Proving $mathbb{E}(S_{tau}|tau)=tau mathbb{E}(xi_1)$ and $mathbb{E}S_{tau}=mathbb{E}tau mathbb{E}xi_1$
$begingroup$
Let $xi_1, x_2,...,xi_n$ and $tau$ be independent random variables in $(Omega, mathcal{F}, mathbb{P})$. $xi_1, x_2,...,xi_n$ are uniformly distributed and $tau={0,1,2,...,n}$. Let $S_{tau}=xi_1+xi_2+...+xi_{tau}$ be a random sum of random variables.
I need to show that
1) $mathbb{E}(S_{tau}|tau)=tau mathbb{E}(xi_1)$ and $mathbb{E}S_{tau}=mathbb{E}tau mathbb{E}xi_1$.
2) $mathbb{D}(S_{tau}|tau)= tau mathbb{D}xi_1$ and $mathbb{D}S_{tau}=mathbb{E}tau mathbb{D}xi_1+mathbb{D}tau (mathbb{E}xi_1)^2$.
So for the first part I think I can use this
$mathbb{E}S_{tau}=mathbb{E}mathbb{E}(S_{tau}|tau)=sum_{ngeq1}mathbb{E}(xi_1+xi_2+...+xi_{tau}|tau)mathbb{P}(tau=n)=sum_{ngeq 1}nmathbb{E}xi_imathbb{P}(tau=n)=mathbb{E}xi_i sum_{ngeq1}mathbb{P}(tau=n)=mathbb{E}xi_1mathbb{E}tau .$
So am I right?
For the second part I am really not sure what to do.
probability probability-theory random-variables conditional-expectation
$endgroup$
add a comment |
$begingroup$
Let $xi_1, x_2,...,xi_n$ and $tau$ be independent random variables in $(Omega, mathcal{F}, mathbb{P})$. $xi_1, x_2,...,xi_n$ are uniformly distributed and $tau={0,1,2,...,n}$. Let $S_{tau}=xi_1+xi_2+...+xi_{tau}$ be a random sum of random variables.
I need to show that
1) $mathbb{E}(S_{tau}|tau)=tau mathbb{E}(xi_1)$ and $mathbb{E}S_{tau}=mathbb{E}tau mathbb{E}xi_1$.
2) $mathbb{D}(S_{tau}|tau)= tau mathbb{D}xi_1$ and $mathbb{D}S_{tau}=mathbb{E}tau mathbb{D}xi_1+mathbb{D}tau (mathbb{E}xi_1)^2$.
So for the first part I think I can use this
$mathbb{E}S_{tau}=mathbb{E}mathbb{E}(S_{tau}|tau)=sum_{ngeq1}mathbb{E}(xi_1+xi_2+...+xi_{tau}|tau)mathbb{P}(tau=n)=sum_{ngeq 1}nmathbb{E}xi_imathbb{P}(tau=n)=mathbb{E}xi_i sum_{ngeq1}mathbb{P}(tau=n)=mathbb{E}xi_1mathbb{E}tau .$
So am I right?
For the second part I am really not sure what to do.
probability probability-theory random-variables conditional-expectation
$endgroup$
$begingroup$
What do you mean by $mathbb{D}(X)$?
$endgroup$
– Lundborg
Jan 3 at 10:55
add a comment |
$begingroup$
Let $xi_1, x_2,...,xi_n$ and $tau$ be independent random variables in $(Omega, mathcal{F}, mathbb{P})$. $xi_1, x_2,...,xi_n$ are uniformly distributed and $tau={0,1,2,...,n}$. Let $S_{tau}=xi_1+xi_2+...+xi_{tau}$ be a random sum of random variables.
I need to show that
1) $mathbb{E}(S_{tau}|tau)=tau mathbb{E}(xi_1)$ and $mathbb{E}S_{tau}=mathbb{E}tau mathbb{E}xi_1$.
2) $mathbb{D}(S_{tau}|tau)= tau mathbb{D}xi_1$ and $mathbb{D}S_{tau}=mathbb{E}tau mathbb{D}xi_1+mathbb{D}tau (mathbb{E}xi_1)^2$.
So for the first part I think I can use this
$mathbb{E}S_{tau}=mathbb{E}mathbb{E}(S_{tau}|tau)=sum_{ngeq1}mathbb{E}(xi_1+xi_2+...+xi_{tau}|tau)mathbb{P}(tau=n)=sum_{ngeq 1}nmathbb{E}xi_imathbb{P}(tau=n)=mathbb{E}xi_i sum_{ngeq1}mathbb{P}(tau=n)=mathbb{E}xi_1mathbb{E}tau .$
So am I right?
For the second part I am really not sure what to do.
probability probability-theory random-variables conditional-expectation
$endgroup$
Let $xi_1, x_2,...,xi_n$ and $tau$ be independent random variables in $(Omega, mathcal{F}, mathbb{P})$. $xi_1, x_2,...,xi_n$ are uniformly distributed and $tau={0,1,2,...,n}$. Let $S_{tau}=xi_1+xi_2+...+xi_{tau}$ be a random sum of random variables.
I need to show that
1) $mathbb{E}(S_{tau}|tau)=tau mathbb{E}(xi_1)$ and $mathbb{E}S_{tau}=mathbb{E}tau mathbb{E}xi_1$.
2) $mathbb{D}(S_{tau}|tau)= tau mathbb{D}xi_1$ and $mathbb{D}S_{tau}=mathbb{E}tau mathbb{D}xi_1+mathbb{D}tau (mathbb{E}xi_1)^2$.
So for the first part I think I can use this
$mathbb{E}S_{tau}=mathbb{E}mathbb{E}(S_{tau}|tau)=sum_{ngeq1}mathbb{E}(xi_1+xi_2+...+xi_{tau}|tau)mathbb{P}(tau=n)=sum_{ngeq 1}nmathbb{E}xi_imathbb{P}(tau=n)=mathbb{E}xi_i sum_{ngeq1}mathbb{P}(tau=n)=mathbb{E}xi_1mathbb{E}tau .$
So am I right?
For the second part I am really not sure what to do.
probability probability-theory random-variables conditional-expectation
probability probability-theory random-variables conditional-expectation
asked Jan 3 at 10:40
AtstovasAtstovas
1159
1159
$begingroup$
What do you mean by $mathbb{D}(X)$?
$endgroup$
– Lundborg
Jan 3 at 10:55
add a comment |
$begingroup$
What do you mean by $mathbb{D}(X)$?
$endgroup$
– Lundborg
Jan 3 at 10:55
$begingroup$
What do you mean by $mathbb{D}(X)$?
$endgroup$
– Lundborg
Jan 3 at 10:55
$begingroup$
What do you mean by $mathbb{D}(X)$?
$endgroup$
– Lundborg
Jan 3 at 10:55
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For the first question, note that
$$
S_tau = sum_{i=1}^n 1_{(tau leq i)} xi_i
$$
therefore
$$
mathbb{E}(S_tau | tau) = mathbb{E}left(sum_{i=1}^n 1_{(tau leq i)} xi_i Big| tau right) = sum_{i=1}^n mathbb{E}left( 1_{(tau leq i)} xi_i big| tau right)
$$
Note now that $1_{(tau leq i)}$ is measurable when given $tau$ and that $xi_i$ are all independent from $tau$ so
$$
mathbb{E}left( 1_{(tau leq i)} xi_i | tau right) = 1_{(tau leq i)} mathbb{E}(xi_i)
$$
In total we get (since $xi_i$ are identically distributed)
$$
mathbb{E}(S_tau | tau) =sum_{i=1}^n mathbb{E}left( 1_{(tau leq i)} xi_i big| tau right) = sum_{i=1}^n 1_{(tau leq i)} mathbb{E}(xi_i) = mathbb{E}(xi_1)sum_{i=1}^n 1_{(tau leq i)} = mathbb{E}(xi_1)tau
$$
Taking expectations on both sides yields the second part of the first question.
Assuming by $mathbb{D}$ you mean the variance, you can perform the exact same calculations as above to get
$$
mathbb{D}(S_{tau}|tau)= tau mathbb{D}xi_1
$$
Then using the law of total variance, we get
$$
mathbb{D}(S_tau) = mathbb{D}(mathbb{E}(S_tau | tau)) + mathbb{E}(mathbb{D}(S_tau | tau)) = mathbb{D}(tau) (mathbb{E}(xi_1))^2 + mathbb{E}(tau) mathbb{D}(xi_1)
$$
as desired.
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
For the first question, note that
$$
S_tau = sum_{i=1}^n 1_{(tau leq i)} xi_i
$$
therefore
$$
mathbb{E}(S_tau | tau) = mathbb{E}left(sum_{i=1}^n 1_{(tau leq i)} xi_i Big| tau right) = sum_{i=1}^n mathbb{E}left( 1_{(tau leq i)} xi_i big| tau right)
$$
Note now that $1_{(tau leq i)}$ is measurable when given $tau$ and that $xi_i$ are all independent from $tau$ so
$$
mathbb{E}left( 1_{(tau leq i)} xi_i | tau right) = 1_{(tau leq i)} mathbb{E}(xi_i)
$$
In total we get (since $xi_i$ are identically distributed)
$$
mathbb{E}(S_tau | tau) =sum_{i=1}^n mathbb{E}left( 1_{(tau leq i)} xi_i big| tau right) = sum_{i=1}^n 1_{(tau leq i)} mathbb{E}(xi_i) = mathbb{E}(xi_1)sum_{i=1}^n 1_{(tau leq i)} = mathbb{E}(xi_1)tau
$$
Taking expectations on both sides yields the second part of the first question.
Assuming by $mathbb{D}$ you mean the variance, you can perform the exact same calculations as above to get
$$
mathbb{D}(S_{tau}|tau)= tau mathbb{D}xi_1
$$
Then using the law of total variance, we get
$$
mathbb{D}(S_tau) = mathbb{D}(mathbb{E}(S_tau | tau)) + mathbb{E}(mathbb{D}(S_tau | tau)) = mathbb{D}(tau) (mathbb{E}(xi_1))^2 + mathbb{E}(tau) mathbb{D}(xi_1)
$$
as desired.
$endgroup$
add a comment |
$begingroup$
For the first question, note that
$$
S_tau = sum_{i=1}^n 1_{(tau leq i)} xi_i
$$
therefore
$$
mathbb{E}(S_tau | tau) = mathbb{E}left(sum_{i=1}^n 1_{(tau leq i)} xi_i Big| tau right) = sum_{i=1}^n mathbb{E}left( 1_{(tau leq i)} xi_i big| tau right)
$$
Note now that $1_{(tau leq i)}$ is measurable when given $tau$ and that $xi_i$ are all independent from $tau$ so
$$
mathbb{E}left( 1_{(tau leq i)} xi_i | tau right) = 1_{(tau leq i)} mathbb{E}(xi_i)
$$
In total we get (since $xi_i$ are identically distributed)
$$
mathbb{E}(S_tau | tau) =sum_{i=1}^n mathbb{E}left( 1_{(tau leq i)} xi_i big| tau right) = sum_{i=1}^n 1_{(tau leq i)} mathbb{E}(xi_i) = mathbb{E}(xi_1)sum_{i=1}^n 1_{(tau leq i)} = mathbb{E}(xi_1)tau
$$
Taking expectations on both sides yields the second part of the first question.
Assuming by $mathbb{D}$ you mean the variance, you can perform the exact same calculations as above to get
$$
mathbb{D}(S_{tau}|tau)= tau mathbb{D}xi_1
$$
Then using the law of total variance, we get
$$
mathbb{D}(S_tau) = mathbb{D}(mathbb{E}(S_tau | tau)) + mathbb{E}(mathbb{D}(S_tau | tau)) = mathbb{D}(tau) (mathbb{E}(xi_1))^2 + mathbb{E}(tau) mathbb{D}(xi_1)
$$
as desired.
$endgroup$
add a comment |
$begingroup$
For the first question, note that
$$
S_tau = sum_{i=1}^n 1_{(tau leq i)} xi_i
$$
therefore
$$
mathbb{E}(S_tau | tau) = mathbb{E}left(sum_{i=1}^n 1_{(tau leq i)} xi_i Big| tau right) = sum_{i=1}^n mathbb{E}left( 1_{(tau leq i)} xi_i big| tau right)
$$
Note now that $1_{(tau leq i)}$ is measurable when given $tau$ and that $xi_i$ are all independent from $tau$ so
$$
mathbb{E}left( 1_{(tau leq i)} xi_i | tau right) = 1_{(tau leq i)} mathbb{E}(xi_i)
$$
In total we get (since $xi_i$ are identically distributed)
$$
mathbb{E}(S_tau | tau) =sum_{i=1}^n mathbb{E}left( 1_{(tau leq i)} xi_i big| tau right) = sum_{i=1}^n 1_{(tau leq i)} mathbb{E}(xi_i) = mathbb{E}(xi_1)sum_{i=1}^n 1_{(tau leq i)} = mathbb{E}(xi_1)tau
$$
Taking expectations on both sides yields the second part of the first question.
Assuming by $mathbb{D}$ you mean the variance, you can perform the exact same calculations as above to get
$$
mathbb{D}(S_{tau}|tau)= tau mathbb{D}xi_1
$$
Then using the law of total variance, we get
$$
mathbb{D}(S_tau) = mathbb{D}(mathbb{E}(S_tau | tau)) + mathbb{E}(mathbb{D}(S_tau | tau)) = mathbb{D}(tau) (mathbb{E}(xi_1))^2 + mathbb{E}(tau) mathbb{D}(xi_1)
$$
as desired.
$endgroup$
For the first question, note that
$$
S_tau = sum_{i=1}^n 1_{(tau leq i)} xi_i
$$
therefore
$$
mathbb{E}(S_tau | tau) = mathbb{E}left(sum_{i=1}^n 1_{(tau leq i)} xi_i Big| tau right) = sum_{i=1}^n mathbb{E}left( 1_{(tau leq i)} xi_i big| tau right)
$$
Note now that $1_{(tau leq i)}$ is measurable when given $tau$ and that $xi_i$ are all independent from $tau$ so
$$
mathbb{E}left( 1_{(tau leq i)} xi_i | tau right) = 1_{(tau leq i)} mathbb{E}(xi_i)
$$
In total we get (since $xi_i$ are identically distributed)
$$
mathbb{E}(S_tau | tau) =sum_{i=1}^n mathbb{E}left( 1_{(tau leq i)} xi_i big| tau right) = sum_{i=1}^n 1_{(tau leq i)} mathbb{E}(xi_i) = mathbb{E}(xi_1)sum_{i=1}^n 1_{(tau leq i)} = mathbb{E}(xi_1)tau
$$
Taking expectations on both sides yields the second part of the first question.
Assuming by $mathbb{D}$ you mean the variance, you can perform the exact same calculations as above to get
$$
mathbb{D}(S_{tau}|tau)= tau mathbb{D}xi_1
$$
Then using the law of total variance, we get
$$
mathbb{D}(S_tau) = mathbb{D}(mathbb{E}(S_tau | tau)) + mathbb{E}(mathbb{D}(S_tau | tau)) = mathbb{D}(tau) (mathbb{E}(xi_1))^2 + mathbb{E}(tau) mathbb{D}(xi_1)
$$
as desired.
answered Jan 3 at 11:07
LundborgLundborg
857617
857617
add a comment |
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$begingroup$
What do you mean by $mathbb{D}(X)$?
$endgroup$
– Lundborg
Jan 3 at 10:55