Martingale of random walk and stopping time
$begingroup$
Let ${S_n}$ be a symmetric random walk such that $S_0 = a$ for some $0 < a < K$. Let $T$ be the stopping time when the walk reaches $0$ or $K$. Show
$$M_n = sum_{i = 0}^n S_i - frac{1}{3}S^3_n $$ is a martingale and deduce that $mathbb{E}bigg[ sum_{i = 0}^T S_i bigg] = frac{1}{3}(K^2-a^2)a + a$.
So far I have tried showing it is both a submartingale and supermartingale and also have tried $mathbb{E}[M_{n+1} - M_n mid X_0,...,X_n] = 0 $ approach. I think the error is in my computing so $S^3_n$ but I am not getting that it is a martingale. The second part just seems to be that I need to apply the martingale convergence theorem/optional stopping time.
I got that $$S_n^3 = sum_{i=0}^n X_i^3 + 6bigg(sum_{i < j}X_i^2X_j + sum_{i < j} X_iX_j^2 + sum_{i < j < k}X_iX_jX_k bigg)$$.
Also I know that ${S_n}$ by itself is a martingale since $p = frac{1}{2}$. So now $$begin{align*} M_{n+1} - M_n &= X_{n+1} + frac{1}{3}(S_{n+1}^3 - S_n^3) \ &= X_{n+1} + frac{1}{3}bigg(X_{n+1}sum_{i = 0}^nX_i + X_{n+1}^2sum_{i = 0}^nX_i + X_{n+1}bigg[sum_{i = 0}^n X_i bigg( sum_{j=i+1}^n X_jbigg)bigg]bigg) end{align*}$$ .
Which seems quite incorrect.
probability-theory martingales random-walk stopping-times
edited Dec 7 '18 at 7:48
saz
79.8k860124
asked Dec 6 '18 at 21:06
all.overall.over
467
$endgroup$
add a comment |
$begingroup$
Let ${S_n}$ be a symmetric random walk such that $S_0 = a$ for some $0 < a < K$. Let $T$ be the stopping time when the walk reaches $0$ or $K$. Show
$$M_n = sum_{i = 0}^n S_i - frac{1}{3}S^3_n $$ is a martingale and deduce that $mathbb{E}bigg[ sum_{i = 0}^T S_i bigg] = frac{1}{3}(K^2-a^2)a + a$.
So far I have tried showing it is both a submartingale and supermartingale and also have tried $mathbb{E}[M_{n+1} - M_n mid X_0,...,X_n] = 0 $ approach. I think the error is in my computing so $S^3_n$ but I am not getting that it is a martingale. The second part just seems to be that I need to apply the martingale convergence theorem/optional stopping time.
I got that $$S_n^3 = sum_{i=0}^n X_i^3 + 6bigg(sum_{i < j}X_i^2X_j + sum_{i < j} X_iX_j^2 + sum_{i < j < k}X_iX_jX_k bigg)$$.
Also I know that ${S_n}$ by itself is a martingale since $p = frac{1}{2}$. So now $$begin{align*} M_{n+1} - M_n &= X_{n+1} + frac{1}{3}(S_{n+1}^3 - S_n^3) \ &= X_{n+1} + frac{1}{3}bigg(X_{n+1}sum_{i = 0}^nX_i + X_{n+1}^2sum_{i = 0}^nX_i + X_{n+1}bigg[sum_{i = 0}^n X_i bigg( sum_{j=i+1}^n X_jbigg)bigg]bigg) end{align*}$$ .
Which seems quite incorrect.
probability-theory martingales random-walk stopping-times
edited Dec 7 '18 at 7:48
saz
79.8k860124
asked Dec 6 '18 at 21:06
all.overall.over
467
$endgroup$
add a comment |
$begingroup$
Let ${S_n}$ be a symmetric random walk such that $S_0 = a$ for some $0 < a < K$. Let $T$ be the stopping time when the walk reaches $0$ or $K$. Show
$$M_n = sum_{i = 0}^n S_i - frac{1}{3}S^3_n $$ is a martingale and deduce that $mathbb{E}bigg[ sum_{i = 0}^T S_i bigg] = frac{1}{3}(K^2-a^2)a + a$.
So far I have tried showing it is both a submartingale and supermartingale and also have tried $mathbb{E}[M_{n+1} - M_n mid X_0,...,X_n] = 0 $ approach. I think the error is in my computing so $S^3_n$ but I am not getting that it is a martingale. The second part just seems to be that I need to apply the martingale convergence theorem/optional stopping time.
I got that $$S_n^3 = sum_{i=0}^n X_i^3 + 6bigg(sum_{i < j}X_i^2X_j + sum_{i < j} X_iX_j^2 + sum_{i < j < k}X_iX_jX_k bigg)$$.
Also I know that ${S_n}$ by itself is a martingale since $p = frac{1}{2}$. So now $$begin{align*} M_{n+1} - M_n &= X_{n+1} + frac{1}{3}(S_{n+1}^3 - S_n^3) \ &= X_{n+1} + frac{1}{3}bigg(X_{n+1}sum_{i = 0}^nX_i + X_{n+1}^2sum_{i = 0}^nX_i + X_{n+1}bigg[sum_{i = 0}^n X_i bigg( sum_{j=i+1}^n X_jbigg)bigg]bigg) end{align*}$$ .
Which seems quite incorrect.
probability-theory martingales random-walk stopping-times
edited Dec 7 '18 at 7:48
saz
79.8k860124
asked Dec 6 '18 at 21:06
all.overall.over
467
$endgroup$
Let ${S_n}$ be a symmetric random walk such that $S_0 = a$ for some $0 < a < K$. Let $T$ be the stopping time when the walk reaches $0$ or $K$. Show
$$M_n = sum_{i = 0}^n S_i - frac{1}{3}S^3_n $$ is a martingale and deduce that $mathbb{E}bigg[ sum_{i = 0}^T S_i bigg] = frac{1}{3}(K^2-a^2)a + a$.
So far I have tried showing it is both a submartingale and supermartingale and also have tried $mathbb{E}[M_{n+1} - M_n mid X_0,...,X_n] = 0 $ approach. I think the error is in my computing so $S^3_n$ but I am not getting that it is a martingale. The second part just seems to be that I need to apply the martingale convergence theorem/optional stopping time.
I got that $$S_n^3 = sum_{i=0}^n X_i^3 + 6bigg(sum_{i < j}X_i^2X_j + sum_{i < j} X_iX_j^2 + sum_{i < j < k}X_iX_jX_k bigg)$$.
Also I know that ${S_n}$ by itself is a martingale since $p = frac{1}{2}$. So now $$begin{align*} M_{n+1} - M_n &= X_{n+1} + frac{1}{3}(S_{n+1}^3 - S_n^3) \ &= X_{n+1} + frac{1}{3}bigg(X_{n+1}sum_{i = 0}^nX_i + X_{n+1}^2sum_{i = 0}^nX_i + X_{n+1}bigg[sum_{i = 0}^n X_i bigg( sum_{j=i+1}^n X_jbigg)bigg]bigg) end{align*}$$ .
Which seems quite incorrect.
probability-theory martingales random-walk stopping-times
probability-theory martingales random-walk stopping-times
edited Dec 7 '18 at 7:48
saz
79.8k860124
asked Dec 6 '18 at 21:06
all.overall.over
467
edited Dec 7 '18 at 7:48
saz
79.8k860124
asked Dec 6 '18 at 21:06
all.overall.over
467
edited Dec 7 '18 at 7:48
saz
79.8k860124
edited Dec 7 '18 at 7:48
saz
79.8k860124
edited Dec 7 '18 at 7:48
saz
79.8k860124
79.8k860124
asked Dec 6 '18 at 21:06
all.overall.over
467
asked Dec 6 '18 at 21:06
all.overall.over
467
asked Dec 6 '18 at 21:06
all.overall.over
467
467
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Clearly,
$$S_{n+1}^3 = (S_n+X_{n+1})^3 = S_{n}^3 +3 S_n X_{n+1}^2 + 3S_n^2 X_{n+1}+X_{n+1}^3$$
which implies ("take out what is known"/"pull out")
$$mathbb{E}(S_{n+1}^3 mid mathcal{F}_n) = S_n^3+3 S_n underbrace{mathbb{E}(X_{n+1}^2 mid mathcal{F}_n)}_{=mathbb{E}(X_{n+1}^2)=1} + 3 S_n^2 underbrace{mathbb{E}(X_{n+1} mid mathcal{F}_n)}_{=mathbb{E}(X_{n+1})=0} + underbrace{mathbb{E}(X_{n+1}^3 mid mathcal{F}_n)}_{=mathbb{E}(X_{n+1}^3)=0}$$
i.e.
$$mathbb{E}(S_{n+1}^3 mid mathcal{F}_n) = S_n^3 + 3 S_n.$$
Using this equation, you shouldn't have much trouble to verify the martingale property of $(M_n)_{n in mathbb{N}}$.
answered Dec 6 '18 at 21:32
sazsaz
79.8k860124
$endgroup$
$begingroup$
I definitely did some calculating wrong. For the second part am I correct that this is an application of optional stopping theorem? It's not quite set up for $mathbb{E}[M_0] = mathbb{E}[M_T]$ But it is $mathbb{E}[M_T + frac{1}{3}S^3_T] $. but here $mathbb{E}[M_0] = a - a^3/3$.
$endgroup$
– all.over
Dec 6 '18 at 21:53
$begingroup$
Sorry made edits to the comment above, hit add before I was done.
$endgroup$
– all.over
Dec 6 '18 at 21:58
$begingroup$
@all.over Well you will need to compute $E (S_T^3) $... to this end use $E (S_T)=0$ to find $P (S_T=0) $ and $P (S_T=K) $
$endgroup$
– saz
Dec 6 '18 at 22:34
$begingroup$
It won't let my at people for some reason. Anyway, isn't $E(S_T) = E(S_0) = a$ by the martingale property?
$endgroup$
– all.over
Dec 6 '18 at 22:48
1
$begingroup$
@all.over Yeah, sorry, that was a typo.
$endgroup$
– saz
Dec 6 '18 at 23:12
add a comment |
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$begingroup$
Clearly,
$$S_{n+1}^3 = (S_n+X_{n+1})^3 = S_{n}^3 +3 S_n X_{n+1}^2 + 3S_n^2 X_{n+1}+X_{n+1}^3$$
which implies ("take out what is known"/"pull out")
$$mathbb{E}(S_{n+1}^3 mid mathcal{F}_n) = S_n^3+3 S_n underbrace{mathbb{E}(X_{n+1}^2 mid mathcal{F}_n)}_{=mathbb{E}(X_{n+1}^2)=1} + 3 S_n^2 underbrace{mathbb{E}(X_{n+1} mid mathcal{F}_n)}_{=mathbb{E}(X_{n+1})=0} + underbrace{mathbb{E}(X_{n+1}^3 mid mathcal{F}_n)}_{=mathbb{E}(X_{n+1}^3)=0}$$
i.e.
$$mathbb{E}(S_{n+1}^3 mid mathcal{F}_n) = S_n^3 + 3 S_n.$$
Using this equation, you shouldn't have much trouble to verify the martingale property of $(M_n)_{n in mathbb{N}}$.
answered Dec 6 '18 at 21:32
sazsaz
79.8k860124
$endgroup$
$begingroup$
I definitely did some calculating wrong. For the second part am I correct that this is an application of optional stopping theorem? It's not quite set up for $mathbb{E}[M_0] = mathbb{E}[M_T]$ But it is $mathbb{E}[M_T + frac{1}{3}S^3_T] $. but here $mathbb{E}[M_0] = a - a^3/3$.
$endgroup$
– all.over
Dec 6 '18 at 21:53
$begingroup$
Sorry made edits to the comment above, hit add before I was done.
$endgroup$
– all.over
Dec 6 '18 at 21:58
$begingroup$
@all.over Well you will need to compute $E (S_T^3) $... to this end use $E (S_T)=0$ to find $P (S_T=0) $ and $P (S_T=K) $
$endgroup$
– saz
Dec 6 '18 at 22:34
$begingroup$
It won't let my at people for some reason. Anyway, isn't $E(S_T) = E(S_0) = a$ by the martingale property?
$endgroup$
– all.over
Dec 6 '18 at 22:48
1
$begingroup$
@all.over Yeah, sorry, that was a typo.
$endgroup$
– saz
Dec 6 '18 at 23:12
add a comment |
$begingroup$
Clearly,
$$S_{n+1}^3 = (S_n+X_{n+1})^3 = S_{n}^3 +3 S_n X_{n+1}^2 + 3S_n^2 X_{n+1}+X_{n+1}^3$$
which implies ("take out what is known"/"pull out")
$$mathbb{E}(S_{n+1}^3 mid mathcal{F}_n) = S_n^3+3 S_n underbrace{mathbb{E}(X_{n+1}^2 mid mathcal{F}_n)}_{=mathbb{E}(X_{n+1}^2)=1} + 3 S_n^2 underbrace{mathbb{E}(X_{n+1} mid mathcal{F}_n)}_{=mathbb{E}(X_{n+1})=0} + underbrace{mathbb{E}(X_{n+1}^3 mid mathcal{F}_n)}_{=mathbb{E}(X_{n+1}^3)=0}$$
i.e.
$$mathbb{E}(S_{n+1}^3 mid mathcal{F}_n) = S_n^3 + 3 S_n.$$
Using this equation, you shouldn't have much trouble to verify the martingale property of $(M_n)_{n in mathbb{N}}$.
answered Dec 6 '18 at 21:32
sazsaz
79.8k860124
$endgroup$
$begingroup$
I definitely did some calculating wrong. For the second part am I correct that this is an application of optional stopping theorem? It's not quite set up for $mathbb{E}[M_0] = mathbb{E}[M_T]$ But it is $mathbb{E}[M_T + frac{1}{3}S^3_T] $. but here $mathbb{E}[M_0] = a - a^3/3$.
$endgroup$
– all.over
Dec 6 '18 at 21:53
$begingroup$
Sorry made edits to the comment above, hit add before I was done.
$endgroup$
– all.over
Dec 6 '18 at 21:58
$begingroup$
@all.over Well you will need to compute $E (S_T^3) $... to this end use $E (S_T)=0$ to find $P (S_T=0) $ and $P (S_T=K) $
$endgroup$
– saz
Dec 6 '18 at 22:34
$begingroup$
It won't let my at people for some reason. Anyway, isn't $E(S_T) = E(S_0) = a$ by the martingale property?
$endgroup$
– all.over
Dec 6 '18 at 22:48
1
$begingroup$
@all.over Yeah, sorry, that was a typo.
$endgroup$
– saz
Dec 6 '18 at 23:12
add a comment |
$begingroup$
Clearly,
$$S_{n+1}^3 = (S_n+X_{n+1})^3 = S_{n}^3 +3 S_n X_{n+1}^2 + 3S_n^2 X_{n+1}+X_{n+1}^3$$
which implies ("take out what is known"/"pull out")
$$mathbb{E}(S_{n+1}^3 mid mathcal{F}_n) = S_n^3+3 S_n underbrace{mathbb{E}(X_{n+1}^2 mid mathcal{F}_n)}_{=mathbb{E}(X_{n+1}^2)=1} + 3 S_n^2 underbrace{mathbb{E}(X_{n+1} mid mathcal{F}_n)}_{=mathbb{E}(X_{n+1})=0} + underbrace{mathbb{E}(X_{n+1}^3 mid mathcal{F}_n)}_{=mathbb{E}(X_{n+1}^3)=0}$$
i.e.
$$mathbb{E}(S_{n+1}^3 mid mathcal{F}_n) = S_n^3 + 3 S_n.$$
Using this equation, you shouldn't have much trouble to verify the martingale property of $(M_n)_{n in mathbb{N}}$.
answered Dec 6 '18 at 21:32
sazsaz
79.8k860124
$endgroup$
Clearly,
$$S_{n+1}^3 = (S_n+X_{n+1})^3 = S_{n}^3 +3 S_n X_{n+1}^2 + 3S_n^2 X_{n+1}+X_{n+1}^3$$
which implies ("take out what is known"/"pull out")
$$mathbb{E}(S_{n+1}^3 mid mathcal{F}_n) = S_n^3+3 S_n underbrace{mathbb{E}(X_{n+1}^2 mid mathcal{F}_n)}_{=mathbb{E}(X_{n+1}^2)=1} + 3 S_n^2 underbrace{mathbb{E}(X_{n+1} mid mathcal{F}_n)}_{=mathbb{E}(X_{n+1})=0} + underbrace{mathbb{E}(X_{n+1}^3 mid mathcal{F}_n)}_{=mathbb{E}(X_{n+1}^3)=0}$$
i.e.
$$mathbb{E}(S_{n+1}^3 mid mathcal{F}_n) = S_n^3 + 3 S_n.$$
Using this equation, you shouldn't have much trouble to verify the martingale property of $(M_n)_{n in mathbb{N}}$.
answered Dec 6 '18 at 21:32
sazsaz
79.8k860124
answered Dec 6 '18 at 21:32
sazsaz
79.8k860124
answered Dec 6 '18 at 21:32
sazsaz
79.8k860124
answered Dec 6 '18 at 21:32
sazsaz
79.8k860124
79.8k860124
$begingroup$
I definitely did some calculating wrong. For the second part am I correct that this is an application of optional stopping theorem? It's not quite set up for $mathbb{E}[M_0] = mathbb{E}[M_T]$ But it is $mathbb{E}[M_T + frac{1}{3}S^3_T] $. but here $mathbb{E}[M_0] = a - a^3/3$.
$endgroup$
– all.over
Dec 6 '18 at 21:53
$begingroup$
Sorry made edits to the comment above, hit add before I was done.
$endgroup$
– all.over
Dec 6 '18 at 21:58
$begingroup$
@all.over Well you will need to compute $E (S_T^3) $... to this end use $E (S_T)=0$ to find $P (S_T=0) $ and $P (S_T=K) $
$endgroup$
– saz
Dec 6 '18 at 22:34
$begingroup$
It won't let my at people for some reason. Anyway, isn't $E(S_T) = E(S_0) = a$ by the martingale property?
$endgroup$
– all.over
Dec 6 '18 at 22:48
1
$begingroup$
@all.over Yeah, sorry, that was a typo.
$endgroup$
– saz
Dec 6 '18 at 23:12
add a comment |
$begingroup$
I definitely did some calculating wrong. For the second part am I correct that this is an application of optional stopping theorem? It's not quite set up for $mathbb{E}[M_0] = mathbb{E}[M_T]$ But it is $mathbb{E}[M_T + frac{1}{3}S^3_T] $. but here $mathbb{E}[M_0] = a - a^3/3$.
$endgroup$
– all.over
Dec 6 '18 at 21:53
$begingroup$
Sorry made edits to the comment above, hit add before I was done.
$endgroup$
– all.over
Dec 6 '18 at 21:58
$begingroup$
@all.over Well you will need to compute $E (S_T^3) $... to this end use $E (S_T)=0$ to find $P (S_T=0) $ and $P (S_T=K) $
$endgroup$
– saz
Dec 6 '18 at 22:34
$begingroup$
It won't let my at people for some reason. Anyway, isn't $E(S_T) = E(S_0) = a$ by the martingale property?
$endgroup$
– all.over
Dec 6 '18 at 22:48
1
$begingroup$
@all.over Yeah, sorry, that was a typo.
$endgroup$
– saz
Dec 6 '18 at 23:12
$begingroup$
I definitely did some calculating wrong. For the second part am I correct that this is an application of optional stopping theorem? It's not quite set up for $mathbb{E}[M_0] = mathbb{E}[M_T]$ But it is $mathbb{E}[M_T + frac{1}{3}S^3_T] $. but here $mathbb{E}[M_0] = a - a^3/3$.
$endgroup$
– all.over
Dec 6 '18 at 21:53
$begingroup$
I definitely did some calculating wrong. For the second part am I correct that this is an application of optional stopping theorem? It's not quite set up for $mathbb{E}[M_0] = mathbb{E}[M_T]$ But it is $mathbb{E}[M_T + frac{1}{3}S^3_T] $. but here $mathbb{E}[M_0] = a - a^3/3$.
$endgroup$
– all.over
Dec 6 '18 at 21:53
$begingroup$
Sorry made edits to the comment above, hit add before I was done.
$endgroup$
– all.over
Dec 6 '18 at 21:58
$begingroup$
Sorry made edits to the comment above, hit add before I was done.
$endgroup$
– all.over
Dec 6 '18 at 21:58
$begingroup$
@all.over Well you will need to compute $E (S_T^3) $... to this end use $E (S_T)=0$ to find $P (S_T=0) $ and $P (S_T=K) $
$endgroup$
– saz
Dec 6 '18 at 22:34
$begingroup$
@all.over Well you will need to compute $E (S_T^3) $... to this end use $E (S_T)=0$ to find $P (S_T=0) $ and $P (S_T=K) $
$endgroup$
– saz
Dec 6 '18 at 22:34
$begingroup$
It won't let my at people for some reason. Anyway, isn't $E(S_T) = E(S_0) = a$ by the martingale property?
$endgroup$
– all.over
Dec 6 '18 at 22:48
$begingroup$
It won't let my at people for some reason. Anyway, isn't $E(S_T) = E(S_0) = a$ by the martingale property?
$endgroup$
– all.over
Dec 6 '18 at 22:48
1
1
$begingroup$
@all.over Yeah, sorry, that was a typo.
$endgroup$
– saz
Dec 6 '18 at 23:12
$begingroup$
@all.over Yeah, sorry, that was a typo.
$endgroup$
– saz
Dec 6 '18 at 23:12
add a comment |
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