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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.

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    $\begingroup$ 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. $\endgroup$ – user8675309 Feb 8 at 21:01
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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

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