Chapter 4, Problem 185SE

a.

To determine

Show that $f\left(y\right)$ is a density function.

Answer to Problem 185SE

The function $f\left(y\right)$ is a density function.

Explanation of Solution

Let Y is a random variable whose density is a combination of two density function and it is defined as follows:

$f\left(y\right)=a{f}_1\left(y\right)+\left(1-a\right)f_2\left(y\right)$

It is given that $f_1\left(y\right)$ and $f_2\left(y\right)$ are the density functions of the random variables $Y_1$ and $Y_2$, respectively. Let ‘a’ is a constant such that $0\le a\le 1$. Therefore, it implies as follows:

${\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{f}_1\left(y\right)dy}=1$, ${\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{f}_2\left(y\right)dy}=1$

One has to show that the function $f\left(y\right)$ is a density function of Y. That is, ${\displaystyle \underset{-\infty }{\overset{\infty }{\int }}f\left(y\right)dy}=1$

Consider, L.H.S

$\begin{array}{c}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}f\left(y\right)dy}={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left(a{f}_1\left(y\right)+\left(1-a\right)f_2\left(y\right)\right)}dy\\ =a{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{f}_1\left(y\right)}dy+\left(1-a\right){\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{f}_2\left(y\right)dy}\\ =a\left(1\right)+\left(1-a\right)\left(1\right)\left\{\because {\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{f}_1\left(y\right)dy}=1,{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{f}_2\left(y\right)dy}=1\right\}\\ =a+1-a\\ =1\end{array}$

Therefore, the function $f\left(y\right)$ is a density function of Y.

b.

To determine

Prove that,

1. i. $E\left(Y\right)=a{\mu }_1+\left(1-a\right){\mu }_2$
2. ii. $V\left(Y\right)=a{\sigma }_1{}^2+\left(1-a\right){\sigma }_2{}^2+a\left(1-a\right){\left[{\mu }_1-{\mu }_2\right]}^2$

Answer to Problem 185SE

It is proved as follows:

1. i. $E\left(Y\right)=a{\mu }_1+\left(1-a\right){\mu }_2$
2. ii. $V\left(Y\right)=a{\sigma }_1{}^2+\left(1-a\right){\sigma }_2{}^2+a\left(1-a\right){\left[{\mu }_1-{\mu }_2\right]}^2$

Explanation of Solution

The random variable Y whose density function is defined as follows:

$f\left(y\right)=a{f}_1\left(y\right)+\left(1-a\right)f_2\left(y\right)$

The expected value of Y is obtained as follows:

$\begin{array}{c}E\left(Y\right)={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}yf\left(y\right)}dy\\ ={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}y\left(a{f}_1\left(y\right)+\left(1-a\right)f_2\left(y\right)\right)}dy\\ =a{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}y}{f}_1\left(y\right)dy+\left(1-a\right){\displaystyle \underset{-\infty }{\overset{\infty }{\int }}y}{f}_2\left(y\right)dy\\ =aE\left(Y_1\right)+\left(1-a\right)E\left(Y_2\right)\\ =a{\mu }_1+\left(1-a\right){\mu }_2\left\{\because E\left(Y_1\right)={\mu }_1,E\left(Y_2\right)={\mu }_2\right\}\end{array}$

Therefore, it is proved that $E\left(Y\right)=a{\mu }_1+\left(1-a\right){\mu }_2$.

Consider,

$\begin{array}{c}E\left(Y^2\right)={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{y}^{2}f\left(y\right)}dy\\ ={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{y}^2\left(a{f}_1\left(y\right)+\left(1-a\right)f_2\left(y\right)\right)}dy\\ =a{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{y}^2}{f}_1\left(y\right)dy+\left(1-a\right){\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{y}^2}{f}_2\left(y\right)dy\\ =aE\left(Y_1{}^2\right)+\left(1-a\right)E\left(Y_2{}^2\right)\\ =a\left({\sigma }_1{}^2+{\mu }_1{}^2\right)+\left(1-a\right)\left({\sigma }_2{}^2+{\mu }_2{}^2\right)\\ \left\{\because E\left(Y_i{}^2\right)=V\left(Y_i\right)+E\left(Y_i{}^2\right)={\sigma }_i{}^2+{\mu }_i{}^2;i=1,2\right\}\end{array}$

The variance of Y is derived as follows:

$\begin{array}{c}V\left(Y\right)=E\left(Y^2\right)-{\left(E\left(Y\right)\right)}^2\\ =a\left({\sigma }_1{}^2+{\mu }_1{}^2\right)+\left(1-a\right)\left({\sigma }_2{}^2+{\mu }_2{}^2\right)-{\left[a{\mu }_1+\left(1-a\right){\mu }_2\right]}^2\\ =a{\sigma }_1{}^2+a{\mu }_1{}^2+{\sigma }_2{}^2+{\mu }_2{}^2-a{\sigma }_2{}^2-a{\mu }_2{}^2-\left[a^2{\mu }_1{}^2+{\left(1-a\right)}^2{\mu }_2{}^2+2a{\mu }_1\left(1-a\right){\mu }_2\right]\\ =a{\sigma }_1{}^2+\left(1-a\right){\sigma }_2{}^2+a{\mu }_1{}^2+{\mu }_2{}^2-a{\mu }_2{}^2-a^2{\mu }_1{}^2-{\left(1-a\right)}^2{\mu }_2{}^2-2a{\mu }_1\left(1-a\right){\mu }_2\\ =\left[\begin{array}{l}a{\sigma }_1{}^2+\left(1-a\right){\sigma }_2{}^2+a{\mu }_1{}^2+{\mu }_2{}^2-a{\mu }_2{}^2-a^2{\mu }_1{}^2-{\mu }_2{}^2\\ -a^2{\mu }_2{}^2+2a{\mu }_2{}^2-2a{\mu }_1\left(1-a\right){\mu }_2\end{array}\right]\\ =a{\sigma }_1{}^2+\left(1-a\right){\sigma }_2{}^2+a\left(1-a\right){\mu }_1{}^2+a\left(1-a\right){\mu }_2{}^2-a\left(1-a\right)2{\mu }_1{\mu }_2\\ =a{\sigma }_1{}^2+\left(1-a\right){\sigma }_2{}^2+a\left(1-a\right)\left[{\mu }_1{}^2+{\mu }_2{}^2-2{\mu }_1{\mu }_2\right]\\ =a{\sigma }_1{}^2+\left(1-a\right){\sigma }_2{}^2+a\left(1-a\right){\left[{\mu }_1+{\mu }_2\right]}^2\end{array}$

Thus, it is shown that $V\left(Y\right)=a{\sigma }_1{}^2+\left(1-a\right){\sigma }_2{}^2+a\left(1-a\right){\left[{\mu }_1+{\mu }_2\right]}^2$.