I am currently trying to figure out a problem and it is using notation that I have never seen before so I am pretty stuck, any suggestions would be greatly appreciated!

Let $X, Y, Z$ be independent exponential random variables with the same mean, $σ$. Find the value of $σ$ so that $\Pr[\max(X,Y,Z)>1]=0.05$.

Any help with how to solve this problem or leading down the right path would be awesome as I am not to sure where to even start!

EDIT: So I have followed what André Nicolas has said and got a $σ$ of $0.25$. However I am still not sure where he got this formula from: The probability they are all $\le a$ is $(1-e^{-a/\sigma})^3$.

I cant find anything like this is my notes or textbook, could any reference where he got this?

  • $\begingroup$ Suggested reading: meta.math.stackexchange.com/questions/17164/… $\endgroup$ – Did Nov 2 '14 at 18:07
  • $\begingroup$ One probability trick very often useful is converting the problem to a complementary probability: $Pr[\max(X,Y,Z) \gt 1] = 1 - Pr[\max(X,Y,Z) \le 1]$. $\endgroup$ – hardmath Nov 2 '14 at 19:20

A start: The probability that $\max(X,Y,Z)\gt 1$ is $1$ minus the probability they are all $\le 1$.

The probability they are all $\le a$ is $(1-e^{-a/\sigma})^3$.

  • $\begingroup$ So basically the max is used instead of putting >=? $\endgroup$ – John Wuang Nov 2 '14 at 18:11
  • $\begingroup$ I am using the fact that for any random variables $X_1,\dots,X_n$, we have $\max(X_1,\dots,X_n)\gt t$ precisely if he $X_i$ are not all $\le t$. Once we have translated the max stuff to more familiar language, we can attack the problem using the usual tools. We want $(1-e^{-1/\sigma})^3=0.95$. To solve for $\sigma$, first take the cube root of both sides, then do some more algebraic manipulation. $\endgroup$ – André Nicolas Nov 2 '14 at 18:16
  • $\begingroup$ Okay I did so and got an answer for σ, however what theorem is this because I cannot find it anywhere in my notes. $\endgroup$ – John Wuang Nov 2 '14 at 18:24
  • $\begingroup$ What theorem is what? An exponential $X$ with mean $\sigma$ has density function $\frac{1}{\sigma}e^{-x/\sigma}$, and then $\Pr(X\le a)$ is the integral from $0$ to $a$ of the density function. This integral turns out to be $1-e^{-a/\sigma}$. $\endgroup$ – André Nicolas Nov 2 '14 at 21:15

$$p(0\le X\le x)=1-e^{-\lambda X}$$
where $\lambda$ is $1/\text{mean}$. When $x=1$, the mean is $a$, and so the equation becomes $1-e^{-1/a}$.


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