Question

Question 2 Let X Pareto(r, 8 = 1) which has pdf: f(x) = 1 , 1 >1 and r > 1 (a) Given a random sample of size n from X show that the mle for r is: r* = 1/7 where Y = SEY and Y = log X (b) Let Y = log X Use the mgf technique (with t <r) to show that: Y Exp(1 = r) [ HINT: My(t) = Eletbox] = E[X“) = * **f(x)dt = ... ] (c) Use the CLT to show an approximate 95% for r is: 95% for r is: (1 + 1 ) | HINT: 0.95 P(-1.96 < < 1.96) with M, o from Y in part (b) ]

Answer 1

(a)

We have sample
X1, X2.... In
from the density $f(x)=\frac{r}{x^{r+1}},\quad x,r>1$

The likelihood can be written as

$L(r,x_1,x_2,...,x_n)=\prod_{i=1}^nf(x_i)\\\\ =\prod_{i=1}^n\frac{r}{x_i^{r+1}}=\frac{r^n}{\prod x_i^{r+1}}\\ \\ \log L=n\log r-(r+1)\sum_{i=1}^n \log x_i$

Since log is a monotone function, Maximising L is equivalent to maximising log L

Now,

$\frac{d}{dr}\log L=\frac{n}{r}-\sum_{i=1}^n \log x_i\\ \frac{d}{dr}\log L=0\\\\ \Rightarrow \frac{n}{r}-\sum_{i=1}^n \log x_i=0\\ or\\ r=\frac{n}{\sum_{i=1}^n \log x_i}\\ \frac{d^2}{dr^2}\log L=\frac{-n}{r^2}<0$

Hence log L and equivalently L is maximum when

$r=\frac{n}{\sum_{i=1}^n \log x_i}\\$

So the mle of r is given by

$r^*=\frac{n}{\sum_{i=1}^n \log X_i}=\frac{n}{\sum_{i=1}^n Y_i}=\frac{1}{\bar Y}$ where

Y; = log X

________________________

(b)

Let $Y=\log X$ where X has the pdf $f(x)=\frac{r}{x^{r+1}}\quad x,r>1$

The moment generating function of Y is given by

My(t) = E(Y) = E(et log x) = E(elos ) = E(X") = 5 ' f(s)die =>[0-0-1 when r>

This is the moment generating function of exponential distribution with mean and since moment generating function unique Y has exponential distribution.

Hence $Y=\log X\sim Exp(\lambda=r)$ . For this exponential distribution mean is $\mu=\frac{1}{r}$ and variance is $\sigma^2=\frac{1}{r^2}$

________________

(c)

Let
X1, X2.... In
be random sample of size n from

Since $Y=\log X\sim Exp(\lambda=r)$

$Y_i=\log X_i\sim Exp(\lambda=r)$and are independent with $\mu=E(Y_i)=\frac{1}{r},\quad \sigma^2=Var(Y_i)=\frac{1}{r^2}$

Hence by Central Limit theorem

$Z=\sqrt{n}\frac{\bar Y-\mu}{\sigma}\sim N(0,1) \quad asymptotically$

That is

$Z=\sqrt{n}\frac{\bar Y-1/r}{1/r}\sim N(0,1) \quad asymptotically$

Since from standard normal table as we have $0.95=P(-1.96<Z<1.96)$

$0.95=P(-1.96<Z<1.96)\approx P(-1.96<\sqrt{n}\frac{\bar Y-\frac{1}{r}}{\frac{1}{r}}<1.96)$

Simplifying the probability on right hand side

$P(-1.96<\sqrt{n}\frac{\bar Y-\frac{1}{r}}{\frac{1}{r}}<1.96)=P(\frac{-1.96}{\sqrt{n}}<\frac{\bar Y-\frac{1}{r}}{\frac{1}{r}}<\frac{1.96}{\sqrt{n}})\\\\= P(\frac{-1.96}{\sqrt{n}}<\frac{\frac{r\bar Y-1}{r}}{\frac{1}{r}}<\frac{1.96}{\sqrt{n}})\\ \\ =P(\frac{-1.96}{\sqrt{n}}<r\bar Y-1<\frac{1.96}{\sqrt{n}})\\\\ =P(1-\frac{1.96}{\sqrt{n}}<r\bar Y<1+\frac{1.96}{\sqrt{n}})\\ \\= P(\frac{1}{\bar Y}(1-\frac{1.96}{\sqrt{n}})<r<\frac{1}{\bar Y}(1+\frac{1.96}{\sqrt{n}}))$

Hence we get $0.95\approx P(\frac{1}{\bar Y}(1-\frac{-1.96}{\sqrt{n}})<r<\frac{1}{\bar Y}(1+\frac{1.96}{\sqrt{n}}))$ or

$\left( \frac{1}{\bar Y}(1-\frac{1.96}{\sqrt{n}}),\quad \frac{1}{\bar Y}(1+\frac{1.96}{\sqrt{n}})\right )$ is an approximate 95% confidence interval for r.

That is approximate 95% confidence interval for r is $\frac{1}{\bar Y}\left( 1\pm \frac{1.96}{\sqrt{n}}\right )$