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.

  • $\begingroup$ How appropriate, modeling fishing as a Poisson Process. $\endgroup$ – Gerry Myerson May 4 '15 at 12:51
  • 1
    $\begingroup$ @GerryMyerson There even are many papers titled "Fishing in Poisson streams", e.g. this one $\endgroup$ – Dilip Sarwate May 4 '15 at 13:03

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*}


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.