Question

1. Suppose that diseased trees are distributed randomly and uniformly throughout a large forest with an average of λ per acre. Let X denote the number of diseased trees in a randomly chosen one-acre plot with range, 0,1,2,.. a) What distribution can we use to model X? Write down its probability mass function (b) Suppose that we observe the number of diseased trees on n randomly chosen one-acre parcels, X1, X2, ..., X The random variables Xi, X2, ...,X, can be assumed to be inde- pendent. Write down the JOINT probability mass function for X1,X2, .., Xn- Simplify this expression which is a function of λ and the X's (c) We are going to use the Method of Maximum Likelihood to estimate λ Write down the Likelihood function LA) (d) Write down the Log-likelihood function (A). (e) Write down the Score Function S(A) (f) Derive the maximum likelihood estimate of λ (g) Write down the Information Function I(A) (h) Use the second derivative test to show that your answer to part(f) is THE absolute maxi- mum (i) Suppose that the numbers of diseased trees observed in 10 randomly chosen one-acre parcels were: 5, 5, 9, 8, 4, 9, 10, 7, 8.4 Compute the maximum likelihood estimate of λ using this data. G) Suppose that the unit of measure was a three-acre plot, i.e. we found the same number of diseased trees in 10 randomly chosen three-acre plots, but A is still the mean number per one acre. What is the maximum likelihood estimate of λ now?

Answer 1

(a)

Poisson distribution can be used to model X.

Probability Mass Function of X :- $f_{X}(x)=P(X=x)=\frac{e^{-\lambda}\lambda^x}{x!}\;\;,\;\;\lambda>0\;\;,\;\;x=0,1,2,.....$

(b)

$\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!f_{X_{i}}(x_{i})=P(X_{i}=x_{i})=\frac{e^{-\lambda}\lambda^{x_{i}}}{x_{i}!}\;\;,\;\;\lambda>0\;\;,\;\;x_{i}=0,1,2,.....\;\;\;\;\forall \;\;i=1.2,....,n$

Joint Probability Mass Function of X₁, X₂, .... , X_n is given by :-

$\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!f_{X_{1},X_{2},...,X_{n}}(x_{1},x_{2},...,x_{n})=P(X_{1}=x_{1},X_{2}=x_{2},.....,X_{n}=x_{n})=\prod_{i=1}^{n}\frac{e^{-\lambda}\lambda^{x_{i}}}{x_{i}!} =\frac{e^{-n\lambda}\lambda^{\sum_{i=1}^{n}x_{i}}}{\prod_{i=1}^{n}x_{i}!}$

(c)

Likelihood function is given by :-    $L(\lambda)=\prod_{i=1}^{n}\frac{e^{-\lambda}\lambda^{x_{i}}}{x_{i}!} =\frac{e^{-n\lambda}\lambda^{\sum_{i=1}^{n}x_{i}}}{\prod_{i=1}^{n}x_{i}!}$

(d)

Log-likelihood function is given by :-

$\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!l(\lambda)=ln\;L(\lambda) =ln\left ( \frac{e^{-n\lambda}\lambda^{\sum_{i=1}^{n}x_{i}}}{\prod_{i=1}^{n}x_{i}!} \right ) =-n\lambda+(ln\;\lambda)\sum_{i=1}^{n}x_{i}-\sum_{i=1}^{n}(ln\;x_{i}!) =-n\lambda+(ln\;\lambda)\sum_{i=1}^{n}x_{i}-\sum_{i=1}^{n}\sum_{j=2}^{x_{i}}(ln\,j)$