How would one solve for a process where a stochastic random variable is divided by a deterministic random...
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I'm currently working on a problem where I am trying to model a stochastic process where a stochastic random variable is divided by a deterministic random variable following a normal distribution. Its been a while wince I had a look at stochastic calc so apologies in advance if my math is a bit rusty!
Consider a simple stochastic process with no drift that satisfies the following SDE
Y(t) = X(t)
dX(t) = $sigma$dB(t) where B(t) is Brownian motion and B(t) ~ N($0$,t).
Solving such a process would give me Y(t) = Y$_0$ + $sigma$B(t)
I'm currently trying to find out what the resultant stochastic process would be if Y(t) = $frac{X(t)}{Z}$ where Z is a Normal distribution with mean $mu_z$ and variance $sigma_z^2$ i.e Z ~ N($mu_z$,$sigma_z^2$).
How would one apply Ito's lemma in such a case?
stochastic-processes random-variables normal-distribution stochastic-calculus stochastic-integrals
asked Jan 9 at 12:11
Hamza JuzerHamza Juzer
61
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add a comment |
$begingroup$
I'm currently working on a problem where I am trying to model a stochastic process where a stochastic random variable is divided by a deterministic random variable following a normal distribution. Its been a while wince I had a look at stochastic calc so apologies in advance if my math is a bit rusty!
Consider a simple stochastic process with no drift that satisfies the following SDE
Y(t) = X(t)
dX(t) = $sigma$dB(t) where B(t) is Brownian motion and B(t) ~ N($0$,t).
Solving such a process would give me Y(t) = Y$_0$ + $sigma$B(t)
I'm currently trying to find out what the resultant stochastic process would be if Y(t) = $frac{X(t)}{Z}$ where Z is a Normal distribution with mean $mu_z$ and variance $sigma_z^2$ i.e Z ~ N($mu_z$,$sigma_z^2$).
How would one apply Ito's lemma in such a case?
stochastic-processes random-variables normal-distribution stochastic-calculus stochastic-integrals
asked Jan 9 at 12:11
Hamza JuzerHamza Juzer
61
$endgroup$
$begingroup$
You already know that $X(t) = X(0) + sigma B(t)$ so $Y(t) = frac{X(0)}{Z} + frac{sigma}{Z} B(t)$.
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– Rhys Steele
Jan 9 at 15:04
$begingroup$
Perfect! That makes total sense! Cheers :)
$endgroup$
– Hamza Juzer
Jan 10 at 22:07
add a comment |
$begingroup$
I'm currently working on a problem where I am trying to model a stochastic process where a stochastic random variable is divided by a deterministic random variable following a normal distribution. Its been a while wince I had a look at stochastic calc so apologies in advance if my math is a bit rusty!
Consider a simple stochastic process with no drift that satisfies the following SDE
Y(t) = X(t)
dX(t) = $sigma$dB(t) where B(t) is Brownian motion and B(t) ~ N($0$,t).
Solving such a process would give me Y(t) = Y$_0$ + $sigma$B(t)
I'm currently trying to find out what the resultant stochastic process would be if Y(t) = $frac{X(t)}{Z}$ where Z is a Normal distribution with mean $mu_z$ and variance $sigma_z^2$ i.e Z ~ N($mu_z$,$sigma_z^2$).
How would one apply Ito's lemma in such a case?
stochastic-processes random-variables normal-distribution stochastic-calculus stochastic-integrals
asked Jan 9 at 12:11
Hamza JuzerHamza Juzer
61
$endgroup$
I'm currently working on a problem where I am trying to model a stochastic process where a stochastic random variable is divided by a deterministic random variable following a normal distribution. Its been a while wince I had a look at stochastic calc so apologies in advance if my math is a bit rusty!
Consider a simple stochastic process with no drift that satisfies the following SDE
Y(t) = X(t)
dX(t) = $sigma$dB(t) where B(t) is Brownian motion and B(t) ~ N($0$,t).
Solving such a process would give me Y(t) = Y$_0$ + $sigma$B(t)
I'm currently trying to find out what the resultant stochastic process would be if Y(t) = $frac{X(t)}{Z}$ where Z is a Normal distribution with mean $mu_z$ and variance $sigma_z^2$ i.e Z ~ N($mu_z$,$sigma_z^2$).
How would one apply Ito's lemma in such a case?
stochastic-processes random-variables normal-distribution stochastic-calculus stochastic-integrals
stochastic-processes random-variables normal-distribution stochastic-calculus stochastic-integrals
asked Jan 9 at 12:11
Hamza JuzerHamza Juzer
61
asked Jan 9 at 12:11
Hamza JuzerHamza Juzer
61
asked Jan 9 at 12:11
Hamza JuzerHamza Juzer
61
asked Jan 9 at 12:11
Hamza JuzerHamza Juzer
61
asked Jan 9 at 12:11
Hamza JuzerHamza Juzer
61
61
$begingroup$
You already know that $X(t) = X(0) + sigma B(t)$ so $Y(t) = frac{X(0)}{Z} + frac{sigma}{Z} B(t)$.
$endgroup$
– Rhys Steele
Jan 9 at 15:04
$begingroup$
Perfect! That makes total sense! Cheers :)
$endgroup$
– Hamza Juzer
Jan 10 at 22:07
add a comment |
$begingroup$
You already know that $X(t) = X(0) + sigma B(t)$ so $Y(t) = frac{X(0)}{Z} + frac{sigma}{Z} B(t)$.
$endgroup$
– Rhys Steele
Jan 9 at 15:04
$begingroup$
Perfect! That makes total sense! Cheers :)
$endgroup$
– Hamza Juzer
Jan 10 at 22:07
$begingroup$
You already know that $X(t) = X(0) + sigma B(t)$ so $Y(t) = frac{X(0)}{Z} + frac{sigma}{Z} B(t)$.
$endgroup$
– Rhys Steele
Jan 9 at 15:04
$begingroup$
You already know that $X(t) = X(0) + sigma B(t)$ so $Y(t) = frac{X(0)}{Z} + frac{sigma}{Z} B(t)$.
$endgroup$
– Rhys Steele
Jan 9 at 15:04
$begingroup$
Perfect! That makes total sense! Cheers :)
$endgroup$
– Hamza Juzer
Jan 10 at 22:07
$begingroup$
Perfect! That makes total sense! Cheers :)
$endgroup$
– Hamza Juzer
Jan 10 at 22:07
add a comment |
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$begingroup$
You already know that $X(t) = X(0) + sigma B(t)$ so $Y(t) = frac{X(0)}{Z} + frac{sigma}{Z} B(t)$.
$endgroup$
– Rhys Steele
Jan 9 at 15:04
$begingroup$
Perfect! That makes total sense! Cheers :)
$endgroup$
– Hamza Juzer
Jan 10 at 22:07