## Problem

Let $X \sim \text{Geom}(p)$.

Find the distribution of $Y = X/(1+X)$.

What is $E(X)$ and $V(X)$?

## Solution

### Finding the distribution for $Y$

This is a pretty straightforward transformation of random variable problem, with $r(x) = /frac{x}{1+x}$.

$F_Y(y) = P(Y \leq y) = P(r(X) \leq y) = P(X \leq r^{-1}(y)) = F_X(r^{-1}(y))$.

$F_X(x)$, the cumulative distribution function for a geometric distribution, is known to be $1-(1-p)^x$.

Let's find $r^-1(x)$ by swapping x and y in $y=x/(1+x)$ and solving for y:

$x = y(1+y) \\ x + xy = y \\ y - xy = x \\ y(1 - x) = x \\ y = r^{-1}(x) = \frac{x}{1-x}$

Finally, $F_Y(y) = F_X(r^{-1}(y)) = 1-(1-p)^{\frac{y}{1-y}}$.

Note that as $x$ goes from $1, 2, ... \infty$, $y$ ranges from $\frac{1}{2}, \frac{2}{3}, ... 1$.

### Finding $E(X)$

This problem is mostly plug and chug, using some tricks in evaluating infinite series.

$E(X) = \sum_{x=1}^{\infty} x f_X(x) = \sum_{x=1}^{\infty} xp(1-p)^{x-1}$

We know the formula for a geometric series: $\sum_{k=a}^{b} r^k = \frac{r^a - r^{b+1}}{1-r}$

The trick is to turn our problem of calculating $\sum_{x=1}^{\infty} xp(1-p)^{x-1}$ into this problem. First, let's set $\alpha = 1-p$ and work in terms of $k$ instead of $x$ to make this look more familiar:

$\sum_{x=1}^{\infty} xp(1-p)^{x-1} = p \sum_{k=1}^{\infty} k \alpha^{k-1} = p \sum_{k=1}^{\infty} \frac{d}{d \alpha} \alpha^k$ = $p \frac{d}{d \alpha} \sum_{k=1}^{\infty} \alpha^k$ = $p \frac{d}{d \alpha} \frac{\alpha}{1-\alpha}$

Applying the quotient rule, $\frac{d}{d \alpha} \frac{\alpha}{1-\alpha} = \frac{1}{(1-\alpha)^2}$

Finally leaving us with $E(X) = \frac{p}{(1-\alpha)^2} = \frac{p}{(1-(1-p))^2} = \frac{p}{p^2} = \frac{1}{p}$.

### Finding $V(X)$

This is a similar plug and chug operation; given that

$V(X) = E(X^2) - (E(X))^2$, and that we've already calculated $E(X)$, this is a matter of calculating $E(X^2)$.

I couldn't bring myself to work it out in as quite as much detail, but here are some steps:

$E(X^2) = \sum_{k=1}^{\infty} k^2 * p(1-p)^{k-1} = \frac{p}{1-p} \sum_{k=1}^{\infty} k^2(1-p)^k = \frac{p}{1-p} \frac{(1-p)(1 + (1-p)}{(1-(1-p))^3} = \frac{2-p}{p^2}.$

So

$V(X) = E(X^2) - (E(X))^2 = \frac{2-p}{p^2} - \frac{1}{p^2} = \frac{1-p}{p^2}$

## Sources

- 36-700 hw4 problem 5