I have to find the distribution of:

$ \epsilon_t + \sum_{i=0}^{N_t}x_{i,t}$


$\epsilon_t$ follows N(0,1)

$N_t$ follows P(0.1)

$x_{i,t}$ i.i.d, follow N(-0.1,0.3)

They are all independent.

How would you calculate that ? (see what I have done in comments)

Edit: I was wondering something. Would it be possible to assume the infinite sum of gaussian is gaussian, so with the characteristic function I can calculate the first two moments and have my answer ?

  • $\begingroup$ The moving upper bound is very limiting for me. The sum of n normal + the $\epsilon $ give: $N(-0.1 N_t,1 + 0.3N_t)$. But I don't know how to put the $N_t$ out. $\endgroup$ Jan 27, 2014 at 0:45
  • $\begingroup$ I have also tried to write down the caracteristic function but I don't know how to find the density. $\endgroup$ Jan 27, 2014 at 0:47

1 Answer 1


You already found the (normal) distribution of $X_n$ (where $n=N_t$): $$ X_n=N(-\frac 1{10}n,1+\frac 3{10}n) $$ Now, we know that $N$ is $P(\frac 1{10})$ (poisson) distributed. From the definition of the normal distribution, we can work out that $$ P(X_n=x)=\frac{e^{-\frac{(n+10 x)^2}{60 n+200}}}{\sqrt{\frac{3 \pi n}{5}+2 \pi }}\, dx $$ Now, we want to find the expected value of $P(X_n=x)$ where $n\sim P(\frac 1{10})$. I put this into Mathematica, and it didn't find a closed expression, so I guess there isn't one. It also doesn't help when I add the $\epsilon$ after evaluating the expected value of the sum.


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