Minimum of two geometric distributions

Question

Let $X,Y$ be independent discrete random variables, both having the geometric distribution, $X$ with parameter $p$ and $Y$ with parameter $r$. How do I show that $U=\min \{X,Y\}$ has the geometric distribution with parameter $p+r-pr$?

Answer 1

Pick $n \in \mathbb{N}$. Then, $\mathbf{P}(U \geq n) = \mathbf{P}(X \geq n, Y \geq n)$ by the definition of $U$. By independance and by the properties of geometric distributions, this is equal to $\mathbf{P}(X \geq n) \mathbf{P}(Y \geq n) = (1 - p)^n(1-r)^n = (1 - r - p + pr)^n$.

Define $q = r+p - pr$ so that $\mathbf{P}(U \geq n) = (1 - q)^n$. You should easily be able to prove now that $U$ follows a geometric law with parameter $q$ (try to calculate $\mathbf{P}(U = n)$ for example).

Note. To me, a geometric r.v. is a r.v. such that $\mathbf{P}(X = k) = (1-p)^k p$ for every $k \in \mathbb{N}$.