Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How can I solve this problem? Is there any formula for this problem

Find the distribution of the random variable $Y$ if

  1. $Y=\min(X_1,X_2)$
  2. $Y=X_1+X_2$

where $X_1$ and $X_2$ are independent exponential random variables with means $\frac 1{\lambda_1}$ and $\frac 1{\lambda_2}$, respectively.

share|improve this question
1  
Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. If this is homework, please add the homework tag; people will still help, so don't worry. –  Did Nov 3 '12 at 12:25

1 Answer 1

Here are two hints that should be sufficient in order to get the problems done:

Problem 1) look at the complementary cumulative distribution, i.e. $$P(\min(X_1, X_2) > x) = P(X_1 > x)\cdot P(X_2 > x).$$ Substitute what you know about $X_1$ and $X_2$ and you should end up with something familiar.

Problem 2) Use moment-generating functions. Recall that the moment-generating function of a exponential random variable with $t<\lambda$ is $$M_X(t) = \frac{\lambda}{\lambda-t}.$$

share|improve this answer
    
Can you give me any examples link about them? –  Horizon Nov 3 '12 at 13:15
    
@Horizon: what do you mean? An example on how to do the proof? What do you know about the cumulative distribution function? Just multiply the two terms in the expression above and the result is immediate. You will find that the minimum is exponentially distributed with parameter $\lambda = \lambda_1 + \lambda_2$. –  M.B. Nov 3 '12 at 13:24

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.