# Expectation of a(n injective) function of a geometric random variable

The setup of this problem is the following.

Tom is playing a game online. He keeps playing until he wins one game. Winning in the $n$th game will give a payout of $\frac{$100}{n}$. Each game is won independently with probability$p$. I'm trying to find the expected winnings. The number of games played before winning one follows a geometric distribution with parameter$p$. So Tom will play$\frac{1}{p}$games, on average, before winning one. I intuitively imagine the winnings then to be$\$100 \times p$. Why exactly is this though?

If we let $X$ be the geometrically distributed random variable representing the number of games played when the first is won, then we can define the variable $Y = f \circ X$ where $f(n) = \frac{$100}{n}$to represent the winnings. Directly trying to find the expectation of$Y$gives $$\operatorname{E}(Y) = \sum_{n=1}^\infty \frac{100}{n} (1-p)^{n-1} p$$ which I am not sure how to evaluate. How should I proceed to find the expectation, formally? I noticed that the function$f$is injective. Does this have a particular significance when trying to find expected values? Is there a more general theorem that relates the expectation of a function$f$of a random variable$X$to the expectation of$X$itself? • hint:$\sum_{k=1}^{\infty} \frac{x^k}{k} = \log \frac{1}{1-x}– Alex Feb 21 '17 at 22:39 ## 2 Answers Hint: Use the Taylor expansion of the logarithm \begin{align} \log(1-x) = -\sum_{n=1}^{\infty}\frac{x^n}{n} &&|x|<1. \end{align} Advanced hint: \begin{align} 100p\sum_{n=1}^{\infty}\frac{(1-p)^{n-1}}{n} = \frac{100p}{1-p}\sum_{n=1}^{\infty}\frac{(1-p)^{n}}{n}. \end{align} • I don't see how I'm supposed to reconcile the fact that exponentn-1$in my situation is different from the denominator$n$. – Jacob Errington Feb 22 '17 at 0:21 Others have already told you how to calculate the expectation, but regarding your injectivity and expectation of$f$questions, the only significance of lack of injectivity is that its possible that$f(X)=a$for more than one value of$a$, so that$\mathbb{P}[f(X)=a]=\sum_{x:f(x)=a}\mathbb{P}[X=x]$rather than just$\mathbb{P}[X=f^{-1}(a)]$. This never complicates the maths though, as you will always write$\mathbb{E}[f(X)]=\sum_{x}f(x)\mathbb{P}[X=x]$, and injectivity or lack thereof of$f$rarely factors into evaluating this sum. • Don't you mean to say that the only significance of injectivity is that given$a$,$f(x) = a$holds for exactly one$x\$? – Jacob Errington Feb 22 '17 at 0:32
• Ah, what I meant to say was "lack of injectivity". Thank you, will edit now. – Pepe Silvia Feb 22 '17 at 1:03