Wald's Equation?
Any help is appreciated!
probability probability-theory expected-value
probability probability-theory expected-value
asked 2 days ago
S. Crim
7810
7810
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Let $ N, X_1, X_2, dots $ independent random variables, $ (X_i) $ identically distributed and integrable with $ mu = mathbb{E}[X_1] $, and $N$ take integer values with $ mathbb{E}[{(mathbb{E}|X_1|)}^N]$ well defined. Then
$$ mathbb{E}left[Pi_{i=1}^NX_iright]=mathbb{E}[mu^N].$$
In fact,
$$mathbb{E}left[Pi_{i=1}^NX_iright]=mathbb{E}left[sum_{n=1}^infty 1_{{N=n}}Pi_{i=1}^N X_iright]= sum_{n=1}^infty mathbb{E}left[1_{{N=n}}Pi_{i=1}^N X_iright]=sum_{n=1}^infty mathbb{E}left[1_{{N=n}}Pi_{i=1}^n X_iright]=sum_{n=1}^infty {mathbb{E}[X_1]}^n mathbb{P}(N=n)
=mathbb{E}[mu^N].$$
Note that in the penultimate equality we require independence and in the last equality we need a hypothesis about $ mu^N $. Initially work with $$|Pi_{i=1}^N X_i|$$
instead of $$Pi_{i=1}^N X_i$$ to show the integratibility.
Is there a proof of this, or a reason that it works?
– S. Crim
2 days ago
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
Let $ N, X_1, X_2, dots $ independent random variables, $ (X_i) $ identically distributed and integrable with $ mu = mathbb{E}[X_1] $, and $N$ take integer values with $ mathbb{E}[{(mathbb{E}|X_1|)}^N]$ well defined. Then
$$ mathbb{E}left[Pi_{i=1}^NX_iright]=mathbb{E}[mu^N].$$
In fact,
$$mathbb{E}left[Pi_{i=1}^NX_iright]=mathbb{E}left[sum_{n=1}^infty 1_{{N=n}}Pi_{i=1}^N X_iright]= sum_{n=1}^infty mathbb{E}left[1_{{N=n}}Pi_{i=1}^N X_iright]=sum_{n=1}^infty mathbb{E}left[1_{{N=n}}Pi_{i=1}^n X_iright]=sum_{n=1}^infty {mathbb{E}[X_1]}^n mathbb{P}(N=n)
=mathbb{E}[mu^N].$$
Note that in the penultimate equality we require independence and in the last equality we need a hypothesis about $ mu^N $. Initially work with $$|Pi_{i=1}^N X_i|$$
instead of $$Pi_{i=1}^N X_i$$ to show the integratibility.
Is there a proof of this, or a reason that it works?
– S. Crim
2 days ago
add a comment |
up vote
0
down vote
Let $ N, X_1, X_2, dots $ independent random variables, $ (X_i) $ identically distributed and integrable with $ mu = mathbb{E}[X_1] $, and $N$ take integer values with $ mathbb{E}[{(mathbb{E}|X_1|)}^N]$ well defined. Then
$$ mathbb{E}left[Pi_{i=1}^NX_iright]=mathbb{E}[mu^N].$$
In fact,
$$mathbb{E}left[Pi_{i=1}^NX_iright]=mathbb{E}left[sum_{n=1}^infty 1_{{N=n}}Pi_{i=1}^N X_iright]= sum_{n=1}^infty mathbb{E}left[1_{{N=n}}Pi_{i=1}^N X_iright]=sum_{n=1}^infty mathbb{E}left[1_{{N=n}}Pi_{i=1}^n X_iright]=sum_{n=1}^infty {mathbb{E}[X_1]}^n mathbb{P}(N=n)
=mathbb{E}[mu^N].$$
Note that in the penultimate equality we require independence and in the last equality we need a hypothesis about $ mu^N $. Initially work with $$|Pi_{i=1}^N X_i|$$
instead of $$Pi_{i=1}^N X_i$$ to show the integratibility.
Is there a proof of this, or a reason that it works?
– S. Crim
2 days ago
add a comment |
up vote
0
down vote
up vote
0
down vote
Let $ N, X_1, X_2, dots $ independent random variables, $ (X_i) $ identically distributed and integrable with $ mu = mathbb{E}[X_1] $, and $N$ take integer values with $ mathbb{E}[{(mathbb{E}|X_1|)}^N]$ well defined. Then
$$ mathbb{E}left[Pi_{i=1}^NX_iright]=mathbb{E}[mu^N].$$
In fact,
$$mathbb{E}left[Pi_{i=1}^NX_iright]=mathbb{E}left[sum_{n=1}^infty 1_{{N=n}}Pi_{i=1}^N X_iright]= sum_{n=1}^infty mathbb{E}left[1_{{N=n}}Pi_{i=1}^N X_iright]=sum_{n=1}^infty mathbb{E}left[1_{{N=n}}Pi_{i=1}^n X_iright]=sum_{n=1}^infty {mathbb{E}[X_1]}^n mathbb{P}(N=n)
=mathbb{E}[mu^N].$$
Note that in the penultimate equality we require independence and in the last equality we need a hypothesis about $ mu^N $. Initially work with $$|Pi_{i=1}^N X_i|$$
instead of $$Pi_{i=1}^N X_i$$ to show the integratibility.
Let $ N, X_1, X_2, dots $ independent random variables, $ (X_i) $ identically distributed and integrable with $ mu = mathbb{E}[X_1] $, and $N$ take integer values with $ mathbb{E}[{(mathbb{E}|X_1|)}^N]$ well defined. Then
$$ mathbb{E}left[Pi_{i=1}^NX_iright]=mathbb{E}[mu^N].$$
In fact,
$$mathbb{E}left[Pi_{i=1}^NX_iright]=mathbb{E}left[sum_{n=1}^infty 1_{{N=n}}Pi_{i=1}^N X_iright]= sum_{n=1}^infty mathbb{E}left[1_{{N=n}}Pi_{i=1}^N X_iright]=sum_{n=1}^infty mathbb{E}left[1_{{N=n}}Pi_{i=1}^n X_iright]=sum_{n=1}^infty {mathbb{E}[X_1]}^n mathbb{P}(N=n)
=mathbb{E}[mu^N].$$
Note that in the penultimate equality we require independence and in the last equality we need a hypothesis about $ mu^N $. Initially work with $$|Pi_{i=1}^N X_i|$$
instead of $$Pi_{i=1}^N X_i$$ to show the integratibility.
edited yesterday
answered 2 days ago
Daniel Camarena Perez
54728
54728
Is there a proof of this, or a reason that it works?
– S. Crim
2 days ago
add a comment |
Is there a proof of this, or a reason that it works?
– S. Crim
2 days ago
Is there a proof of this, or a reason that it works?
– S. Crim
2 days ago
Is there a proof of this, or a reason that it works?
– S. Crim
2 days ago
add a comment |
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