Chapter 2 Section 3 Question 8
Question
Let $W$ ~ $Poisson(\lambda )$. For what value of $\lambda$ is $P(W = 11)$ maximized?
Answer
Author: Mohammad-Ali Bandzar| Date:Oct 15 2020
From Example 2.3.6 in the book we know: $$p\scriptsize{Y}\normalsize{(y)=P(Y=y)=\frac{\lambda^y}{y!}e^{-\lambda}}$$
we shall now plug in w=11 $$P(W=11)=\frac{\lambda^{11}}{11!}e^{-\lambda}$$ we can now differentiate the function to try and locate our peak
start by factoring constant $$P'(W=11)=\frac{1}{11!}\lambda^{11}*e^{-\lambda}$$ Apply the product rule and differentiate $$P'(W=11)=\frac{1}{11!}(11\lambda^{10}*e^{-\lambda}+\lambda^{11}*(-e^{-\lambda}))$$ we can now factor out $e^{-\lambda}$ $$P'(W=11)=\frac{e^{-\lambda}}{11!}(11\lambda^{10}-\lambda^{11})$$ we can now factor out $\lambda^{10}$ $$P'(W=11)=\frac{e^{-\lambda}}{11!}(\lambda^{10}(11-\lambda))$$ we can now set out factors equal to zero $$e^{-\lambda}=0$$ this has no real solutions $$\lambda^{10}=0$$ $$\lambda=0$$ or: $$11-\lambda=0$$ $$\lambda=11$$ By the definition of Poisson distribution given to us in page 46 of the textbook, Example 2.3.6 $\lambda > 0$ therefore we only have one solution $\lambda=11$.