# Solved: The number of breakdowns per week for a type of minicomputer is a random variable $Y$ with a Poisson distribution and me

#### ByDr. Raju Chaudhari

Oct 15, 2020

The number of breakdowns per week for a type of minicomputer is a random variable $Y$ with a Poisson distribution and mean $\lambda$. A random sample $Y_1, Y_2, \cdots, Y_n$ of observations on the weekly number of breakdowns is available.

(a) Suggest an unbiased estimator for $\lambda$.
(b) The weekly cost of repairing these breakdowns is $C = 3Y + Y^2$. Show that $E(C) = 4\lambda + \lambda^2$.
(c) Find a function of $Y_1, Y_2,\cdots, Y_n$ that is an unbiased estimator of $E(C)$.

### Solution

The number of breakdowns per week for a type of minicomputer is a random variable $Y$ with a Poisson distribution and mean $\lambda$. That is $Y\sim P(\lambda)$.

(a) $E(Y) =\lambda$. So for a random sample of size $n$ $E(\overline{Y}) = \lambda$. Thus an unbiased estimator for $\lambda$ is $\overline{Y}$.

(b) The weekly cost of repairing these breakdowns is $C = 3Y + Y^2$.

 \begin{aligned} E(C) &= E(3Y+Y^2)\\ &= 3E(Y) + E(Y^2)\\ &= 3\lambda + [V(Y)+E(Y)^2]\\ &=3\lambda + [\lambda +\lambda^2]\\ &= 4\lambda + \lambda^2. \end{aligned}
(c) As $E(\overline{Y}) =\lambda$ and $V(\overline{Y}) = \frac{\lambda}{n}$, we have

 \begin{aligned} E(\overline{Y}^2) &= V(\overline{Y})+[E(\overline{Y})]^2\\ &=\frac{\lambda}{n}+\lambda^2 \end{aligned}

Let the unbiased estimator of $E(C)$ be $a\overline{Y}+b\overline{Y}^2$.

 \begin{aligned} E(a\overline{Y}+b\overline{Y}^2)&=aE(\overline{Y})+bE(\overline{Y}^2)\\ &= a\lambda + b\big(\frac{\lambda}{n}+\lambda^2\big)\\ &= \big(a+\frac{b}{n}\big)\lambda + \lambda^2 \end{aligned}

Comparing RHS with $4\lambda+\lambda^2$, we get $b=1$ and $a+\frac{b}{n}=4\Rightarrow a=4-\frac{1}{n}$.

Thus the unbiased estimator of $E(C)$ is $\big(4-\frac{1}{n}\big)\overline{Y} + \overline{Y}^2$.