What is wrong with this proof of wald's identity?
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When I first saw the wald's identity, I think proof is very simple just like below. But in my textbook or wikipedia page, the proof is much more complicated. So I think there's a huge mistake. What is wrong with my proof?
begin{gather*}
E(X_1+X_2+cdots+X_tau)
\=sum_{n=0}^{infty} E(X_1+X_2+cdots+X_n)P(tau=n)
\=E(X_1)sum_{n=0}^{infty} nP(tau=n)
\=E(X_1)E(tau)
end{gather*}
probability probability-theory
asked Dec 6 '18 at 7:34
Lee.HWLee.HW
1137
$endgroup$
add a comment |
$begingroup$
When I first saw the wald's identity, I think proof is very simple just like below. But in my textbook or wikipedia page, the proof is much more complicated. So I think there's a huge mistake. What is wrong with my proof?
begin{gather*}
E(X_1+X_2+cdots+X_tau)
\=sum_{n=0}^{infty} E(X_1+X_2+cdots+X_n)P(tau=n)
\=E(X_1)sum_{n=0}^{infty} nP(tau=n)
\=E(X_1)E(tau)
end{gather*}
probability probability-theory
asked Dec 6 '18 at 7:34
Lee.HWLee.HW
1137
$endgroup$
$begingroup$
You are assuming that $E(X_1+X_2+cdots+X_{tau})$ exists. This is not obvious.
$endgroup$
– Kavi Rama Murthy
Dec 6 '18 at 7:40
2
$begingroup$
Why does the first "=" hold? Note that the expression on the first line equals $$E left( sum_{n=0}^{infty} (X_1+ldots+X_n) 1_{{tau=n}} right)$$ and you will need some result to justify that you can pull the infinite sum outside the expectation. (In particular, you need to know that the expectation is finite.)
$endgroup$
– saz
Dec 6 '18 at 7:40
add a comment |
$begingroup$
When I first saw the wald's identity, I think proof is very simple just like below. But in my textbook or wikipedia page, the proof is much more complicated. So I think there's a huge mistake. What is wrong with my proof?
begin{gather*}
E(X_1+X_2+cdots+X_tau)
\=sum_{n=0}^{infty} E(X_1+X_2+cdots+X_n)P(tau=n)
\=E(X_1)sum_{n=0}^{infty} nP(tau=n)
\=E(X_1)E(tau)
end{gather*}
probability probability-theory
asked Dec 6 '18 at 7:34
Lee.HWLee.HW
1137
$endgroup$
When I first saw the wald's identity, I think proof is very simple just like below. But in my textbook or wikipedia page, the proof is much more complicated. So I think there's a huge mistake. What is wrong with my proof?
begin{gather*}
E(X_1+X_2+cdots+X_tau)
\=sum_{n=0}^{infty} E(X_1+X_2+cdots+X_n)P(tau=n)
\=E(X_1)sum_{n=0}^{infty} nP(tau=n)
\=E(X_1)E(tau)
end{gather*}
probability probability-theory
probability probability-theory
asked Dec 6 '18 at 7:34
Lee.HWLee.HW
1137
asked Dec 6 '18 at 7:34
Lee.HWLee.HW
1137
asked Dec 6 '18 at 7:34
Lee.HWLee.HW
1137
asked Dec 6 '18 at 7:34
Lee.HWLee.HW
1137
asked Dec 6 '18 at 7:34
Lee.HWLee.HW
1137
1137
$begingroup$
You are assuming that $E(X_1+X_2+cdots+X_{tau})$ exists. This is not obvious.
$endgroup$
– Kavi Rama Murthy
Dec 6 '18 at 7:40
2
$begingroup$
Why does the first "=" hold? Note that the expression on the first line equals $$E left( sum_{n=0}^{infty} (X_1+ldots+X_n) 1_{{tau=n}} right)$$ and you will need some result to justify that you can pull the infinite sum outside the expectation. (In particular, you need to know that the expectation is finite.)
$endgroup$
– saz
Dec 6 '18 at 7:40
add a comment |
$begingroup$
You are assuming that $E(X_1+X_2+cdots+X_{tau})$ exists. This is not obvious.
$endgroup$
– Kavi Rama Murthy
Dec 6 '18 at 7:40
2
$begingroup$
Why does the first "=" hold? Note that the expression on the first line equals $$E left( sum_{n=0}^{infty} (X_1+ldots+X_n) 1_{{tau=n}} right)$$ and you will need some result to justify that you can pull the infinite sum outside the expectation. (In particular, you need to know that the expectation is finite.)
$endgroup$
– saz
Dec 6 '18 at 7:40
$begingroup$
You are assuming that $E(X_1+X_2+cdots+X_{tau})$ exists. This is not obvious.
$endgroup$
– Kavi Rama Murthy
Dec 6 '18 at 7:40
$begingroup$
You are assuming that $E(X_1+X_2+cdots+X_{tau})$ exists. This is not obvious.
$endgroup$
– Kavi Rama Murthy
Dec 6 '18 at 7:40
2
2
$begingroup$
Why does the first "=" hold? Note that the expression on the first line equals $$E left( sum_{n=0}^{infty} (X_1+ldots+X_n) 1_{{tau=n}} right)$$ and you will need some result to justify that you can pull the infinite sum outside the expectation. (In particular, you need to know that the expectation is finite.)
$endgroup$
– saz
Dec 6 '18 at 7:40
$begingroup$
Why does the first "=" hold? Note that the expression on the first line equals $$E left( sum_{n=0}^{infty} (X_1+ldots+X_n) 1_{{tau=n}} right)$$ and you will need some result to justify that you can pull the infinite sum outside the expectation. (In particular, you need to know that the expectation is finite.)
$endgroup$
– saz
Dec 6 '18 at 7:40
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
See comments for mistakes in the argument. However everything you have done can easily be justified if $X_i geq 0$. By writing $X_i$ as $X_i^{+}-X_i^{-}$ you can get proof for the general case.
answered Dec 6 '18 at 7:44
Kavi Rama MurthyKavi Rama Murthy
65k42766
$endgroup$
$begingroup$
Thank you! Now I'm understand what I'm wrong!
$endgroup$
– Lee.HW
Dec 6 '18 at 8:30
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
See comments for mistakes in the argument. However everything you have done can easily be justified if $X_i geq 0$. By writing $X_i$ as $X_i^{+}-X_i^{-}$ you can get proof for the general case.
answered Dec 6 '18 at 7:44
Kavi Rama MurthyKavi Rama Murthy
65k42766
$endgroup$
$begingroup$
Thank you! Now I'm understand what I'm wrong!
$endgroup$
– Lee.HW
Dec 6 '18 at 8:30
add a comment |
$begingroup$
See comments for mistakes in the argument. However everything you have done can easily be justified if $X_i geq 0$. By writing $X_i$ as $X_i^{+}-X_i^{-}$ you can get proof for the general case.
answered Dec 6 '18 at 7:44
Kavi Rama MurthyKavi Rama Murthy
65k42766
$endgroup$
$begingroup$
Thank you! Now I'm understand what I'm wrong!
$endgroup$
– Lee.HW
Dec 6 '18 at 8:30
add a comment |
$begingroup$
See comments for mistakes in the argument. However everything you have done can easily be justified if $X_i geq 0$. By writing $X_i$ as $X_i^{+}-X_i^{-}$ you can get proof for the general case.
answered Dec 6 '18 at 7:44
Kavi Rama MurthyKavi Rama Murthy
65k42766
$endgroup$
See comments for mistakes in the argument. However everything you have done can easily be justified if $X_i geq 0$. By writing $X_i$ as $X_i^{+}-X_i^{-}$ you can get proof for the general case.
answered Dec 6 '18 at 7:44
Kavi Rama MurthyKavi Rama Murthy
65k42766
answered Dec 6 '18 at 7:44
Kavi Rama MurthyKavi Rama Murthy
65k42766
answered Dec 6 '18 at 7:44
Kavi Rama MurthyKavi Rama Murthy
65k42766
answered Dec 6 '18 at 7:44
Kavi Rama MurthyKavi Rama Murthy
65k42766
65k42766
$begingroup$
Thank you! Now I'm understand what I'm wrong!
$endgroup$
– Lee.HW
Dec 6 '18 at 8:30
add a comment |
$begingroup$
Thank you! Now I'm understand what I'm wrong!
$endgroup$
– Lee.HW
Dec 6 '18 at 8:30
$begingroup$
Thank you! Now I'm understand what I'm wrong!
$endgroup$
– Lee.HW
Dec 6 '18 at 8:30
$begingroup$
Thank you! Now I'm understand what I'm wrong!
$endgroup$
– Lee.HW
Dec 6 '18 at 8:30
add a comment |
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$begingroup$
You are assuming that $E(X_1+X_2+cdots+X_{tau})$ exists. This is not obvious.
$endgroup$
– Kavi Rama Murthy
Dec 6 '18 at 7:40
2
$begingroup$
Why does the first "=" hold? Note that the expression on the first line equals $$E left( sum_{n=0}^{infty} (X_1+ldots+X_n) 1_{{tau=n}} right)$$ and you will need some result to justify that you can pull the infinite sum outside the expectation. (In particular, you need to know that the expectation is finite.)
$endgroup$
– saz
Dec 6 '18 at 7:40