suppose we have three independent Poisson random variables $X_1$ and $X_2$ and $X_3$ with the same $\lambda$. We want to have the expected time we need to wait so all of three of them be more than zero ($ X_1, X_2, X_3 > 0 $). What is the solution?

$T^k_x$ = the random variable of wait time to see the kth event on random variable X

P.S: I did this: $P(T^1_{X_1} > t) = e^{-\lambda t}$ and three of them are the same and independent so $P(T^1_{all} > t) = P(T^1_{X_1} > t) \times P(T^2_{X_1} > t) \times P(T^3_{X_1} > t) = e^{-\lambda t} \times e^{-\lambda t} \times e^{-\lambda t} = e^{-3\lambda t}$ but the problem is expected of this is $\frac{1}{3\lambda}$ wich is lower than $\frac{1}{\lambda}$ so it is wrong! the expected time to have all three > 0 must be bigger that just one > 0. What is my mistake?

thanks in advance.

  • $\begingroup$ Your mistake is that what you calculated is (correctly) the expression for "any of them more than zero". $\endgroup$ – Lee David Chung Lin Oct 20 '18 at 1:53
  • $\begingroup$ @LeeDavidChungLin Why "any of them?" I multiply them. that doesn't mean "all of them together?" $\endgroup$ – Peyman mohseni kiasari Oct 20 '18 at 2:05
  • $\begingroup$ @LeeDavidChungLin sorry I just ask how I aks my friends when I'm confused so much. I'll edit it. $\endgroup$ – Peyman mohseni kiasari Oct 20 '18 at 2:06
  • $\begingroup$ I'm a beginner so I could be wrong... if $X_1$ is a Poisson RVs then the wait time until the first event (that is $X_1$ is more than zero) is an exponential RV, call it $Y_1$, so you are looking for $E[\max(Y_1,Y_2,Y_3)]$ and you can leverage this answer math.stackexchange.com/questions/146973/… $\endgroup$ – HJ_beginner Oct 20 '18 at 2:14

Denote the random variable $T_i$ as the waiting time length for $X_i$ to "be more than zero".

\begin{align} &\phantom{{}={}}P(~ \text{all three more than zero before}~t~) \\ &= P(~ T_1 < t~~\&~~T_2 < t~~\&~~T_3 < t) \\ &= P(~ T_1 < t~) \cdot P(~T_2 < t~)\cdot P(~T_3 < t) \qquad \because X_i \quad \text{mutually independent}\\ &= \left( 1 - e^{-\lambda_1 t} \right)\left( 1 - e^{-\lambda_2 t} \right)\left( 1 - e^{-\lambda_3 t} \right) \\ &= \left( 1 - e^{-\lambda t} \right)^3 \end{align}

That is your desired probability. Note that this does NOT take the form of an exponential distribution.

The keyword you should be looking for is order statistics. Here you are dealing with the maximum of a set of iid exponential distribution, while your previous erroneous calculation happens to be the minimum.

  • $\begingroup$ Thank you I got my mistake! sorry, I'm new and I can't upvote you. Chinese are always genius, good genes :D $\endgroup$ – Peyman mohseni kiasari Oct 20 '18 at 2:23

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