# $\mathbb{E}(W_1N_1(W_2N_2)^2)+\mathbb{P}(N_4>N_3>2, W_2<W_3<2)=?$

$$W$$ - Wiener process, $$N$$- Poisson Process with parameter $$\lambda$$ and $$W$$ and $$N$$ are independent. Compute: $$\mathbb{E}(W_1N_1(W_2N_2)^2)+\mathbb{P}(N_4>N_3>2, W_2

$$\mathbb{E}(W_1N_1(W_2N_2)^2)=\mathbb{E}(W_1W_2^2)\cdot \mathbb{E}(N_1N_2^2)=0$$

Could I break down the expected value like that? I used independence.

$$\mathbb{P}(N_4>N_3>2, W_2N_3>2)\cdot \mathbb{P}(W_2 and here i have problem, i dont know how to calculate this :( Can you help me?

• Could you please explain what you mean by "$N_1$" and "$N_2$" etc.? Are these processes defined for real times or--as suggested by the notation--just integral times? – whuber Jan 16 at 23:58
• $\left\{N_t\right\}$ is a Poisson Process where $t$ is from the set ${0,1,2,3,...}$. $\left\{W_t\right\}$ is a Wiener process defined for $t\in[0,\infty)$ – Piszczu Jan 17 at 0:06
• The Poisson probability can be expressed in terms of the modified Bessel function of the first kind, $I_0.$ The Wiener process probability requires a double integral of a Bivariate Normal density over a sector. This question looks artificial and one wonders what its purpose might be--calculations like this are worth doing only if they have some point. Could you explain its statistical application or interest? – whuber Jan 17 at 3:59
• that's the question I had to test. I think there is a trick here that makes it easy to count – Piszczu Jan 17 at 9:09
• $P(W_2<W_3<2)=P(0<W_3-W_2<2-W_2)$ and how now to calculate this? – Piszczu Jan 17 at 10:19