Show that $E(Z)=E(K)E(X)$

Question

Let ${X_n, n\in N}$ be an iid sequence of psitive rrvs and let $K$ be a rrv independent of this sequence and taking its values in $N$ with $P(K=k)=p_k$. Consider the rrv $Z=\sum_{n=1}^{K} X_n$. Suppose that $E(X_n) <\infty$. Then show $E(Z)=E(K)E(X_n)$

My attempt is that since $K$ is independent of $X_n$ then how can i apply independence formula for expectation?

Answer 1

$EZ=\sum_k E(Z|K=k)P(K=k)$ $=\sum_k (kEX_1) P(K=k)=E(X_1) \sum_k kP(Z=k)=EX_1EK$. Note that $EX_n$ doe not depend on $n$ since $(X_I)$ is i.i.d..

Answer 2

Note that $$ Z=\sum_{n=1}^\infty X_nI(K\geq n). $$ Since the $X_n$ are non-negative the monotone convergence theorem together with independence of $X_n$ and $K$ imply that $$ EZ=\sum_{n=1}^\infty (EX_n)P(K\geq n)=\sum_{n=1}^\infty EX_1 P(K\geq n)=EX_1\sum_{n=1}^\infty P(K\geq n)=EX_1 EK $$ where in the final step we use the tail formula for expectation.