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?

  • 1
    $\begingroup$ Of the two current answers, I think it is Graham Kemp/s that it is better to accept, because of the "Another approach" part. $\endgroup$ Nov 27 '14 at 6:24
  • $\begingroup$ indeeds it helps me to understand. @André Nicolas $\endgroup$
    – usengec
    Nov 27 '14 at 6:27
  • 1
    $\begingroup$ There are several posts of exactly the same question. Judging by the content, it's hard to decide which one should be deemed the original and the others duplicate. In chronological order: 90782, 845706, 1056296, 1169142, and 1207241. $\endgroup$ Mar 21 '18 at 1:17

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

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

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

  • $\begingroup$ I like the second approach, conceptual not computational. $\endgroup$ Nov 27 '14 at 4:57
  • $\begingroup$ @AndréNicolas Indeed. It would have been nice if it had occurred to me before wading through the computations, but that's hindsight for you. $\endgroup$ Nov 27 '14 at 5:06
  • $\begingroup$ Happens fairly often, the computed answer shouts that one is missing a better way. Unlike you I had earplugs on. $\endgroup$ Nov 27 '14 at 5:11
  • $\begingroup$ I liked this post a lot--the "physical" solution was immediately obvious to me but the algebraic solution stumped me. Never occurred to me to take the probability at a point as a difference of two inequalities. Thanks! $\endgroup$
    – Addem
    Mar 3 '19 at 21:32

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 $p=1-(1-\alpha)(1-\beta)=\alpha+\beta-\alpha\beta$.


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