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.
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.