91Ó°ÊÓ

There are three random variables \({X_1},{X_2}\;and\;{X_3}\) from the uniform distribution on the interval [0,1].

The random variables \({X_i}\;\left( {i = 1,2,3} \right)\) has a uniform distribution on the interval [0,1]. So, the mean of the distribution is \(E\left( {{X_i}} \right) = \int_0^1 {{x_i}dx} = \frac{1}{2}\) and the variance is\(\begin{array}{c}Var\left( {{X_i}} \right) = E\left( {X_i^2} \right) - {\left( {E\left( {{X_i}} \right)} \right)^2}\\ = \int_0^1 {x_i^2dx} - \frac{1}{4}\\ = \frac{1}{3} - \frac{1}{4} = \frac{1}{{12}}\end{array}\)

Let us consider\(Z = \left( {{X_1} - 2{X_2} + {X_3}} \right)\). As,\({X_i}'s\)are independent,

So,

\(\begin{array}{c}E\left( Z \right) = E\left( {{X_1}} \right) - 2E\left( {{X_2}} \right) + E\left( {{X_3}} \right)\\ = \frac{1}{2} - 1 + \frac{1}{2} = 0\end{array}\)

\(\begin{array}{c}Var\left( Z \right) = Var\left( {{X_1}} \right) + 4Var\left( {{X_2}} \right) + Var\left( {{X_3}} \right)\\ = \frac{1}{{12}} + \frac{1}{3} + \frac{1}{{12}} = \frac{1}{2}\end{array}\)

Therefore,

\(\begin{array}{c}E\left( {{Z^2}} \right) = E\left[ {\left( {{X_1} - 2{X_2} + {X_3}} \right)} \right]\\ = Var\left( Z \right) + \left[ {E{{\left( Z \right)}^2}} \right]\\ = \frac{1}{2} + 0\\ = \frac{1}{2}\end{array}\)

Thus, the required value is\(\frac{1}{2}\).