Find the probability mass function of $V=\min(X,Y)$ Let $X$ and $Y$ be independent $\mathrm {Geom}(p)$ random variables.
Find the probability mass function of $V=\min(X,Y).$
I don't understand what I am supposed to find the probability for. I know that $X\sim \mathrm{Geom}(p)$ for $p\in[0,1]$ if $\mathbb{P}(X=k)=(1-p)^{k-1}p$. But how would I use this to find a p.m.f?
 A: The probability that the minimum of X, Y is k is the probability that they're both k, X is k and Y is larger, or Y is k and X is larger. Fortunately, the latter two are the same because the distributions are the same.
$$\begin{split}P(V=k)&=P(X=k,Y>k)+P(Y=k,X>k)+P(X=k,Y=k)\\
&=2P(X=k,Y>k)+P(X=k,Y=k)\\
&=2\left((1-p)^{k-1}p\left[1-\sum_{i=0}^{k-1}(1-p)^ip\right]\right)+\left((1-p)^{k-1}p\right)^2\end{split}$$
Use $\sum_{k=0}^n r^k=\frac{1-r^{n+1}}{1-r}$ to get $\sum_{i=0}^{k-1}(1-p)^i=\frac{1-(1-p)^{k}}{p}$
$$\begin{split}2\left((1-p)^{k-1}p\left[1-\left(1-(1-p)^{k}\right)\right]\right)+\left((1-p)^{k-1}p\right)^2\\
=2(1-p)^{k-1}p(1-p)^{k}+\left((1-p)^{k-1}p\right)^2\\
=2(1-p)^{2k-1}p+(1-p)^{2k-2}p^2\\
=(1-p)^{2k-2}p(2-p)\end{split}$$
A: $$P(V=k)=P(X=k, Y \geq k)+P(Y=k, X \geq k)$$ $$=2P(X=k, Y \geq k)-P(X=k,Y=k)$$ $$= \sum\limits_{i=k}^{\infty}(1-p)^{k-1}p (1-p)^{i-1}p-(1-p)^{k-1}p (1-p)^{k-1}p.$$
Use the formula for a geometric sum to finish the calculation.
A: Hint: $P(V=k) = P(\min(X, Y)=k)$ is equal to $$P(X=k\cap Y>k)+P(X>k\cap Y=k)+P(X=k\cap Y=k)$$
A: 
I don't understand what I am supposed to find the probability for. I know that $X\sim \mathrm{Geom}(p)$ for $p\in[0,1]$ if $\mathbb{P}(X=k)=(1-p)^{k-1}p$. But how would I use this to find a p.m.f?

Such a random variable is the count of trials in a sequence of iid Bernoulli trials until a success occurs.   Such as, for example, the count of flips of a biased coin until the first head occurs.
If both $X,Y$ as identical and independent geometric random variables, then the minimum is the count of trials in two parallel sequences of iid Bernoulli trials until the first success occurs.   Such as, for example, the count of flips of two biased coins until at least one head occurs.
What kind of random variable is $\min(X,Y)$, and what parameter(s) will it have?

 $\min(X,Y)\sim\mathcal{Geo}_1( 1-(1-p)^2)$ $$\mathsf P(\min(X,Y){=}k)=(1-(1-p)^2)(1-p)^{2k-2}\mathbf 1_{k\in\Bbb N^+}$$

A: Start with:
$$\begin{align}P(V\geq k) &=P(X\geq k \land Y\geq k)\\&=P(X\geq k)\cdot P(Y\geq k)\end{align}$$
and
$$P(V=k)=P(V\geq k)-P(V\geq k+1)$$
So you just need to show $$P(X\geq k)=P(Y\geq k)=(1-p)^{k-1}.$$
Then you get:
$$\begin{align}P(V=k)&=(1-p)^{2k-2}-(1-p)^{2k}\\
&=(1-p)^{2k-2}p(2-p)\end{align}$$
So $V \sim\mathrm{Geom}(2p-p^2).$

You can see that last relationship to $\mathrm{Geom}(2p-p^2)$ by remembering the most common model of $X.$
If you have a coin with probability $p$ For heads, and $X$ is the number of tosses before you get your first heads, then $X\sim\mathrm{Geom}(p).$
Then $V$ can be seen as the number of tosses of a pair of such coins, until at least one of them comes up heads. But this is the same as one coin with the probability of a heads being $1-(1-p)^2=2p-p^2.$
More generally, let $V=\min(X_1,\dots,X_n),$ where the $X_i$ are independent and $X_i\sim \mathrm{Geom}(p_i).$
Then $V\sim \mathrm{Geom}\left(1-(1-p_1)(1-p_2)\cdots(1-p_n)\right).$
