David is fishing together with two friends. Each of them catches fish independently of the others according to a Possion process with rate 2 per hour. What is the expected time until everyone has caught at least one fish?
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Let $X_{i}$ be the time until the first catch for $i=1,2,3$. Hence, $ X_{i} \sim \exp(2), \,\, i=1,2,3$. You are interested in the time when everyone of them caught a fish, namely, in the distribution of $X_{(3)} = \max\{ X_{1}, X_{2}, X_{3}\}$. You can show that $P(X_{(3)}\le x) = [F_{X}(x)]^3 = (1-e^{-2x})^3$. Now to find $E[X_{(3)}]$ you can use $ \int tf_{X_{(3)}}(t)dt$ or $E[X_{(3)}] = \int(1-F(t))dt $ for non-negative random variables, or alternatively, utilize some properties of the exponential distribution, i.e., \begin{align} E[X_{(3)}] &= E[min\{X_1, X_2, X_3\}] + A\\ &=1/6 + E[A], \end{align} where the first summand stems from the exponential distribution of the minimum and $A$ is the minimum among the two others who left, hence using the memoryless property of the exponential distribution you get $$ E[X_{(3)}] = 1/6 + 1/4 + E[X_1]= 1/6 + 1/4 +1/2. $$
V. Vancak
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Let $T_1, T_2, T_3$ be the time of catching the first fish by person 1, 2 and 3 respectively. Let $T = max(T_1, T_2, T_3)$. Then
$P(T\le t) = P(T_1\le t) P(T_2\le t) P(T_3\le t) = (1-e^{-\lambda t})^3$,
from this we can drive the pdf of $T$ and the expectation.
Jay.H
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