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Determine $Bbb E[langle S rangle]$ where $S$ is simple symmetric random walk
Clash Royale CLAN TAG#URR8PPP
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Let $(Z_k)_k in Bbb N$ be i.i.d. with $Bbb P[Z_1 = 1] = Bbb P[Z_1 = -1] = 1 over 2$. Consider the process $S = (S_n)_n in Bbb N_0$ with $S_n = sum_k=1^n Z_k$. I want to determine $Bbb E[langle S rangle]$ for $n ge 0$, where $langle S rangle := (A_n)_n in Bbb N$ which is given by the Doob decomposition $X_n = M_n + A_n$ with $M_n$ a martingal, $A_n$ is predictable and $X_n := S_n²$ (quadratic variation process).
stochastic-processes martingales
asked Jul 21 at 17:51
Pazu
359213
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0
down vote
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Let $(Z_k)_k in Bbb N$ be i.i.d. with $Bbb P[Z_1 = 1] = Bbb P[Z_1 = -1] = 1 over 2$. Consider the process $S = (S_n)_n in Bbb N_0$ with $S_n = sum_k=1^n Z_k$. I want to determine $Bbb E[langle S rangle]$ for $n ge 0$, where $langle S rangle := (A_n)_n in Bbb N$ which is given by the Doob decomposition $X_n = M_n + A_n$ with $M_n$ a martingal, $A_n$ is predictable and $X_n := S_n²$ (quadratic variation process).
stochastic-processes martingales
asked Jul 21 at 17:51
Pazu
359213
2
Any thoughts on the problem? What do you get if you compute $mathbbE(S_n^2 mid mathcalF_n-1)$?
– saz
Jul 21 at 18:40
I think that I get $M_n-1 + A_n$.
– Pazu
Jul 21 at 18:58
@saz If you know how the solution works, please share it with me!
– Pazu
Jul 22 at 0:20
1
I do know the answer, I simply have other things on my mind on a saturday evening. Re your 1st comment: That's correct, but it doesn't help you. Try to use the very definition of $S_n$, i.e. $S_n = sum_k=1^n Z_k$. Write $$S_n^2 = (S_n-S_n-1+S_n-1)^2 = (S_n-S_n-1)^2 + 2 S_n-1 (S_n-S_n-1) + S_n-1^2$$ plug this into $mathbbE(S_n^2 mid mathcalF_n-1)$, and then compute each of the terms separately using $S_n = sum_k=1^n Z_k$.
– saz
Jul 22 at 6:07
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $(Z_k)_k in Bbb N$ be i.i.d. with $Bbb P[Z_1 = 1] = Bbb P[Z_1 = -1] = 1 over 2$. Consider the process $S = (S_n)_n in Bbb N_0$ with $S_n = sum_k=1^n Z_k$. I want to determine $Bbb E[langle S rangle]$ for $n ge 0$, where $langle S rangle := (A_n)_n in Bbb N$ which is given by the Doob decomposition $X_n = M_n + A_n$ with $M_n$ a martingal, $A_n$ is predictable and $X_n := S_n²$ (quadratic variation process).
stochastic-processes martingales
asked Jul 21 at 17:51
Pazu
359213
Let $(Z_k)_k in Bbb N$ be i.i.d. with $Bbb P[Z_1 = 1] = Bbb P[Z_1 = -1] = 1 over 2$. Consider the process $S = (S_n)_n in Bbb N_0$ with $S_n = sum_k=1^n Z_k$. I want to determine $Bbb E[langle S rangle]$ for $n ge 0$, where $langle S rangle := (A_n)_n in Bbb N$ which is given by the Doob decomposition $X_n = M_n + A_n$ with $M_n$ a martingal, $A_n$ is predictable and $X_n := S_n²$ (quadratic variation process).
stochastic-processes martingales
asked Jul 21 at 17:51
Pazu
359213
edited Jul 21 at 19:39
asked Jul 21 at 17:51
Pazu
359213
asked Jul 21 at 17:51
Pazu
359213
asked Jul 21 at 17:51
Pazu
359213
359213
2
Any thoughts on the problem? What do you get if you compute $mathbbE(S_n^2 mid mathcalF_n-1)$?
– saz
Jul 21 at 18:40
I think that I get $M_n-1 + A_n$.
– Pazu
Jul 21 at 18:58
@saz If you know how the solution works, please share it with me!
– Pazu
Jul 22 at 0:20
1
I do know the answer, I simply have other things on my mind on a saturday evening. Re your 1st comment: That's correct, but it doesn't help you. Try to use the very definition of $S_n$, i.e. $S_n = sum_k=1^n Z_k$. Write $$S_n^2 = (S_n-S_n-1+S_n-1)^2 = (S_n-S_n-1)^2 + 2 S_n-1 (S_n-S_n-1) + S_n-1^2$$ plug this into $mathbbE(S_n^2 mid mathcalF_n-1)$, and then compute each of the terms separately using $S_n = sum_k=1^n Z_k$.
– saz
Jul 22 at 6:07
add a comment |Â
2
Any thoughts on the problem? What do you get if you compute $mathbbE(S_n^2 mid mathcalF_n-1)$?
– saz
Jul 21 at 18:40
I think that I get $M_n-1 + A_n$.
– Pazu
Jul 21 at 18:58
@saz If you know how the solution works, please share it with me!
– Pazu
Jul 22 at 0:20
1
I do know the answer, I simply have other things on my mind on a saturday evening. Re your 1st comment: That's correct, but it doesn't help you. Try to use the very definition of $S_n$, i.e. $S_n = sum_k=1^n Z_k$. Write $$S_n^2 = (S_n-S_n-1+S_n-1)^2 = (S_n-S_n-1)^2 + 2 S_n-1 (S_n-S_n-1) + S_n-1^2$$ plug this into $mathbbE(S_n^2 mid mathcalF_n-1)$, and then compute each of the terms separately using $S_n = sum_k=1^n Z_k$.
– saz
Jul 22 at 6:07
2
2
Any thoughts on the problem? What do you get if you compute $mathbbE(S_n^2 mid mathcalF_n-1)$?
– saz
Jul 21 at 18:40
Any thoughts on the problem? What do you get if you compute $mathbbE(S_n^2 mid mathcalF_n-1)$?
– saz
Jul 21 at 18:40
I think that I get $M_n-1 + A_n$.
– Pazu
Jul 21 at 18:58
I think that I get $M_n-1 + A_n$.
– Pazu
Jul 21 at 18:58
@saz If you know how the solution works, please share it with me!
– Pazu
Jul 22 at 0:20
@saz If you know how the solution works, please share it with me!
– Pazu
Jul 22 at 0:20
1
1
I do know the answer, I simply have other things on my mind on a saturday evening. Re your 1st comment: That's correct, but it doesn't help you. Try to use the very definition of $S_n$, i.e. $S_n = sum_k=1^n Z_k$. Write $$S_n^2 = (S_n-S_n-1+S_n-1)^2 = (S_n-S_n-1)^2 + 2 S_n-1 (S_n-S_n-1) + S_n-1^2$$ plug this into $mathbbE(S_n^2 mid mathcalF_n-1)$, and then compute each of the terms separately using $S_n = sum_k=1^n Z_k$.
– saz
Jul 22 at 6:07
I do know the answer, I simply have other things on my mind on a saturday evening. Re your 1st comment: That's correct, but it doesn't help you. Try to use the very definition of $S_n$, i.e. $S_n = sum_k=1^n Z_k$. Write $$S_n^2 = (S_n-S_n-1+S_n-1)^2 = (S_n-S_n-1)^2 + 2 S_n-1 (S_n-S_n-1) + S_n-1^2$$ plug this into $mathbbE(S_n^2 mid mathcalF_n-1)$, and then compute each of the terms separately using $S_n = sum_k=1^n Z_k$.
– saz
Jul 22 at 6:07
add a comment |Â
1 Answer
1
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Elaborating on @saz's hint, we can write
beginalign
S_n^2 &= (S_n-S_n-1)^2 + 2S_n-1(S_n-S_n-1) + S_n-1^2\
&= Z_n^2 + 2S_n-1Z_n + S_n-1^2.
endalign
Letting $mathcal F_n$ be the filtration generated by $Z_n$, it is clear that $Z_n^2$ is independent of $mathcal F_n-1$ and $S_n-1$, $S_n-1^2$ are $mathcal F_n-1$ measurable, hence
beginalign
mathbb E[X_nmidmathcal F_n-1] &= mathbb E[S_n^2midmathcal F_n-1]\
&= mathbb E[Z_n^2 + 2S_n-1Z_n + S_n-1^2mid mathcal F_n-1]\
&= mathbb E[Z_n^2] + 2S_n-1mathbb E[Z_n] + S_n-1^2\
&= 1 + S_n-1^2.
endalign
It follows that the (predictable) quadratic variation of $X_n$ is
beginalign
A_n &= sum_k=1^n (mathbb E[X_kmidmathcal F_k-1] - X_k-1)\
&= sum_k=1^n (1 + S_n-1^2-S_n-1^2)\
&= n.
endalign
It follows that $M_n=S_n^2-n$ as expected (since $S_n$ is a martingale).
answered Jul 24 at 2:24
Math1000
18.4k31544
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Elaborating on @saz's hint, we can write
beginalign
S_n^2 &= (S_n-S_n-1)^2 + 2S_n-1(S_n-S_n-1) + S_n-1^2\
&= Z_n^2 + 2S_n-1Z_n + S_n-1^2.
endalign
Letting $mathcal F_n$ be the filtration generated by $Z_n$, it is clear that $Z_n^2$ is independent of $mathcal F_n-1$ and $S_n-1$, $S_n-1^2$ are $mathcal F_n-1$ measurable, hence
beginalign
mathbb E[X_nmidmathcal F_n-1] &= mathbb E[S_n^2midmathcal F_n-1]\
&= mathbb E[Z_n^2 + 2S_n-1Z_n + S_n-1^2mid mathcal F_n-1]\
&= mathbb E[Z_n^2] + 2S_n-1mathbb E[Z_n] + S_n-1^2\
&= 1 + S_n-1^2.
endalign
It follows that the (predictable) quadratic variation of $X_n$ is
beginalign
A_n &= sum_k=1^n (mathbb E[X_kmidmathcal F_k-1] - X_k-1)\
&= sum_k=1^n (1 + S_n-1^2-S_n-1^2)\
&= n.
endalign
It follows that $M_n=S_n^2-n$ as expected (since $S_n$ is a martingale).
answered Jul 24 at 2:24
Math1000
18.4k31544
add a comment |Â
up vote
0
down vote
Elaborating on @saz's hint, we can write
beginalign
S_n^2 &= (S_n-S_n-1)^2 + 2S_n-1(S_n-S_n-1) + S_n-1^2\
&= Z_n^2 + 2S_n-1Z_n + S_n-1^2.
endalign
Letting $mathcal F_n$ be the filtration generated by $Z_n$, it is clear that $Z_n^2$ is independent of $mathcal F_n-1$ and $S_n-1$, $S_n-1^2$ are $mathcal F_n-1$ measurable, hence
beginalign
mathbb E[X_nmidmathcal F_n-1] &= mathbb E[S_n^2midmathcal F_n-1]\
&= mathbb E[Z_n^2 + 2S_n-1Z_n + S_n-1^2mid mathcal F_n-1]\
&= mathbb E[Z_n^2] + 2S_n-1mathbb E[Z_n] + S_n-1^2\
&= 1 + S_n-1^2.
endalign
It follows that the (predictable) quadratic variation of $X_n$ is
beginalign
A_n &= sum_k=1^n (mathbb E[X_kmidmathcal F_k-1] - X_k-1)\
&= sum_k=1^n (1 + S_n-1^2-S_n-1^2)\
&= n.
endalign
It follows that $M_n=S_n^2-n$ as expected (since $S_n$ is a martingale).
answered Jul 24 at 2:24
Math1000
18.4k31544
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Elaborating on @saz's hint, we can write
beginalign
S_n^2 &= (S_n-S_n-1)^2 + 2S_n-1(S_n-S_n-1) + S_n-1^2\
&= Z_n^2 + 2S_n-1Z_n + S_n-1^2.
endalign
Letting $mathcal F_n$ be the filtration generated by $Z_n$, it is clear that $Z_n^2$ is independent of $mathcal F_n-1$ and $S_n-1$, $S_n-1^2$ are $mathcal F_n-1$ measurable, hence
beginalign
mathbb E[X_nmidmathcal F_n-1] &= mathbb E[S_n^2midmathcal F_n-1]\
&= mathbb E[Z_n^2 + 2S_n-1Z_n + S_n-1^2mid mathcal F_n-1]\
&= mathbb E[Z_n^2] + 2S_n-1mathbb E[Z_n] + S_n-1^2\
&= 1 + S_n-1^2.
endalign
It follows that the (predictable) quadratic variation of $X_n$ is
beginalign
A_n &= sum_k=1^n (mathbb E[X_kmidmathcal F_k-1] - X_k-1)\
&= sum_k=1^n (1 + S_n-1^2-S_n-1^2)\
&= n.
endalign
It follows that $M_n=S_n^2-n$ as expected (since $S_n$ is a martingale).
answered Jul 24 at 2:24
Math1000
18.4k31544
Elaborating on @saz's hint, we can write
beginalign
S_n^2 &= (S_n-S_n-1)^2 + 2S_n-1(S_n-S_n-1) + S_n-1^2\
&= Z_n^2 + 2S_n-1Z_n + S_n-1^2.
endalign
Letting $mathcal F_n$ be the filtration generated by $Z_n$, it is clear that $Z_n^2$ is independent of $mathcal F_n-1$ and $S_n-1$, $S_n-1^2$ are $mathcal F_n-1$ measurable, hence
beginalign
mathbb E[X_nmidmathcal F_n-1] &= mathbb E[S_n^2midmathcal F_n-1]\
&= mathbb E[Z_n^2 + 2S_n-1Z_n + S_n-1^2mid mathcal F_n-1]\
&= mathbb E[Z_n^2] + 2S_n-1mathbb E[Z_n] + S_n-1^2\
&= 1 + S_n-1^2.
endalign
It follows that the (predictable) quadratic variation of $X_n$ is
beginalign
A_n &= sum_k=1^n (mathbb E[X_kmidmathcal F_k-1] - X_k-1)\
&= sum_k=1^n (1 + S_n-1^2-S_n-1^2)\
&= n.
endalign
It follows that $M_n=S_n^2-n$ as expected (since $S_n$ is a martingale).
answered Jul 24 at 2:24
Math1000
18.4k31544
answered Jul 24 at 2:24
Math1000
18.4k31544
answered Jul 24 at 2:24
Math1000
18.4k31544
answered Jul 24 at 2:24
Math1000
18.4k31544
18.4k31544
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2
Any thoughts on the problem? What do you get if you compute $mathbbE(S_n^2 mid mathcalF_n-1)$?
– saz
Jul 21 at 18:40
I think that I get $M_n-1 + A_n$.
– Pazu
Jul 21 at 18:58
@saz If you know how the solution works, please share it with me!
– Pazu
Jul 22 at 0:20
1
I do know the answer, I simply have other things on my mind on a saturday evening. Re your 1st comment: That's correct, but it doesn't help you. Try to use the very definition of $S_n$, i.e. $S_n = sum_k=1^n Z_k$. Write $$S_n^2 = (S_n-S_n-1+S_n-1)^2 = (S_n-S_n-1)^2 + 2 S_n-1 (S_n-S_n-1) + S_n-1^2$$ plug this into $mathbbE(S_n^2 mid mathcalF_n-1)$, and then compute each of the terms separately using $S_n = sum_k=1^n Z_k$.
– saz
Jul 22 at 6:07