Question

Exam.(Jan 15) Circle out your Class Mon&Wed or Mon.Evening /uestion 7. Suppose that Xi, distribution with the following p.d.f. Aio is a simple random sample from the , f(x8) 0x01,for0sxsi 0, otherwise where 0 0, a random sample of size 10 yields data 0.92 0.79 0.9 0.65 0.86 0.47 0.73 0.97 0.94 0.77 1) (6 points) Get the moment estimator of 0, and compute the estimate for this data;

Answer 1

here E(X)=$\int_{0}^{1}$ x f(x) dx =$\int_{0}^{1}$x$\theta$x^$\theta$-1 dx =$\int_{0}^{1}$$\theta$x^$\theta$ dx =x^$\theta$+1/($\theta$+1)|¹₀ =$\theta$/($\theta$+1) =Xbar

($\theta$+1)/$\theta$ =1/Xbar

1/$\theta$ =1/Xbar-1

moment estimate of $\theta$ =Xbar/(1-Xbar)

from above Xbar =(0.92+0.79+0.9+0.65+0.86+0.47+0.73+0.97+0.94+0.77)/10=0.8

therefore estimate for $\theta$ =0.8/(1-0.8)=4

2)

here L($\theta$ )=f(x1)*f(x2)*...f(xn) =$\theta$ⁿ*(x1^$\theta$-1+x2^$\theta$-1+x3^$\theta$-1+...xn^$\theta$-1)

therefore Ln(L($\theta$)) =n*ln($\theta$)+($\theta$-1)*$\sum$ln(Xi)

(d/d$\theta$)Ln(L($\theta$)) =n/$\theta$+$\sum$ln(Xi)

putting above equal to 0;

maximum likelihood estimate of $\theta$ =-n/($\sum$ln(Xi) )

from above data n=10

and $\sum$ln(Xi) =ln(0.92)+ln(0.79)+..+ln(0.77)=-2.4295

hence estimate of $\theta$ =-10/(-2.4295)=4.11607