Question

For the conditions of Exercise 1, find the p.d.f. of the

average \(\frac{{\left( {{{\bf{X}}_{\bf{1}}}{\bf{ + }}{{\bf{X}}_{\bf{2}}}} \right)}}{{\bf{2}}}\)

Short answer

The p.d.f of \({Y_1} = \frac{{\left( {{X_1} + {X_2}} \right)}}{2}\) is

\({f_{{y_1}}}\left( {{y_1}} \right) = 2,0 \le {y_1} \le 1\)

Step-by-step solution

Given information

\({X_1},{X_2}\) is a random variable following uniform distribution on a given interval [0,1], that is \({X_i} \sim U[0,1],\;i = 1,2\)

Define p.d.f of X1 and X2

The pdf of a uniform distribution is obtained by using the formula: \(\frac{1}{{b - a}};a \le x \le b\).

Here, \(a = 0,b = 1\).

Therefore, the PDF \({X_i} \sim U[0,1],\;i = 1,2\) is expressed as,

\({f_{{x_i}}} = \left\{ \begin{array}{l}\frac{1}{{1 - 0}} = 1\;\;\;\;\;\;\;\;\;\;0 \le {x_i} \le 1\\0;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;otherwise\end{array} \right.\)

Since both the random variables are independent, the joint density function of both the random variable is :

\(\begin{aligned}{f_{{x_1}}}_{{x_2}}\left( {{x_1},{x_2}} \right) &= {f_{{x_1}}}\left( {{x_1}} \right) \cdot {f_{{x_2}}}\left( {{x_2}} \right)\\ &= 1 \cdot 1\\ &= 1,\;0 \le {x_1} \le 1,\;0 \le {x_2} \le 1\end{aligned}\)

Define new variables and perform Jacobian Transformation

Let \(\begin{aligned}{Y_1} &= \frac{{\left( {{X_1} + {X_2}} \right)}}{2}\\{Y_2} &= \frac{{\left( {{X_1} - {X_2}} \right)}}{2}\end{aligned}\)

Therefore, \(\begin{aligned}{x_1} &= {y_1} + {y_2}\\{x_2} &= {y_1} - {y_2}\end{aligned}\)

Now the Jacobian of the transformed variable:

\(\begin{aligned}J &= \left| {\begin{aligned}{}{\frac{{\partial {x_1}}}{{\partial {y_1}}}}&{\frac{{\partial {x_2}}}{{\partial {y_1}}}}\\{\frac{{\partial {x_1}}}{{\partial {y_2}}}}&{\frac{{\partial {x_2}}}{{\partial {y_2}}}}\end{aligned}} \right|\\ &= \left| {\begin{aligned}{}1& 1\\1&{ - 1}\end{aligned}} \right|\\ &= \left| { - 2} \right|\end{aligned}\)

The joint density function of the variables

\(\begin{aligned}{f_{{y_1}}}_{{y_2}}\left( {{y_1},{y_2}} \right) &= {f_{{x_1}}}\left( {{x_1}} \right) \cdot {f_{{x_2}}}\left( {{x_2}} \right) \cdot \left| J \right|\\ &= 1 \cdot 2\\ &= 2\end{aligned}\)

Find the marginal of Y1

\({f_{{y_1}}}\left( {{y_1}} \right) = \int\limits_{{y_2}} {{f_{{y_1}}}_{{y_2}}\left( {{y_1},{y_2}} \right)d{y_2}} \)

The range \({Y_2} = \frac{{\left( {{X_1} - {X_2}} \right)}}{2}\) lies in the interval \(\left[ { - \frac{1}{2},\frac{1}{2}} \right]\) since \({X_i} \sim U[0,1],\;i = 1,2\)

\(\begin{aligned}{f_{{y_1}}}\left( {{y_1}} \right) &= \int_{\frac{{ - 1}}{2}}^{\frac{1}{2}} 2 d{y_2}\\ &= 2,0 \le {y_1} \le 1\end{aligned}\)

Therefore, the p.d.f of the average \({Y_1} = \frac{{\left( {{X_1} + {X_2}} \right)}}{2}\) is \({f_{{y_1}}}\left( {{y_1}} \right) = 2,0 \le {y_1} \le 1\).