Question

Mary is a waitress in a city centre restaurant. She receives tips from customers at an average rate of λ per hour. During a particular eight a (2 marks) Let X be the number of tips Mary receives in an 8 hour period. If tips are received according to a Poisson process with rate λ tips per hour, state the distribution of X, with parameters b (5 marks) Write down the likelihood function, L(X; 5) for this problem. Remember to state the range of values of λ for which the likelihood is defined. Hence find the maximumlikelihood estimate of λ. You should make reference to the graph of the likelihood function, shown in Figure 1. Show all working in your answer. c (2 marks) In general, if Mary receives X tips in an 8 hour period, what is the maximum likelihood estimator for λ in terms of X? d (4 marks) Let λ be the maximum likelihood estimator for λ. Find expressions for the variance and estimated variance of λ: Var(A) and Var(X). Use the result that you have obtained in part (b) for computing Var(A)

Answer 1

The distribution of number of tips Mary receives per hour is Poisson with parameter $\lambda$ .

Poisson PMF is

.. λ (A)一:n = 0,1,2,3,...-

a) The Poisson rate for 8 hours is $8\lambda$ . The distribution is

X Poisson (8A)

b) The likelihood function,

${\color{Blue} L\left (\lambda ;5\right )=e^{-8\lambda }\frac{\left (8\lambda \right ) ^5}{5!};\lambda >0}$

The log likelihood is

$l\left (\lambda ;5\right )=\ln L\left (\lambda ;5\right )\\ l\left (\lambda ;5\right )=-8\lambda +5\ln \left (8\lambda \right )+\ln 5!$

Differentiating and equating to 0,

0 λ) λλ し8

c) In general the loglikelihood is

$l\left (\lambda ;X\right )=\ln L\left (\lambda ;X\right )\\ l\left (\lambda ;X\right )=-8\lambda +X\ln \left (8\lambda \right )+\ln X!$\

Differentiating and equating to 0,

$\frac{\partial }{\partial \lambda }l\left (\lambda ;X\right )=0\\ \ln L\left (\lambda ;X\right )\\ -8+\frac{X}{\lambda }=0\\ {\color{Blue} \widehat{\lambda }=\frac{X}{8}}$

d) The variance of the MLE is

Var(X) _ Var(X Var (λ)= 82 var (5) = 82 var(5)=8

The estimated variance using the estimate in part (b) is

5/8 Var (λ) 8 5 var (5) .. 64