# Thread: Help: 2 exponential problem

1. ## Help: 2 exponential problem

Interesting title, huh?

I am having trouble solving this problem. Assume X and Y are independent exponential random variables with means 1/x and 1/y, respectively. If Z=min(X,Y): Is Z exponentially distributed as well (if so, how do you know)? What is the expectation of Z? What is the probability that x < y?

Lastly, with the information from above, show how a M/M/1 queue could be represented as a Markov chain that is continuous-time with transition rates Qn,n+1=L and Qn,n-1=U, n=0,1,2,... M/M/1 queue=Arrival is Poisson/Service is Exponential/1 server (with an infinite buffer size).

2. Do X and Y have the same mean?
If so, then the min of these two is another EXP rv.

Let the density of X be

$\displaystyle {1\over \beta}e^{-x/\beta}$ where x>0

and $\displaystyle P(X>a)=\int_a^{\infty}{1\over \beta}e^{-x/\beta}=e^{-a/\beta}$

Let Z be the min of X and Y.

Then $\displaystyle P(Z\le a)=1-P(Z>a)=1-P(X>a)P(Y>a)$

And I'll stop here, not because it's 2:12 am,
but I need to know if you have the same parameter for
both X and Y or not.

3. Hey thanks for replying. X and Y don't have the same means. The means are 1/x and 1/y, respectively.

4. I'm not sure how to compute the Markov chain transition rates either. Is there a base equation you work with and go from there?