Confusion about given proof of the compensated Poisson process being a Martingale?
$begingroup$
Given the following proof of the compensated Poisson process being a Martingale
Why does the proof start with $E[N(t)-lambda t|N(s)]$ when the question asks to prove that $X(t)$ is a Martingale? Shouldn't it start with $E[N(t)-lambda t|N(s)-lambda s]$?
I see that the solution shows that when you plug in the 2nd line of the solution into the left side of $E[N(s)|N(t)]-lambda t$ you will get $N(s)-lambda s$ which is equivalent to $X(s)$ but I'm still confused as to why the proof is starting with, essentially, $E[X(t)|X(s)+lambda s]$ to prove the Martingale when the given definition seems to imply you should start with $E[X(t)|X(s)]$ and then try to get to a final answer of $X(s)$
stochastic-processes markov-chains conditional-expectation
asked Dec 4 '18 at 21:48
user3491700user3491700
535
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add a comment |
$begingroup$
Given the following proof of the compensated Poisson process being a Martingale
Why does the proof start with $E[N(t)-lambda t|N(s)]$ when the question asks to prove that $X(t)$ is a Martingale? Shouldn't it start with $E[N(t)-lambda t|N(s)-lambda s]$?
I see that the solution shows that when you plug in the 2nd line of the solution into the left side of $E[N(s)|N(t)]-lambda t$ you will get $N(s)-lambda s$ which is equivalent to $X(s)$ but I'm still confused as to why the proof is starting with, essentially, $E[X(t)|X(s)+lambda s]$ to prove the Martingale when the given definition seems to imply you should start with $E[X(t)|X(s)]$ and then try to get to a final answer of $X(s)$
stochastic-processes markov-chains conditional-expectation
asked Dec 4 '18 at 21:48
user3491700user3491700
535
$endgroup$
add a comment |
$begingroup$
Given the following proof of the compensated Poisson process being a Martingale
Why does the proof start with $E[N(t)-lambda t|N(s)]$ when the question asks to prove that $X(t)$ is a Martingale? Shouldn't it start with $E[N(t)-lambda t|N(s)-lambda s]$?
I see that the solution shows that when you plug in the 2nd line of the solution into the left side of $E[N(s)|N(t)]-lambda t$ you will get $N(s)-lambda s$ which is equivalent to $X(s)$ but I'm still confused as to why the proof is starting with, essentially, $E[X(t)|X(s)+lambda s]$ to prove the Martingale when the given definition seems to imply you should start with $E[X(t)|X(s)]$ and then try to get to a final answer of $X(s)$
stochastic-processes markov-chains conditional-expectation
asked Dec 4 '18 at 21:48
user3491700user3491700
535
$endgroup$
Given the following proof of the compensated Poisson process being a Martingale
Why does the proof start with $E[N(t)-lambda t|N(s)]$ when the question asks to prove that $X(t)$ is a Martingale? Shouldn't it start with $E[N(t)-lambda t|N(s)-lambda s]$?
I see that the solution shows that when you plug in the 2nd line of the solution into the left side of $E[N(s)|N(t)]-lambda t$ you will get $N(s)-lambda s$ which is equivalent to $X(s)$ but I'm still confused as to why the proof is starting with, essentially, $E[X(t)|X(s)+lambda s]$ to prove the Martingale when the given definition seems to imply you should start with $E[X(t)|X(s)]$ and then try to get to a final answer of $X(s)$
stochastic-processes markov-chains conditional-expectation
stochastic-processes markov-chains conditional-expectation
asked Dec 4 '18 at 21:48
user3491700user3491700
535
asked Dec 4 '18 at 21:48
user3491700user3491700
535
asked Dec 4 '18 at 21:48
user3491700user3491700
535
asked Dec 4 '18 at 21:48
user3491700user3491700
535
asked Dec 4 '18 at 21:48
user3491700user3491700
535
535
add a comment |
add a comment |
1 Answer
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Conditioning on $X(s)+lambda s$ is the same as conditioning on $X(s)$ since $lambda s$ is not random. I do not know your background on probability theory, but in an elementary explanation, the information $X(s)+lambda s$ and $X(s)$ carry is exactly the same since the two differ by a deterministic quantity, so knowing one can recover another. In measure-theoretic probability terms. the sigma-fields generated by the two are exactly the same.
answered Dec 4 '18 at 22:00
UchihaUchiha
9310
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1 Answer
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1 Answer
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$begingroup$
Conditioning on $X(s)+lambda s$ is the same as conditioning on $X(s)$ since $lambda s$ is not random. I do not know your background on probability theory, but in an elementary explanation, the information $X(s)+lambda s$ and $X(s)$ carry is exactly the same since the two differ by a deterministic quantity, so knowing one can recover another. In measure-theoretic probability terms. the sigma-fields generated by the two are exactly the same.
answered Dec 4 '18 at 22:00
UchihaUchiha
9310
$endgroup$
add a comment |
$begingroup$
Conditioning on $X(s)+lambda s$ is the same as conditioning on $X(s)$ since $lambda s$ is not random. I do not know your background on probability theory, but in an elementary explanation, the information $X(s)+lambda s$ and $X(s)$ carry is exactly the same since the two differ by a deterministic quantity, so knowing one can recover another. In measure-theoretic probability terms. the sigma-fields generated by the two are exactly the same.
answered Dec 4 '18 at 22:00
UchihaUchiha
9310
$endgroup$
add a comment |
$begingroup$
Conditioning on $X(s)+lambda s$ is the same as conditioning on $X(s)$ since $lambda s$ is not random. I do not know your background on probability theory, but in an elementary explanation, the information $X(s)+lambda s$ and $X(s)$ carry is exactly the same since the two differ by a deterministic quantity, so knowing one can recover another. In measure-theoretic probability terms. the sigma-fields generated by the two are exactly the same.
answered Dec 4 '18 at 22:00
UchihaUchiha
9310
$endgroup$
Conditioning on $X(s)+lambda s$ is the same as conditioning on $X(s)$ since $lambda s$ is not random. I do not know your background on probability theory, but in an elementary explanation, the information $X(s)+lambda s$ and $X(s)$ carry is exactly the same since the two differ by a deterministic quantity, so knowing one can recover another. In measure-theoretic probability terms. the sigma-fields generated by the two are exactly the same.
answered Dec 4 '18 at 22:00
UchihaUchiha
9310
answered Dec 4 '18 at 22:00
UchihaUchiha
9310
answered Dec 4 '18 at 22:00
UchihaUchiha
9310
answered Dec 4 '18 at 22:00
UchihaUchiha
9310
9310
add a comment |
add a comment |
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