We define $S(t) = \sum\limits_{i = 1}^{N(t)} {{e^{ - r{U_i}}}{X_i}}$ with $N(t)$ a poisson process, $U_i$ is a sequence of i.i.d. random variables having $U(0,1)$ distribution, $X_i$ are iid.
$X_i$ and $U_i$ are independant
I have to compute $E(S(t))$ and $var(S(t))$
This is what i did:
$E(S(t)) = E\left( {E\left( {\sum\limits_{i = 1}^{N(t)} {{e^{ - r{T_i}}}{X_i}} \left| {N(t) = n} \right.} \right)} \right)$
I am stuck here
What you have done is a very good start. You just need to use linearity of expectation and using i.i.d. assumption. Suppose that $\overline{X^2}$ is the second moment of $X$ and $\mu_X$ its mean. Suppose that $\phi(t)$ is the moment generating function of $U$.
First see that the conditional expectation is evaluated as : $$ E\left( {\sum\limits_{i = 1}^{N(t)} {{e^{ - r{U_i}}}{X_i}} \left| {N(t) = n} \right.} \right)=nE(e^{-r{U_i}}{X_i})=n\mu_X\phi(-r) $$ where $N(t)=n$ is then entered into expectation: $$ E(S(t)) = \mu_X\phi(-r)E(N(t))=\mu_X\phi(-r)\lambda t. $$ Similarly, second moment of $S(t)$ can be calculated: $$ E(S(t)^2|N(t)=n) = n\overline{X^2}\phi(-2r)+n(n-1) \mu_X^2\phi^2(-r) $$ Therefore, the second moment is obtained as: $$ E(S(t)^2)=\overline{X^2}\phi(-2r)\lambda t+(\lambda t)^2 \mu_X^2\phi^2(-r). $$