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) =

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Answer #1

Answer:-

Given That:-

Let X_1, X_2, ............,X_n 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

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

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}]

F_X(x)=\1-(\frac{\beta }{x})^3

Therefore CDF of X is

F_X(x)=\left\{\begin{matrix} 0 &x \leq \beta \\ 1-(\frac{\beta }{x})^x& x \geq \beta\end{matrix}\right.

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

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

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

=1-P|X_{(1)}> 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}

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