Let $X\sim \text{Pois}(0.35)$ and set $Y=e^X$. Let $Y_1,\ldots,Y_{83}$ be i.i.d with the same distribution as $Y$ and set $W=Y_1+\cdots+Y_{83}$. Find the mean and variance of $Y$.
I recently learned the Central Limit Theorem. Since $Y=e ^X$, I consider $Y$ to be a sum of $n$ i.i.d. $\text{Expo}(λ)$ r.v.s.
$Y \sim\gamma(n, λ)$
For large $n$, $Y \sim N(n/λ, n/λ^2)$
$n =83$
Then, which one is $λ$? Is $λ=0.35$?
Thanks.
$e^X$ where $X$ is Poisson is not an exponential RV. For one, it has a discrete distribution, taking only the values $1,e,e^2\ldots$ whereas an exponential has a continuous distribution.
Finding the mean and variance of sums of i.i.d. random variables is easy. The mean of the sum is just the sum of the means (this is true regardless of independence) and the variance is the sum of the variances (true for uncorrelated RVs and thus for independent RVs). So we have $$ E(\sum_{i=1}^n Y_i) = \sum_{i=1}^n E(Y_i) = n E(Y)\\ Var(\sum_{i=1}^n Y_i) = \sum_{i=1}^n Var(Y_i) = n Var(Y).$$
So all you need to do is caluclate $E(e^X)$ and $Var(e^X).$ Notice since the pmf of the Poisson is $$ P(X=k) = e^{-\lambda}\frac{\lambda^k}{k!}$$ we have $$ E(Y)=E(e^X) = \sum_{k=0}^\infty e^k e^{-\lambda}\frac{\lambda^k}{n!} = e^{-\lambda}\sum_{k=0}^\infty \frac{(e\lambda)^k}{k!}$$ and $$ E(Y^2) = E(e^{2X}) = \sum_{k=0}^\infty e^{2k} e^{-\lambda} \frac{\lambda^k}{n!}.$$
Hopefully that's enough for you to calculate from there. (Everything can be done in closed form.)