Let $X,Y$ be independent geometric random variables with parameters $\lambda$ and $\mu$. If $Z=\min(X,Y)$. Show that $Z$ is geometric and find its parameter. (Answer $\lambda\mu$)

$\displaystyle P(Z=z)=P(\min(X,Y)=z)=P(\min(X,Y)\le z)-P(\min(X,Y)\le z-1)$

$=(1-P(\min(X,Y)> z)-(1-P(\min(X,Y)> z-1))$

$=P(\min(X,Y)> z-1)-P(\min(X,Y)> z)$






but this is not true, because if I set $z=0$, it should give $\lambda\mu$, but my formula gives something else, where did I do a mistake ?

Thanks for your help.

  • $\begingroup$ 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, 1040620, 1056296, 1169142, and 1207241. $\endgroup$ – Lee David Chung Lin Mar 21 '18 at 1:11

If the answer is to be $\lambda\mu$, then the "parameter" being referred to is the probability of failure. This is in fact a common choice for the parameter of the geometric. You used as parameter the probability of success.

Your computations can be much shortened. For note that $$\Pr(\min(X,Y)\gt t)=\Pr((X\gt t)\cap (Y\gt t))=\lambda^t\mu^t=(\lambda\mu)^t.$$

  • $\begingroup$ but Is $\Pr(X>t)$ not $\lambda^{t+1}$, what is your $X$, number of failure before reaching a success ? $\endgroup$ – OBDA Jun 24 '14 at 11:33
  • 1
    $\begingroup$ It depends on the definition of what $X$ stands for, the number of failures until the first success, or the number of trials. If you look for example at Wikipedia, geometric distribution, you will see that this is a standard use of the term "geometric distribution." So is the other one, and the "trials" version is more common in elementary books. In the failures version, it would be$\lambda^t$, in the trials version it would be $\lambda^{t-1}$. $\endgroup$ – André Nicolas Jun 24 '14 at 11:47

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