Comoare expectations of $S$ and $T$ and Variance of $S$ and $T$

Question

Let, $X\sim Bin(n,p)$ and $Y\sim Poisson(\lambda )$. Let $$T=X_1+X_2+...+X_Y$$ with $X_i$'s i. i. d. Bin(n,p) (and independent to Y) and $$S=Y_1+Y_2+...+Y_X$$ with $Y_i$'s i. i. d. Poisson($\lambda$) (and independent to X). Compare expectation of T and S and Variance of T and S.

My attempt :

$X_i$'s i. i. d. Bin(n,p) then $E(X_i) =np \ and \ Var(X_i) =npq$

$Y_i$'s i. i. d. Poisson($\lambda$) then $E(Y_i) =\lambda=Var(Y_i) $

Then $E(T) =Ynp$ and $Var(T) =Ynpq$ Also $E(S) =X\lambda$ and $Var(T) =X\lambda$ Then I can't get any relation between them. Can anyone give me a suggestion..

Answer 1

Note $X, Y$ are random variable. For example,

If $Y=1, T=X_1, $ with probability $P(Y=1)=\lambda e^{-\lambda}$

If $Y=2, T=X_1+X_2, $ with probability $P(Y=2)=\frac{\lambda^2}2 e^{-\lambda}$

If $Y=k, T=X_1+X_2+\dots+X_k, $ with probability $P(Y=k)=\frac{\lambda^k}{k!} e^{-\lambda}$

So the expectation for $T$ is:

$$E(T)=\sum_{k=1}^\infty knp\cdot \frac{\lambda^k}{k!} e^{-\lambda}=\lambda np$$

you can compute other cases by the similar way.