The distribution of the minimum of two independent geometric random variables

Question

Let $X$ and $Y$ be independent geometric random variables. What is the distribution of $Z=\min(X,Y)$?

The probability mass functions are $\operatorname{Pr}(X=k)=(1-p)^{k-1}p$ and $\operatorname{Pr}(Y=k)=(1-q)^{k-1}q$. And the event $(Z=k)$ is the union of

• $(X=k)$ and $(Y\ge k)$
• $(Y=k)$ and $(X\ge k)$

But these are not disjoint. Is there a better way to approach this problem?

Answer 1

Let $X\sim\mathcal{Geo}(p), Y\sim\mathcal{Geo}(q), X\perp Y$

$\begin{align} \Pr(X\geq k) & = (1-p)^{k-1} & \impliedby X\sim \mathcal{Geo}(p) \tag{1} \\[2ex] \Pr(Y\geq k) & = (1-q)^{k-1}& \impliedby Y\sim \mathcal{Geo}(q)\tag{2} \\[2ex] \Pr(\min(X,Y)\geq k) & = \Pr(X\geq k,Y\geq k) \\[1ex] & = \Pr(X\geq k)\Pr(Y\geq k) & \impliedby X\perp Y \\[1ex] & = (1-p)^{k-1}(1-q)^{k-1} & \impliedby (1)\wedge (2) \tag{3} \\[2ex] \Pr(\min(X,Y)= k) & = \Pr(\min(X,Y)\geq k) - \Pr(\min(X,Y)\geq k+1) \\[1ex] & = (1-p)^{k-1}(1-q)^{k-1} - (1-p)^{k}(1-q)^{k} \\[1ex] & = (p+q-pq)((1-p)(1-q))^{k-1} \\[1ex] & = (p+q-pq)(1-(p+q-pq))^{k-1} \end{align}$

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Another approach.

$X$ is the number of trials until a success with trial probability $p$, and $Y$ is the number of trials until a success with trial probability $q$, the $\min(X,Y)$ is the number of trials until either success; so it is geometric with trial probability $p+q-pq$ (the probability of the union).

Then $\min(X,Y) \sim\mathcal{Geo}(p+q-pq)$

Answer 2

Let the parameters of the two geometrics be $\alpha$ and $\beta$. So these are the probabilities of "success," and the geometrics give the number of trials until the first success. Let $Z=\min(X,Y)$.We have $Z\ge z$ if and only if $X\ge z$ and $Y\ge z$.

The probability that $X$ is $\ge z$ is the probability of $z-1$ "failures" in a row. This probability is $(1-\alpha)^{z-1}$. Similarly, $\Pr(Y\ge z)=(1-\beta)^{z-1}$.

It follows that $\Pr(Z\ge z)=((1-\alpha)(1-\beta))^{z-1}$.

Thus $Z$ has geometric distribution. For $$\Pr(Z=z)=\Pr(Z\ge z)-\Pr(Z\ge z+1)=((1-\alpha)(1-\beta))^{z-1}-((1-\alpha)(1-\beta))^z.$$ This simplifies to $p(1-p)^{z-1}$, where $p=1-(1-\alpha)(1-\beta)=\alpha+\beta-\alpha\beta$.