Probability & Statistics Ch.2 Section 3 Question 8

Solutions for “Probability and Statistics: The Science of Uncertainty” (Second Edition). These are solutions I have come up with; I offer no guarantee of accuracy.

Question

Let $W \sim \text{Poisson}(\lambda)$ . For what value of $\lambda$ is $P(W = 11)$ maximized?

Solution

From Example 2.3.6 in the textbook, we know:

p_Y(y) = P(Y=y) = \frac{\lambda^y}{y!}e^{-\lambda}

We plug in w = 11:

P(W=11) = \frac{\lambda^{11}}{11!}e^{-\lambda}

We differentiate to locate the peak. Start by factoring the constant:

P(W=11) = \frac{1}{11!}\lambda^{11}e^{-\lambda}

Apply the product rule:

\frac{d}{d\lambda}P(W=11) = \frac{1}{11!}\left(11\lambda^{10}e^{-\lambda} + \lambda^{11}(-e^{-\lambda})\right)

Factor out $e^{-\lambda}$ :

= \frac{e^{-\lambda}}{11!}(11\lambda^{10} - \lambda^{11})

Factor out $\lambda^{10}$ :

= \frac{e^{-\lambda}}{11!}\lambda^{10}(11 - \lambda)

Set our factors equal to zero:

$e^{-\lambda} = 0$ has no real solutions
$\lambda^{10} = 0 \Rightarrow \lambda = 0$
$11 - \lambda = 0 \Rightarrow \lambda = 11$

By the definition of Poisson distribution (Example 2.3.6), $\lambda > 0$ , so $\lambda = 0$ is not valid.

Therefore: $\lambda = 11$