Why is $(S_n)^2 - n(q-p)$ a martingale for a random walk
2
$begingroup$
If we have a random walk such that
$$P(S_{n+1} = S_n + 1|S_n)=p$$ and $$P(S_{n+1} = S_n - 1|S_n)= 1-p=q$$
then why is $$(S_n)^2 - n(q-p)$$ a martingale.
I understand that $(S_n)^2$ is a sub-martingale but why do we take away $n(q-p)$ to get a martingale?
In the symmetric random walk case I understand this would be $(S_n)^2 - n$, but again I can't understand this inutition behind this.
stochastic-processes martingales random-walk
asked Dec 3 '18 at 0:35
mathsexam2013mathsexam2013
242
$endgroup$
add a comment |
2
$begingroup$
If we have a random walk such that
$$P(S_{n+1} = S_n + 1|S_n)=p$$ and $$P(S_{n+1} = S_n - 1|S_n)= 1-p=q$$
then why is $$(S_n)^2 - n(q-p)$$ a martingale.
I understand that $(S_n)^2$ is a sub-martingale but why do we take away $n(q-p)$ to get a martingale?
In the symmetric random walk case I understand this would be $(S_n)^2 - n$, but again I can't understand this inutition behind this.
stochastic-processes martingales random-walk
asked Dec 3 '18 at 0:35
mathsexam2013mathsexam2013
242
$endgroup$
$begingroup$
Actually, $(S_n)^2 - n(q-p)$ is not a martingale.
$endgroup$
– Did
Dec 3 '18 at 7:13
$begingroup$
You'd first need to take away the correct mean of $S_n^2$, which is $n(q-p)+npq$ (the mean of $S_n$ plus the variance of $S_n$). This is because a martingale has constant expectation. It is then an interesting question whether just subtracting off the deterministic mean is enough to get a martingale. It's difficult to get intuition for why this should be the case; I usually find it easier to just do the calculation.
$endgroup$
– Ian
Dec 3 '18 at 14:53
$begingroup$
@Did. Yep sorry, I was tired and made a mistake in my question.
$endgroup$
– mathsexam2013
Dec 4 '18 at 12:18
$begingroup$
@Ian thank you very much for your answer. That helped me understand it
$endgroup$
– mathsexam2013
Dec 4 '18 at 12:19
add a comment |
2
2
2
$begingroup$
If we have a random walk such that
$$P(S_{n+1} = S_n + 1|S_n)=p$$ and $$P(S_{n+1} = S_n - 1|S_n)= 1-p=q$$
then why is $$(S_n)^2 - n(q-p)$$ a martingale.
I understand that $(S_n)^2$ is a sub-martingale but why do we take away $n(q-p)$ to get a martingale?
In the symmetric random walk case I understand this would be $(S_n)^2 - n$, but again I can't understand this inutition behind this.
stochastic-processes martingales random-walk
asked Dec 3 '18 at 0:35
mathsexam2013mathsexam2013
242
$endgroup$
If we have a random walk such that
$$P(S_{n+1} = S_n + 1|S_n)=p$$ and $$P(S_{n+1} = S_n - 1|S_n)= 1-p=q$$
then why is $$(S_n)^2 - n(q-p)$$ a martingale.
I understand that $(S_n)^2$ is a sub-martingale but why do we take away $n(q-p)$ to get a martingale?
In the symmetric random walk case I understand this would be $(S_n)^2 - n$, but again I can't understand this inutition behind this.
stochastic-processes martingales random-walk
stochastic-processes martingales random-walk
asked Dec 3 '18 at 0:35
mathsexam2013mathsexam2013
242
asked Dec 3 '18 at 0:35
mathsexam2013mathsexam2013
242
asked Dec 3 '18 at 0:35
mathsexam2013mathsexam2013
242
asked Dec 3 '18 at 0:35
mathsexam2013mathsexam2013
242
asked Dec 3 '18 at 0:35
mathsexam2013mathsexam2013
242
242
$begingroup$
Actually, $(S_n)^2 - n(q-p)$ is not a martingale.
$endgroup$
– Did
Dec 3 '18 at 7:13
$begingroup$
You'd first need to take away the correct mean of $S_n^2$, which is $n(q-p)+npq$ (the mean of $S_n$ plus the variance of $S_n$). This is because a martingale has constant expectation. It is then an interesting question whether just subtracting off the deterministic mean is enough to get a martingale. It's difficult to get intuition for why this should be the case; I usually find it easier to just do the calculation.
$endgroup$
– Ian
Dec 3 '18 at 14:53
$begingroup$
@Did. Yep sorry, I was tired and made a mistake in my question.
$endgroup$
– mathsexam2013
Dec 4 '18 at 12:18
$begingroup$
@Ian thank you very much for your answer. That helped me understand it
$endgroup$
– mathsexam2013
Dec 4 '18 at 12:19
add a comment |
$begingroup$
Actually, $(S_n)^2 - n(q-p)$ is not a martingale.
$endgroup$
– Did
Dec 3 '18 at 7:13
$begingroup$
You'd first need to take away the correct mean of $S_n^2$, which is $n(q-p)+npq$ (the mean of $S_n$ plus the variance of $S_n$). This is because a martingale has constant expectation. It is then an interesting question whether just subtracting off the deterministic mean is enough to get a martingale. It's difficult to get intuition for why this should be the case; I usually find it easier to just do the calculation.
$endgroup$
– Ian
Dec 3 '18 at 14:53
$begingroup$
@Did. Yep sorry, I was tired and made a mistake in my question.
$endgroup$
– mathsexam2013
Dec 4 '18 at 12:18
$begingroup$
@Ian thank you very much for your answer. That helped me understand it
$endgroup$
– mathsexam2013
Dec 4 '18 at 12:19
$begingroup$
Actually, $(S_n)^2 - n(q-p)$ is not a martingale.
$endgroup$
– Did
Dec 3 '18 at 7:13
$begingroup$
Actually, $(S_n)^2 - n(q-p)$ is not a martingale.
$endgroup$
– Did
Dec 3 '18 at 7:13
$begingroup$
You'd first need to take away the correct mean of $S_n^2$, which is $n(q-p)+npq$ (the mean of $S_n$ plus the variance of $S_n$). This is because a martingale has constant expectation. It is then an interesting question whether just subtracting off the deterministic mean is enough to get a martingale. It's difficult to get intuition for why this should be the case; I usually find it easier to just do the calculation.
$endgroup$
– Ian
Dec 3 '18 at 14:53
$begingroup$
You'd first need to take away the correct mean of $S_n^2$, which is $n(q-p)+npq$ (the mean of $S_n$ plus the variance of $S_n$). This is because a martingale has constant expectation. It is then an interesting question whether just subtracting off the deterministic mean is enough to get a martingale. It's difficult to get intuition for why this should be the case; I usually find it easier to just do the calculation.
$endgroup$
– Ian
Dec 3 '18 at 14:53
$begingroup$
@Did. Yep sorry, I was tired and made a mistake in my question.
$endgroup$
– mathsexam2013
Dec 4 '18 at 12:18
$begingroup$
@Did. Yep sorry, I was tired and made a mistake in my question.
$endgroup$
– mathsexam2013
Dec 4 '18 at 12:18
$begingroup$
@Ian thank you very much for your answer. That helped me understand it
$endgroup$
– mathsexam2013
Dec 4 '18 at 12:19
$begingroup$
@Ian thank you very much for your answer. That helped me understand it
$endgroup$
– mathsexam2013
Dec 4 '18 at 12:19
add a comment |
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$begingroup$
Actually, $(S_n)^2 - n(q-p)$ is not a martingale.
$endgroup$
– Did
Dec 3 '18 at 7:13
$begingroup$
You'd first need to take away the correct mean of $S_n^2$, which is $n(q-p)+npq$ (the mean of $S_n$ plus the variance of $S_n$). This is because a martingale has constant expectation. It is then an interesting question whether just subtracting off the deterministic mean is enough to get a martingale. It's difficult to get intuition for why this should be the case; I usually find it easier to just do the calculation.
$endgroup$
– Ian
Dec 3 '18 at 14:53
$begingroup$
@Did. Yep sorry, I was tired and made a mistake in my question.
$endgroup$
– mathsexam2013
Dec 4 '18 at 12:18
$begingroup$
@Ian thank you very much for your answer. That helped me understand it
$endgroup$
– mathsexam2013
Dec 4 '18 at 12:19