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Three fishermen are fishing, we model the fishing as a Poisson Process of rate $2.5$ fish/hour. The fishermen leave only when each of them them has caught at least 3 fish, we call this leaving time $T$. Calculate $\mathbb{E}[T]$.

My attempt:

I started out calculating the distribution $F_T$:

$F_T(t)=\mathbb{P}(X_1(t)\ge 3, X_2(t)\ge 3, X_3(t)\ge 3)=(\mathbb{P}(X_1(t)\ge 3))^3,$ where I have assumed independence of the Poisson processes corresponding to each fisherman $X_i$. Then I integrate $(1-F_T)$ to get the expectation.

Question: Is my reasoning correct? The numerical calculation yields an expected time of 3.56 hours. Also, is there any easier way to do this using arrival waiting or interarrival times?

Thanks in advance.

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  • $\begingroup$ How appropriate, modeling fishing as a Poisson Process. $\endgroup$ – Gerry Myerson May 4 '15 at 12:51
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    $\begingroup$ @GerryMyerson There even are many papers titled "Fishing in Poisson streams", e.g. this one $\endgroup$ – Dilip Sarwate May 4 '15 at 13:03
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I think you started off OK in finding $F_T(t)$. To find $E(T)$, you need to differentiate $F_T(t)$ to get $f_T(t)$ and use that to obtain $E(T)$.

\begin{eqnarray*} f_T(t) &=& F_T^{'}(t) \\ &=& \dfrac{d}{dt} \left( 1-e^{-\lambda t} - \lambda te^{-\lambda t} - \dfrac{(\lambda t)^2}{2}e^{-\lambda t} \right)^3 \\ &=& \dfrac{3}{2} \lambda^3 t^2 e^{-\lambda t} \left( 1-e^{-\lambda t} - \lambda te^{-\lambda t} - \dfrac{(\lambda t)^2}{2}e^{-\lambda t} \right)^2. \end{eqnarray*}

$$\\$$

\begin{eqnarray*} E(T) &=& \int_{t=0}^{\infty}{tf_T(t)\;dt} \\ &=& \int_{t=0}^{\infty}{\dfrac{3}{2} \lambda^3 t^3 e^{-\lambda t} \left( 1-e^{-\lambda t} - \lambda te^{-\lambda t} - \dfrac{(\lambda t)^2}{2}e^{-\lambda t} \right)^2\;dt} \\ &=& \bigg[ \dfrac{1}{3888\lambda} e^{-3\lambda t}\left[ -5832 e^{2\lambda t} (\lambda^3t^3 + 3\lambda^2t^2 + 6\lambda t + 6) + 729e^{\lambda t} (4\lambda^5t^5 + 18\lambda^4t^4 + 44\lambda^3t^3 + 66\lambda^2t^2 + 66\lambda t + 33) -2(243\lambda^7t^7 + 1539\lambda^6t^6 + 5022\lambda^5t^5 + 10314\lambda^4t^4 + 14724\lambda^3t^3 + 14724\lambda^2t^2 + 9816\lambda t + 3272) \right] \bigg]_0^{\infty} \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\text{(using WolframAlpha)} \\ &=& \dfrac{-1}{3888\lambda} \left( -5832\times 6 + 729 \times 33 - 2\times 3272 \right) \\ &=& \dfrac{17579}{9720} \\ &\approx& 1.81 \text{ hours}. \\ \end{eqnarray*}

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