# expected time for happening the first events of three independent Poisson random variables

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?

• Your mistake is that what you calculated is (correctly) the expression for "any of them more than zero". – Lee David Chung Lin Oct 20 '18 at 1:53
• @LeeDavidChungLin Why "any of them?" I multiply them. that doesn't mean "all of them together?" – Peyman mohseni kiasari Oct 20 '18 at 2:05
• @LeeDavidChungLin sorry I just ask how I aks my friends when I'm confused so much. I'll edit it. – Peyman mohseni kiasari Oct 20 '18 at 2:06
• 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/… – 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}