Question

Let X1, X2, ..., Xn denote a random sample of size n from a population whose density fucntion is given by 383x-4 f S x f(x) = 0 elsewhere where ß > 0 is unknown. Consider the estimator ß = min(X1, X2, ...,Xn). Derive the bias of the estimator ß.

Answer 1

Answer:-

Given That:-

Let denote a random sample of size n from a population whose density function is given by

$f(x)=\left\{\begin{matrix} 3\beta^3x^{-4} &\beta \leq x \\ 0& else\ where \end{matrix}\right.$

CDF of X is

Fx(2) = 3834-4 •dt 8

$F_X(x)=3\beta ^3\int_{\beta }^{x}t^{-4}dt$

$F_X(x)=3\beta ^3[\frac{t^{-3}}{3}]_{\beta }^{x}$

$F_X(x)=\beta ^3[\beta ^{-3}-x^{-3}]$

Fx(x) = 1-()

Therefore CDF of X is

Fx() = 0 <B 1-)" x > B

$\hat{\beta }=X_{(1)}=T$

$F_T(t)=P|T\leq t|$

$=P|X_{(1)}\leq t|$

$=1-\begin{Bmatrix} P|X_{(1)}> t| \end{Bmatrix}^n$

$=1-\begin{Bmatrix} 1-F_X(t) \end{Bmatrix}^n$

$=1-(\frac{\beta }{t})^{3n}$

$f_T(t)=\frac{3x\beta ^{(n*3)}}{t^{3n+1}}x\geq \beta$

pdf of X

$E(\hat{\beta })=E[X_{(1)}]=\int_{\beta }^{\infty }\frac{x3n\beta ^{(3n)}}{x^{3n+1}}dx$

$=3n\beta ^{3n}\int_{\beta }^{\infty }\frac{1}{x^{3n}}dx$

$=3n\beta ^{3n}[\frac{x^{-3n+1}}{-3n+1}]_{\beta }^{\infty }$

$=\frac{{3n}}{-3n+1}\beta ^{3n}[0-\beta ^{-3n+1}]$

$=\frac{{3n}}{3n-1}\beta ^{3n}\beta ^{-3n+1}$

$=\frac{{3n}}{3n-1}\beta$

Therefore,

Bias of the estimator $\hat{\beta }=X_{(1)}$ is $E(X_{(1)})-\beta$

$=\frac{{3n}}{3n-1}\beta -\beta$

$=\beta (\frac{{3n}}{3n-1}-1)$

$=\beta (\frac{{1}}{3n-1})$

$=\frac{{\beta }}{3n-1}$