# A solution's variance and mean

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 ?

Best regards

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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

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 $$\mathrm E(X_t^2)=at+\tfrac12k^2\lambda \frac{t^{2c+1}}{2c+1}.$$