91Ó°ÊÓ

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

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

(a)

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

Define the mean of X as:

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

Therefore, mean is \(E\left( X \right) = \frac{2}{3}\)

Hence, \(E\left( {{{\left( {X - d} \right)}^2}} \right)\) is minimized when \(d = \frac{2}{3}\).

(b)

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

Define the median of X as:

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

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

Hence, \(E\left( {\left| {X - d} \right|} \right)\) is minimized when \(d = \frac{1}{{\sqrt 2 }}\).