# If $X,Y$ are independent and geometric, then $Z=\min(X,Y)$ is also geometric

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 $\lambda\mu$)

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

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

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

$P(X>z-1)P(Y>z-1)-P(X>z)P(Y>z)$

Thus;

$\displaystyle\bigg(\sum_{j=z}^{\infty}(1-\lambda)^j\lambda\bigg)\bigg(\sum_{j=z}^{\infty}(1-\mu)^j\mu\bigg)-\bigg(\sum_{j=z+1}^{\infty}(1-\lambda)^j\lambda\bigg)\bigg(\sum_{j=z+1}^{\infty}(1-\mu)^j\mu\bigg)$

$=\displaystyle\lambda\mu\frac{(1-\lambda)^z}{1-(1-\lambda)}\frac{(1-\mu)^z}{1-(1-\mu)}-\lambda\mu\frac{(1-\lambda)^{z+1}}{1-(1-\lambda)}\frac{(1-\mu)^{z+1}}{1-(1-\mu)}$

$=(1-\lambda)^z(1-\mu)^z-(1-\lambda)^{z+1}(1-\mu)^{z+1}$

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 ?

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 $$\Pr(\min(X,Y)\gt t)=\Pr((X\gt t)\cap (Y\gt t))=\lambda^t\mu^t=(\lambda\mu)^t.$$
• but Is $\Pr(X>t)$ not $\lambda^{t+1}$, what is your $X$, number of failure before reaching a success ? – OBDA Jun 24 '14 at 11:33
• 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 $\lambda^{t-1}$. – André Nicolas Jun 24 '14 at 11:47