91Ó°ÊÓ

A random variable X has a continuous distribution for which the pdf is as follows:

\(f\left( x \right) = \left\{ {\begin{aligned}{{}{}}{x + \frac{1}{2}}&{{\rm{for}}\;0 \le x \le 1}\\0&{{\rm{otherwise}}}\end{aligned}} \right.\)

(a)

\(M.S.E = E\left( {{{\left( {X - d} \right)}^2}} \right)\)is minimized when d takes the value of the mean of X.

Define mean of X as:

\(\begin{aligned}{}E\left( X \right) &= \int\limits_0^1 {xf\left( x \right)dx} \\ &= \int\limits_0^1 {x\left( {x + \frac{1}{2}} \right)dx} \\ &= \int\limits_0^1 {{x^2} + \frac{x}{2}dx} \\ &= \frac{1}{3}\left( {{x^3}} \right)_0^1 + \frac{1}{4}\left( {{x^2}} \right)_0^1\\ &= \frac{1}{3} + \frac{1}{4}\\ &= \frac{7}{{12}}.\end{aligned}\)

Therefore, mean is \(E\left( X \right) = \frac{7}{{12}}.\)

Hence, the \(M.S.E\) is minimized when the predicted value of X is \(d = \frac{7}{{12}}\).

(b)

\(M.A.E = E\left( {\left| {X - d} \right|} \right)\)is minimized when d takes the value of the median of X.

Define the median of X as:

\(\begin{aligned}{}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Pr \left( {X \le m} \right) &= \frac{1}{2}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow \int\limits_0^m {f\left( x \right)dx} &= \frac{1}{2}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow \int\limits_0^m {x + \frac{1}{2}dx} &= \frac{1}{2}\\ \Rightarrow \frac{1}{2}\left( {{x^2}} \right)_0^m + \frac{1}{2}\left( x \right)_0^m = \frac{1}{2}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow {m^2} + m &= 1\\\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow {m^2} + m - 1 = 0 \ldots \ldots ....... \ldots \left( 1 \right)\end{aligned}\)

Solving equation (1), we get:

\(\begin{aligned}{}m &= \frac{{ - 1 \pm \sqrt {1 + 4} }}{2}\\ &= \frac{{ - 1 \pm \sqrt 5 }}{2}\\ &= \frac{{ - 1 - \sqrt 5 }}{2},\frac{{\sqrt 5 - 1}}{2}.\end{aligned}\)

Now, since \(0 \le X \le 1\), \(m = \frac{{\sqrt 5 - 1}}{2}.\)

Then, the median is \(m = \frac{{\sqrt 5 - 1}}{2}\).

Hence, the M.A.E is minimized when the predicted value of X is \(d = \frac{{\sqrt 5 - 1}}{2}\).