The random variables $X_1$ and $X_2$ are i.i.d, with probability functions The random variables $X_1$ and $X_2$ are i.i.d, with probability functions \begin{equation}f(x)=
\begin{cases}
 (1−p)^{x−1}p;\qquad &x=1,2,...\\ 
  0; & \text{otherwise}
\end{cases}
\end{equation}
The random variable $Y$ is defined as the minimum of $X_1$ and $X_2$. Find the probability generating
function of $Y$.
Attempt:
By using the definition of generating functions we have
$$E(z^Y)=\sum_{x_1=1}^{\infty}\sum_{x_2=1}^{\infty}z^{(\min{x_1,x_2})}f(x_1)f(x_2)$$
However, I have no idea where I should go from here. should I make the substitution for $x_1$ and $x_2$? Then the summation is ridiculous. 
Also, can I assume that $(\min{x_1,x_2})=1$?
Please can someone point me in the right direction with this problem.
 A: For simplicity of notation, write $X = X_1$, $Y = X_2$, $W = \min(X,Y)$.  Then the direct calculation proceeds as follows:  $$\begin{align*} \operatorname{E}[t^W] &= p^2 \sum_{y=1}^\infty \sum_{x=1}^\infty t^{\min(x,y)} (1-p)^{x-1} (1-p)^{y-1} \\ 
&= p^2 t \sum_{y=0}^\infty \sum_{x=0}^\infty t^{\min(x,y)} (1-p)^x (1-p)^y \\ 
&= p^2 t \sum_{y=0}^\infty \left( (1-p)^y \sum_{x=0}^{y-1} t^x (1-p)^x + (1-p)^y t^y \sum_{x=y}^\infty (1-p)^x \right) \\ 
&= p^2 t \sum_{y=0}^\infty \left( (1-p)^y \frac{1 - (t(1-p))^y}{1-t(1-p)} + t^y \frac{(1-p)^{2y}}{1-(1-p)} \right) \\ 
&= p^2 t \left( \frac{1}{1-t(1-p)} \sum_{y=0}^\infty (1-p)^y - (t(1-p)^2)^y + \frac{1}{p} \sum_{y=0}^\infty (t(1-p)^2)^y \right) \\ 
&= p^2 t \left( \frac{1}{1-t(1-p)}\left(\frac{1}{1-(1-p)} - \frac{1}{1-t(1-p)^2}\right) + \frac{1}{p} \cdot \frac{1}{1-t(1-p)^2}\right) \\ 
&= \frac{p(2-p)t}{1-t(1-p)^2} \\ &= \frac{(1-(1-p)^2)t}{1-(1-p)^2t}. \end{align*}$$  As you can see, it is tedious and not particularly illuminating, but it is computationally feasible.  The more intuitive approach is to recognize that the minimum order statistic of $n$ IID geometric random variables is itself geometric (along the same lines as the exponential distribution), owing to the fact that both are memoryless:  specifically, if $$X_1, \ldots, X_n \overset{\text{iid}}{\sim} \operatorname{Geometric}(p),$$ then $$X_{(1)} \sim \operatorname{Geometric}(1-(1-p)^n),$$ a fact that I leave to the reader as an exercise.
A: Alicia tosses a coin that has probability $p$ of landing head, until she gets a head. Let $X$ be the number of tosses until the first head. Then $X$ has geometric distribution, and $\Pr(X=x)=(1-p)^{x-1}p$.
Beti also tosses a coin that has probability $p$ of landing head, until she gets a head. Let $Y$ be the number of tosses until the first head. Then $\Pr(Y=y)=(1-p)^{y-1}p$.
Let $Z=\min(X,Y)$. Then $Z=z$ if Alicia and Beti each get $z-1$ tails in a row, and then one or both of Alicia and Beti gets a head on the next toss.
The probability that Alicia and Beti each get $z-1$ tails in a row is $((1-p)^{2})^{z-1}$. Given that this has happened, the probability at least one of them gets a head is $1-(1-p)^2$, that is, $2p-p^2$. 
Thus $\Pr(Z=z)=((1-p)^{2})^{z-1}(2p-p^2)$. Now the probability generating function should be straightforward.
