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Problem on the exponential martingale with the Brownian motion
Clash Royale CLAN TAG#URR8PPP
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Let $S$ be a stochastic process defined by $S_t$ = exp$(int_0^t$ $z_s$ $dW_s$ $- frac12int_0^t$$vert z_s vert^2$ $ds$), where $W$ is a standard brownian motion and $z$ is deterministic with $int_0^T vert z_s vert^2 ds$ $lt$ $infty$
I want to show that $S$ is a martingale:
I think the integrability follows from Hölder's inequality using $mathbbE$[exp($nu$)] = exp($fracsigma^22$), if $nu$ ~ $mathcalN$($0$, $sigma^2$).
Furthermore, I think adaptability is trivially given since the brownian motion is also a martingale.
However, I am unable to show the decisive martingale property $mathbbE$[$S_t+1$|$mathcalF_t$] = $S_t$. Is the independence of the increments $int_0^t z_t dW_t - int_0^s z_s dW_s$ needed?
I am grateful for any tip and piece of advice
stochastic-processes stochastic-calculus brownian-motion martingales stochastic-integrals
asked Jul 20 at 15:41
fidel castro
1738
add a comment |Â
up vote
1
down vote
favorite
Let $S$ be a stochastic process defined by $S_t$ = exp$(int_0^t$ $z_s$ $dW_s$ $- frac12int_0^t$$vert z_s vert^2$ $ds$), where $W$ is a standard brownian motion and $z$ is deterministic with $int_0^T vert z_s vert^2 ds$ $lt$ $infty$
I want to show that $S$ is a martingale:
I think the integrability follows from Hölder's inequality using $mathbbE$[exp($nu$)] = exp($fracsigma^22$), if $nu$ ~ $mathcalN$($0$, $sigma^2$).
Furthermore, I think adaptability is trivially given since the brownian motion is also a martingale.
However, I am unable to show the decisive martingale property $mathbbE$[$S_t+1$|$mathcalF_t$] = $S_t$. Is the independence of the increments $int_0^t z_t dW_t - int_0^s z_s dW_s$ needed?
I am grateful for any tip and piece of advice
stochastic-processes stochastic-calculus brownian-motion martingales stochastic-integrals
asked Jul 20 at 15:41
fidel castro
1738
Do you know Ito's lemma?
– Rhys Steele
Jul 20 at 15:48
No I unfortunately do not - is there a way to solve the problem without using the lemma?
– fidel castro
Jul 20 at 15:56
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $S$ be a stochastic process defined by $S_t$ = exp$(int_0^t$ $z_s$ $dW_s$ $- frac12int_0^t$$vert z_s vert^2$ $ds$), where $W$ is a standard brownian motion and $z$ is deterministic with $int_0^T vert z_s vert^2 ds$ $lt$ $infty$
I want to show that $S$ is a martingale:
I think the integrability follows from Hölder's inequality using $mathbbE$[exp($nu$)] = exp($fracsigma^22$), if $nu$ ~ $mathcalN$($0$, $sigma^2$).
Furthermore, I think adaptability is trivially given since the brownian motion is also a martingale.
However, I am unable to show the decisive martingale property $mathbbE$[$S_t+1$|$mathcalF_t$] = $S_t$. Is the independence of the increments $int_0^t z_t dW_t - int_0^s z_s dW_s$ needed?
I am grateful for any tip and piece of advice
stochastic-processes stochastic-calculus brownian-motion martingales stochastic-integrals
asked Jul 20 at 15:41
fidel castro
1738
Let $S$ be a stochastic process defined by $S_t$ = exp$(int_0^t$ $z_s$ $dW_s$ $- frac12int_0^t$$vert z_s vert^2$ $ds$), where $W$ is a standard brownian motion and $z$ is deterministic with $int_0^T vert z_s vert^2 ds$ $lt$ $infty$
I want to show that $S$ is a martingale:
I think the integrability follows from Hölder's inequality using $mathbbE$[exp($nu$)] = exp($fracsigma^22$), if $nu$ ~ $mathcalN$($0$, $sigma^2$).
Furthermore, I think adaptability is trivially given since the brownian motion is also a martingale.
However, I am unable to show the decisive martingale property $mathbbE$[$S_t+1$|$mathcalF_t$] = $S_t$. Is the independence of the increments $int_0^t z_t dW_t - int_0^s z_s dW_s$ needed?
I am grateful for any tip and piece of advice
stochastic-processes stochastic-calculus brownian-motion martingales stochastic-integrals
asked Jul 20 at 15:41
fidel castro
1738
asked Jul 20 at 15:41
fidel castro
1738
asked Jul 20 at 15:41
fidel castro
1738
asked Jul 20 at 15:41
fidel castro
1738
1738
Do you know Ito's lemma?
– Rhys Steele
Jul 20 at 15:48
No I unfortunately do not - is there a way to solve the problem without using the lemma?
– fidel castro
Jul 20 at 15:56
add a comment |Â
Do you know Ito's lemma?
– Rhys Steele
Jul 20 at 15:48
No I unfortunately do not - is there a way to solve the problem without using the lemma?
– fidel castro
Jul 20 at 15:56
Do you know Ito's lemma?
– Rhys Steele
Jul 20 at 15:48
Do you know Ito's lemma?
– Rhys Steele
Jul 20 at 15:48
No I unfortunately do not - is there a way to solve the problem without using the lemma?
– fidel castro
Jul 20 at 15:56
No I unfortunately do not - is there a way to solve the problem without using the lemma?
– fidel castro
Jul 20 at 15:56
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Hint: Because $z$ is deterministic, the stochastic integral $int_t^t+u z_s,dW_s$ is independent of $mathcal F_t$, for each $t>0, u>0$. Now write $int_0^t+u z_s,dW_s$ as $int_0^t z_s,dW_s+int_t^t+u z_s,dW_s$.
answered Jul 20 at 16:15
John Dawkins
12.5k1917
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Hint: Because $z$ is deterministic, the stochastic integral $int_t^t+u z_s,dW_s$ is independent of $mathcal F_t$, for each $t>0, u>0$. Now write $int_0^t+u z_s,dW_s$ as $int_0^t z_s,dW_s+int_t^t+u z_s,dW_s$.
answered Jul 20 at 16:15
John Dawkins
12.5k1917
add a comment |Â
up vote
2
down vote
accepted
Hint: Because $z$ is deterministic, the stochastic integral $int_t^t+u z_s,dW_s$ is independent of $mathcal F_t$, for each $t>0, u>0$. Now write $int_0^t+u z_s,dW_s$ as $int_0^t z_s,dW_s+int_t^t+u z_s,dW_s$.
answered Jul 20 at 16:15
John Dawkins
12.5k1917
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Hint: Because $z$ is deterministic, the stochastic integral $int_t^t+u z_s,dW_s$ is independent of $mathcal F_t$, for each $t>0, u>0$. Now write $int_0^t+u z_s,dW_s$ as $int_0^t z_s,dW_s+int_t^t+u z_s,dW_s$.
answered Jul 20 at 16:15
John Dawkins
12.5k1917
Hint: Because $z$ is deterministic, the stochastic integral $int_t^t+u z_s,dW_s$ is independent of $mathcal F_t$, for each $t>0, u>0$. Now write $int_0^t+u z_s,dW_s$ as $int_0^t z_s,dW_s+int_t^t+u z_s,dW_s$.
answered Jul 20 at 16:15
John Dawkins
12.5k1917
answered Jul 20 at 16:15
John Dawkins
12.5k1917
answered Jul 20 at 16:15
John Dawkins
12.5k1917
answered Jul 20 at 16:15
John Dawkins
12.5k1917
12.5k1917
add a comment |Â
add a comment |Â
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Do you know Ito's lemma?
– Rhys Steele
Jul 20 at 15:48
No I unfortunately do not - is there a way to solve the problem without using the lemma?
– fidel castro
Jul 20 at 15:56