# Expected Total Discounted Reward

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.

• if I'm reading this correctly, your setup is essentially the same as that as for proving the Wald Equation (as done in renewal theory). I.e. first consider the non-negative case $R_k':= \big \vert R_k\big \vert$ so $E(\sum_{k=1}^{\infty}β^ {k−1} R_k') = \sum_{k=1}^{\infty}β^ {k−1} E[R_k'] = E[R_1'] \cdot \sum_{k=1}^{\infty}β^ {k−1}=E[R_1'] E[T]$ where the interchange of limit and expectation is justified on monotone convergence. Then re-run the argument justifying the interchange on dominated convergence. That's the nice approach. There are uglier ones but it depends on what you know. – user8675309 Feb 8 at 21:01

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