Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have a stochastic differential equation of this type : $$ \ dX(t) = a dB(t)+k c t^{c-1} \cos(\theta(t))dP(t). $$ with

$$ \begin{align} a,k &:\text{ constants }\\ B(t) &:\text{ Brownian process}\\ P(t) &: \text{A Poisson process with parameter }\lambda\\ \theta(t) &\text{: A uniform distribution on }[0,2\pi]\\ c &:\text{ a constant }(0<c<2) \end{align} $$ I tried to calculate the variance and the mean of the solution using the Ito general formula but it was too complicated. How can I find these information please ?

Thanks for your help

Best regards

share|improve this question
    
Got something from my answer below? –  Did Jul 25 '12 at 16:31
    
This helped me a lot Thank you did –  Samatix Jul 30 '12 at 9:44
    
Sorry I'm a new user. I didn't know how to accept the solution :D –  Samatix Jul 30 '12 at 9:46
    
But now you even know how to unaccept them... Why did you unaccept this one? –  Did Aug 31 '12 at 7:22
    
Sorry did, Is it okay now ? –  Samatix Oct 8 '12 at 13:57

1 Answer 1

up vote 2 down vote accepted

In other words, assuming that $X_0=0$, one asks that $$ X_t=aB_t+k\sum_{n=1}^{+\infty}T_n^c\cdot\cos(\theta_n)\cdot[T_n\leqslant t], $$ where the sequence $(T_n)_{n\geqslant1}$ enumerates the points of a Poisson process with intensity $\lambda$ and counting process $(N_t)_{t\geqslant0}$, and where I guess that one should also assume that $(\theta_n)_{n\geqslant1}$ is i.i.d. with the prescribed distribution and that the processes $(B_t)_{t\geqslant0}$, $(\theta_n)_{n\geqslant1}$ and $(T_n)_{n\geqslant1}$ are independent.

(Note that I replaced your $ct^{c-1}$, which I believe is a mistake, by $t^c$. If I am wrong, one can adapt the following to $ct^{c-1}$.)

Since every $B_t$ is centered and every $\cos(\theta_n)$ is centered and independent on $T_n$, $$ \mathrm E(X_t)=0. $$ Since $B_t$ is independent of everything else and centered, since $\mathrm E(B_t^2)=t$ and $\mathrm E(\cos(\theta_n)^2)=\frac12$, and since the sequence $(\cos(\theta_n))_n$ is independent of everything else and centered, $$ \mathrm E(X_t^2)=at+\tfrac12k^2\sum_{n=1}^{+\infty}\mathrm E(T_n^{2c}:T_n\leqslant t). $$ Conditionally on $N_t=k$, the set $\{T_n\mid n\leqslant k\}$ is distributed like an i.i.d. sample of size $n$ uniformly distributed on $(0,t)$. Let $U_t$ denote a random variable uniformly distributed on $(0,t)$. Hence, $$ \mathrm E\left(\sum_{n=1}^{N_t}T_n^{2c}\,\bigg|\, N_t=k\right)=k\cdot\mathrm E(U_t^{2c})=\frac{kt^{2c}}{2c+1}. $$ Since $\mathrm E(N_t)=\lambda t$, this yields $$ \mathrm E(X_t^2)=at+\tfrac12k^2\lambda \frac{t^{2c+1}}{2c+1}. $$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.