Question

Suppose that\({{\bf{X}}_{\bf{1}}}{\bf{,}}{{\bf{X}}_{\bf{2}}}{\bf{,}}...{\bf{,}}{{\bf{X}}_{\bf{n}}}\)form a random sample from the uniform distribution on the interval\(\left( {{\bf{0,1}}} \right)\), and let\({\bf{W}}\)denote the range of the sample, as defined in Example 3.9.7. Also, let\({{\bf{g}}_{\bf{n}}}\left( {\bf{x}} \right)\)denote the p.d.f of the random

variable\({\bf{2n}}\left( {{\bf{1 - W}}} \right)\), and let\({\bf{g}}\left( {\bf{x}} \right)\)denote the p.d.f of the\({\chi ^{\bf{2}}}\)distribution with four degrees of freedom. Show that

\(\mathop {{\bf{lim}}}\limits_{{\bf{n}} \to \infty } {{\bf{g}}_{\bf{n}}}\left( {\bf{x}} \right){\bf{ = g}}\left( {\bf{x}} \right)\) for\({\bf{x > 0}}\).

Short answer

\(\mathop {\lim }\limits_{n \to \infty } {g_n}\left( x \right) = g\left( x \right)\) for \(x > 0\)

Step-by-step solution

Given information 

\({X_1},{X_2},...,{X_n}\) be a uniform random sample.

\(W\)denote the range of the sample and \({g_n}\left( x \right)\)denote the p.d.f of\(2n\left( {1 - W} \right)\).

\(g\left( x \right)\) denote the p.d.f of the \({\chi ^2}\)distribution with degrees of freedom 4.

Determine the p.d.f of \({\bf{W}}\) 

The p.d.f of \(W\)is

\({h_1}\left( w \right) = n\left( {n - 1} \right){w^{n - 2}}\left( {1 - w} \right)\)for\(0 < w < 1\)

Let,\(X = 2n\left( {1 - W} \right)\)

\(\begin{align} \Rightarrow W &= \frac{{1 - X}}{{2n}}\\ \Rightarrow \frac{{dW}}{{dx}} &= - \frac{1}{{2n}}\end{align}\)

Determine the p.d.f of \({{\bf{g}}_{\bf{n}}}\left( {\bf{x}} \right)\) 

The p.d.f of \({g_n}\left( x \right)\)is

\(\begin{align}{g_n}\left( x \right) &= {h_1}\left( {1 - \frac{x}{{2n}}} \right)\left| {\frac{{dW}}{{dx}}} \right|\\ &= n\left( {n - 1} \right){\left( {1 - \frac{x}{{2n}}} \right)^{n - 2}}\left( {\frac{x}{{2n}}} \right)\left( {\frac{1}{{2n}}} \right)\\ &= \frac{1}{4}\left( {\frac{{n - 1}}{n}} \right)x{\left( {1 - \frac{x}{{2n}}} \right)^{ - 2}}{\left( {1 - \frac{x}{{2n}}} \right)^n}\end{align}\) for \(0 < x < 2n\)

As\(n \to \infty \),

\(\begin{align}\frac{{n - 1}}{n} \to 1\\{\left( {1 - \frac{x}{{2n}}} \right)^{ - 2}} \to 1\end{align}\)

For any real number\(t\),

\({\left( {\frac{{1 + t}}{n}} \right)^n} \to \exp \left( t \right)\)

So,\({\left( {1 - \frac{x}{{2n}}} \right)^n} \to \exp \left( { - \frac{x}{2}} \right)\)

Hence, for\(x > 0\),

\({g_n}\left( x \right) \to \frac{1}{4}x\exp \left( { - \frac{x}{2}} \right)\)

Hence,\(\mathop {\lim }\limits_{n \to \infty } {g_n}\left( x \right) = g\left( x \right)\)for\(x > 0\).

Hence, proved.