5.5.2 The Case of Normal Random Variables

When the $X_i$ ’s form a random sample from a normal distribution, $\bar{X}$ and $T_o$ are both normally distributed. Here is a more general result concerning linear combinations.

Proposition

If $X_1,X_2,\dots ,X_n$ are independent, normally distributed rv’s (with possibly different means and/or variances), then any linear combination of the $X_i$ ’s also has a normal distribution. In particular, the difference $X_1-X_2$ between two independent, normally distributed variables is itself normally distributed.

EXAMPLE 5.32 (Example 5.30 continued)

The CLT can also be generalized so it applies to certain linear combinations. Roughly speaking, if $n$ is large and no individual term is likely to contribute too much to the overall value, then $Y$ has approximately a normal distribution.

Proofs for the Case $n=2$

For the result concerning expected values, suppose that $X_1$ and $X_2$ are continuous with joint $\mathrm{pdf}f\left(x_1,x_2\right)$ . Then

$$\begin{array}{rl}& E\left(a_1{X}_1+a_2{X}_2\right)\\ & ={\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }\left(a_1{x}_1+a_2{x}_2\right)f\left(x_1,x_2\right)d{x}_{1}d{x}_2\\ & =a_1{\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }{x}_{1}f\left(x_1,x_2\right)d{x}_{2}d{x}_1\\ & +a_2{\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }{x}_{2}f\left(x_1,x_2\right)d{x}_{1}d{x}_2\\ & =a_1{\int }_{-\infty }^{\infty }{x}_1{f}_{X_1}\left(x_1\right)d{x}_1+a_2{\int }_{-\infty }^{\infty }{x}_2{f}_{X_2}\left(x_2\right)d{x}_2\\ & =a_{1}E\left(X_1\right)+a_{2}E\left(X_2\right)\end{array}$$

Summation replaces integration in the discrete case. The argument for the variance result does not require specifying whether either variable is discrete or continuous. Recalling that $V\left(Y\right)=E\left[{\left(Y-{\mu }_Y\right)}^2\right]$ ,

$$\begin{array}{rl}V\left(a_1{X}_1+a_2{X}_2\right)& =E\left\{{\left[a_1{X}_1+a_2{X}_2-\left(a_1{\mu }_1+a_2{\mu }_2\right)\right]}^2\right\}\\ & =E\{a_1^2{\left(X_1-{\mu }_1\right)}^2+a_2^2{\left(X_2-{\mu }_2\right)}^2\\ & +2a_1{a}_2\left(X_1-{\mu }_1\right)\left(X_2-{\mu }_2\right)\}\end{array}$$

The expression inside the braces is a linear combination of the variables $Y_1=$ ${\left(X_1-{\mu }_1\right)}^2,Y_2={\left(X_2-{\mu }_2\right)}^2$ , and $Y_3=\left(X_1-{\mu }_1\right)\left(X_2-{\mu }_2\right)$ , so carrying the $E$ operation through to the three terms gives $a_1^{2}V\left(X_1\right)+a_2^{2}V\left(X_2\right)+2a_1{a}_2\mathrm{Cov}\left(X_1,X_2\right)$ as required.