Show that this processes are martingales
$begingroup$
I have to show that the following two processes are martingales
- $M_t=g(t)B_t-int_0^tg'(s)B_sds$
- $X_t=expleft(e^tB_t-int^t_0e^sB_sds-frac{e^{2t}}{4}+frac{1}{4}right)$
Where $g(t)$ is a real valued continuously differentiable function and $B_t$ is a brownian motion.
I tried the first part straight forward but without success. I guess I can not simply use Fubini here to switch expectation and integration. So I believe it has something to do with Ito. But i really don't know how to apply it here to show that 1. is a martingale.
For the second process: I know that the first two terms in the exponent are 1. with $g(t)=e^t$ so this is a martingale. I assume I also have to apply Ito here aswell, but like in 1., I'm not sure how.
stochastic-processes brownian-motion martingales
edited Dec 12 '18 at 9:39
Larry
2,39131129
asked Dec 12 '18 at 9:24
ToxxiqqToxxiqq
173
$endgroup$
|
show 7 more comments
$begingroup$
I have to show that the following two processes are martingales
- $M_t=g(t)B_t-int_0^tg'(s)B_sds$
- $X_t=expleft(e^tB_t-int^t_0e^sB_sds-frac{e^{2t}}{4}+frac{1}{4}right)$
Where $g(t)$ is a real valued continuously differentiable function and $B_t$ is a brownian motion.
I tried the first part straight forward but without success. I guess I can not simply use Fubini here to switch expectation and integration. So I believe it has something to do with Ito. But i really don't know how to apply it here to show that 1. is a martingale.
For the second process: I know that the first two terms in the exponent are 1. with $g(t)=e^t$ so this is a martingale. I assume I also have to apply Ito here aswell, but like in 1., I'm not sure how.
stochastic-processes brownian-motion martingales
edited Dec 12 '18 at 9:39
Larry
2,39131129
asked Dec 12 '18 at 9:24
ToxxiqqToxxiqq
173
$endgroup$
1
$begingroup$
For the first one you can use Fubini's theorem for conditional expectations. Alternatively, you can apply Itô's formula to $f(t,x) := g(t) cdot x$.
$endgroup$
– saz
Dec 12 '18 at 9:53
$begingroup$
I tried using Fubini and I got for $u<t$:$mathbb{E}[M_t|mathcal{F}_u]=g(t)B_u-int^u_0g'(s)B_sds$. I don't see where the first $g(t)$ gets to a $g(u)$
$endgroup$
– Toxxiqq
Dec 12 '18 at 10:34
$begingroup$
You went wrong when you calculated the conditional expectation of the integral. Compute $mathbb{E}(B_s mid mathcal{F}_u)$ separately for $s leq u$ and $s>u$. What do you get?
$endgroup$
– saz
Dec 12 '18 at 10:42
$begingroup$
I did separate it and I got $int_0^ug'(s)mathbb{E}[B_s|mathcal{F}_u]ds+int^t_ug'(s)mathbb{E}[B_s|mathcal{F}_u]$. In the first integral we have $sleq u$ and therefore, all $B_s$ are $mathcal{F}_u$ measurable, i.e. we just get $int^u_0g'(s)B_sds$. In the second integral we just get $int_u^tg'(s)mathbb{E}[B_s]ds$ which is zero since $mathbb{E}[B_s]=0$. Or am I wrong with that?
$endgroup$
– Toxxiqq
Dec 12 '18 at 11:21
1
$begingroup$
There is no "dt" in the 2nd term on the right-hand side.. it should read $langle X rangle_t$ (or, perhaps, $(dX_t)^2$ ... depending on which notation you are using).
$endgroup$
– saz
Dec 12 '18 at 13:56
|
show 7 more comments
$begingroup$
I have to show that the following two processes are martingales
- $M_t=g(t)B_t-int_0^tg'(s)B_sds$
- $X_t=expleft(e^tB_t-int^t_0e^sB_sds-frac{e^{2t}}{4}+frac{1}{4}right)$
Where $g(t)$ is a real valued continuously differentiable function and $B_t$ is a brownian motion.
I tried the first part straight forward but without success. I guess I can not simply use Fubini here to switch expectation and integration. So I believe it has something to do with Ito. But i really don't know how to apply it here to show that 1. is a martingale.
For the second process: I know that the first two terms in the exponent are 1. with $g(t)=e^t$ so this is a martingale. I assume I also have to apply Ito here aswell, but like in 1., I'm not sure how.
stochastic-processes brownian-motion martingales
edited Dec 12 '18 at 9:39
Larry
2,39131129
asked Dec 12 '18 at 9:24
ToxxiqqToxxiqq
173
$endgroup$
I have to show that the following two processes are martingales
- $M_t=g(t)B_t-int_0^tg'(s)B_sds$
- $X_t=expleft(e^tB_t-int^t_0e^sB_sds-frac{e^{2t}}{4}+frac{1}{4}right)$
Where $g(t)$ is a real valued continuously differentiable function and $B_t$ is a brownian motion.
I tried the first part straight forward but without success. I guess I can not simply use Fubini here to switch expectation and integration. So I believe it has something to do with Ito. But i really don't know how to apply it here to show that 1. is a martingale.
For the second process: I know that the first two terms in the exponent are 1. with $g(t)=e^t$ so this is a martingale. I assume I also have to apply Ito here aswell, but like in 1., I'm not sure how.
stochastic-processes brownian-motion martingales
stochastic-processes brownian-motion martingales
edited Dec 12 '18 at 9:39
Larry
2,39131129
asked Dec 12 '18 at 9:24
ToxxiqqToxxiqq
173
edited Dec 12 '18 at 9:39
Larry
2,39131129
asked Dec 12 '18 at 9:24
ToxxiqqToxxiqq
173
edited Dec 12 '18 at 9:39
Larry
2,39131129
edited Dec 12 '18 at 9:39
Larry
2,39131129
edited Dec 12 '18 at 9:39
Larry
2,39131129
2,39131129
asked Dec 12 '18 at 9:24
ToxxiqqToxxiqq
173
asked Dec 12 '18 at 9:24
ToxxiqqToxxiqq
173
asked Dec 12 '18 at 9:24
ToxxiqqToxxiqq
173
173
1
$begingroup$
For the first one you can use Fubini's theorem for conditional expectations. Alternatively, you can apply Itô's formula to $f(t,x) := g(t) cdot x$.
$endgroup$
– saz
Dec 12 '18 at 9:53
$begingroup$
I tried using Fubini and I got for $u<t$:$mathbb{E}[M_t|mathcal{F}_u]=g(t)B_u-int^u_0g'(s)B_sds$. I don't see where the first $g(t)$ gets to a $g(u)$
$endgroup$
– Toxxiqq
Dec 12 '18 at 10:34
$begingroup$
You went wrong when you calculated the conditional expectation of the integral. Compute $mathbb{E}(B_s mid mathcal{F}_u)$ separately for $s leq u$ and $s>u$. What do you get?
$endgroup$
– saz
Dec 12 '18 at 10:42
$begingroup$
I did separate it and I got $int_0^ug'(s)mathbb{E}[B_s|mathcal{F}_u]ds+int^t_ug'(s)mathbb{E}[B_s|mathcal{F}_u]$. In the first integral we have $sleq u$ and therefore, all $B_s$ are $mathcal{F}_u$ measurable, i.e. we just get $int^u_0g'(s)B_sds$. In the second integral we just get $int_u^tg'(s)mathbb{E}[B_s]ds$ which is zero since $mathbb{E}[B_s]=0$. Or am I wrong with that?
$endgroup$
– Toxxiqq
Dec 12 '18 at 11:21
1
$begingroup$
There is no "dt" in the 2nd term on the right-hand side.. it should read $langle X rangle_t$ (or, perhaps, $(dX_t)^2$ ... depending on which notation you are using).
$endgroup$
– saz
Dec 12 '18 at 13:56
|
show 7 more comments
1
$begingroup$
For the first one you can use Fubini's theorem for conditional expectations. Alternatively, you can apply Itô's formula to $f(t,x) := g(t) cdot x$.
$endgroup$
– saz
Dec 12 '18 at 9:53
$begingroup$
I tried using Fubini and I got for $u<t$:$mathbb{E}[M_t|mathcal{F}_u]=g(t)B_u-int^u_0g'(s)B_sds$. I don't see where the first $g(t)$ gets to a $g(u)$
$endgroup$
– Toxxiqq
Dec 12 '18 at 10:34
$begingroup$
You went wrong when you calculated the conditional expectation of the integral. Compute $mathbb{E}(B_s mid mathcal{F}_u)$ separately for $s leq u$ and $s>u$. What do you get?
$endgroup$
– saz
Dec 12 '18 at 10:42
$begingroup$
I did separate it and I got $int_0^ug'(s)mathbb{E}[B_s|mathcal{F}_u]ds+int^t_ug'(s)mathbb{E}[B_s|mathcal{F}_u]$. In the first integral we have $sleq u$ and therefore, all $B_s$ are $mathcal{F}_u$ measurable, i.e. we just get $int^u_0g'(s)B_sds$. In the second integral we just get $int_u^tg'(s)mathbb{E}[B_s]ds$ which is zero since $mathbb{E}[B_s]=0$. Or am I wrong with that?
$endgroup$
– Toxxiqq
Dec 12 '18 at 11:21
1
$begingroup$
There is no "dt" in the 2nd term on the right-hand side.. it should read $langle X rangle_t$ (or, perhaps, $(dX_t)^2$ ... depending on which notation you are using).
$endgroup$
– saz
Dec 12 '18 at 13:56
1
1
$begingroup$
For the first one you can use Fubini's theorem for conditional expectations. Alternatively, you can apply Itô's formula to $f(t,x) := g(t) cdot x$.
$endgroup$
– saz
Dec 12 '18 at 9:53
$begingroup$
For the first one you can use Fubini's theorem for conditional expectations. Alternatively, you can apply Itô's formula to $f(t,x) := g(t) cdot x$.
$endgroup$
– saz
Dec 12 '18 at 9:53
$begingroup$
I tried using Fubini and I got for $u<t$:$mathbb{E}[M_t|mathcal{F}_u]=g(t)B_u-int^u_0g'(s)B_sds$. I don't see where the first $g(t)$ gets to a $g(u)$
$endgroup$
– Toxxiqq
Dec 12 '18 at 10:34
$begingroup$
I tried using Fubini and I got for $u<t$:$mathbb{E}[M_t|mathcal{F}_u]=g(t)B_u-int^u_0g'(s)B_sds$. I don't see where the first $g(t)$ gets to a $g(u)$
$endgroup$
– Toxxiqq
Dec 12 '18 at 10:34
$begingroup$
You went wrong when you calculated the conditional expectation of the integral. Compute $mathbb{E}(B_s mid mathcal{F}_u)$ separately for $s leq u$ and $s>u$. What do you get?
$endgroup$
– saz
Dec 12 '18 at 10:42
$begingroup$
You went wrong when you calculated the conditional expectation of the integral. Compute $mathbb{E}(B_s mid mathcal{F}_u)$ separately for $s leq u$ and $s>u$. What do you get?
$endgroup$
– saz
Dec 12 '18 at 10:42
$begingroup$
I did separate it and I got $int_0^ug'(s)mathbb{E}[B_s|mathcal{F}_u]ds+int^t_ug'(s)mathbb{E}[B_s|mathcal{F}_u]$. In the first integral we have $sleq u$ and therefore, all $B_s$ are $mathcal{F}_u$ measurable, i.e. we just get $int^u_0g'(s)B_sds$. In the second integral we just get $int_u^tg'(s)mathbb{E}[B_s]ds$ which is zero since $mathbb{E}[B_s]=0$. Or am I wrong with that?
$endgroup$
– Toxxiqq
Dec 12 '18 at 11:21
$begingroup$
I did separate it and I got $int_0^ug'(s)mathbb{E}[B_s|mathcal{F}_u]ds+int^t_ug'(s)mathbb{E}[B_s|mathcal{F}_u]$. In the first integral we have $sleq u$ and therefore, all $B_s$ are $mathcal{F}_u$ measurable, i.e. we just get $int^u_0g'(s)B_sds$. In the second integral we just get $int_u^tg'(s)mathbb{E}[B_s]ds$ which is zero since $mathbb{E}[B_s]=0$. Or am I wrong with that?
$endgroup$
– Toxxiqq
Dec 12 '18 at 11:21
1
1
$begingroup$
There is no "dt" in the 2nd term on the right-hand side.. it should read $langle X rangle_t$ (or, perhaps, $(dX_t)^2$ ... depending on which notation you are using).
$endgroup$
– saz
Dec 12 '18 at 13:56
$begingroup$
There is no "dt" in the 2nd term on the right-hand side.. it should read $langle X rangle_t$ (or, perhaps, $(dX_t)^2$ ... depending on which notation you are using).
$endgroup$
– saz
Dec 12 '18 at 13:56
|
show 7 more comments
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1
$begingroup$
For the first one you can use Fubini's theorem for conditional expectations. Alternatively, you can apply Itô's formula to $f(t,x) := g(t) cdot x$.
$endgroup$
– saz
Dec 12 '18 at 9:53
$begingroup$
I tried using Fubini and I got for $u<t$:$mathbb{E}[M_t|mathcal{F}_u]=g(t)B_u-int^u_0g'(s)B_sds$. I don't see where the first $g(t)$ gets to a $g(u)$
$endgroup$
– Toxxiqq
Dec 12 '18 at 10:34
$begingroup$
You went wrong when you calculated the conditional expectation of the integral. Compute $mathbb{E}(B_s mid mathcal{F}_u)$ separately for $s leq u$ and $s>u$. What do you get?
$endgroup$
– saz
Dec 12 '18 at 10:42
$begingroup$
I did separate it and I got $int_0^ug'(s)mathbb{E}[B_s|mathcal{F}_u]ds+int^t_ug'(s)mathbb{E}[B_s|mathcal{F}_u]$. In the first integral we have $sleq u$ and therefore, all $B_s$ are $mathcal{F}_u$ measurable, i.e. we just get $int^u_0g'(s)B_sds$. In the second integral we just get $int_u^tg'(s)mathbb{E}[B_s]ds$ which is zero since $mathbb{E}[B_s]=0$. Or am I wrong with that?
$endgroup$
– Toxxiqq
Dec 12 '18 at 11:21
1
$begingroup$
There is no "dt" in the 2nd term on the right-hand side.. it should read $langle X rangle_t$ (or, perhaps, $(dX_t)^2$ ... depending on which notation you are using).
$endgroup$
– saz
Dec 12 '18 at 13:56