Let random variable $R_k$ denote the revenue received in the kth period. Suppose that $R_1, R_2, . . .$ are independent and identically distributed. The quantity $Q = \sum_{k=1}^{\infty}\beta^{k-1}R_k$ denotes the total discounted revenue with discount factor β. Let T denote a geometric random variable with success probability 1 − β and T takes values 1, 2, . . .. That is, $P(T = k) = β^ {k−1} (1 − β)$, k = 1, 2, . . . . We further assume that T, R1, R2, . . . are independent. (3 marks) Show that the expected total discount revenue is equal to the expected total (undiscounted) reward received by time T. In other words, show that
$$E(\sum_{k=1}^{\infty}β^ {k−1} R_k)=E(\sum_{k=1}^T R_k)$$
I am unsure where to start or how to identify what to do.
First let the mean of $R_k$ be $\mu$.
Then you have to identify that the RHS is the expectation of a random sum, which evaluates to the product of mean of $R_k$ and mean of T, so the RHS = $E(T)E(R_k)$. Since T is a geometric RV, the RHS is ultimately $\mu / (1-\beta)$
Then to prove that the LHS is also equal to $\mu/(1-\beta)$, you have to bring the expectation in on the $R_k$ then just sum up the converging series using the sum of gp formula on the left