Let, X~Bin(n,p) and Y~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..
To compare the expectation and variance of T and S, let's first find their respective formulas.
Recall that the expectation of a sum is the sum of the expectations, and the expectation of a random variable multiplied by a constant is the constant times the expectation of the random variable.
For T: $$ E(T) = E(X_1 + X_2 + \cdots + X_Y) = E(Y) \cdot E(X) $$ Since $X \sim \operatorname{Bin}(n, p)$, $E(X) = np$. And since $Y \sim \operatorname{Poisson}(\lambda)$, $E(Y) = \lambda$. So, $$ E(T) = \lambda(np) = \lambda np $$
For S: $$ E(S) = E(Y_1 + Y_2 + \cdots + Y_X) = E(X) \cdot E(Y) $$ Since $E(X) = np$ and $E(Y) = \lambda$, we have: $$ E(S) = np \cdot \lambda = \lambda np $$
Thus, $E(T) = E(S) = \lambda np$.
For T, we use the formula for the variance of the sum of independent random variables:
$$ Var(T) = E(Y) \cdot Var(X) + Var(Y) \cdot (E(X))^2 $$
Since $X \sim \operatorname{Bin}(n, p)$, $Var(X) = np(1-p)$. And since $Y \sim \operatorname{Poisson}(\lambda)$, $Var(Y) = \lambda$. So, $$ Var(T) = \lambda(np(1-p)) + \lambda(np)^2 = \lambda np(1-p) + \lambda n^2p^2 $$
For S, we use the same formula:
$$ Var(S) = E(X) \cdot Var(Y) + Var(X) \cdot (E(Y))^2 $$
Since $Var(X) = np(1-p)$ and $Var(Y) = \lambda$, we have: $$ Var(S) = np(\lambda) + np(1-p)(\lambda)^2 = \lambda np + \lambda^2 np(1-p) $$
In conclusion, we have: