Question

If X and Y are independent random variables with variances σ2X = 5 and σ2Y = 3, find the variance of the random variable Z = −2X +4Y − 3.
Repeat Exercise 4.62 if X and Y are not inde- pendent and σXY =1.

Answer 1

If $X_1, X_2, ..., X_n$ are independent random variables having normal distributions with means $\mu_1, \mu_2,..., \mu_n$ and variances $\sigma_1^2, \sigma_2^2 , ..., \sigma_n ^2$ respectively, then the random variable

$Y=a_1X_1+a_2X_2+...+a_nX_n$ has a normal distribution with mean $\mu_Y=a_1\mu_1+a_2\mu_2+...+a_n\mu_n$ and variance $\sigma_Y^2=a_1^2\sigma_1^2+a_2^2\sigma_2^2+...+a_n^2\sigma_n^2$

Then

$$\sigma_Z^2=\sigma_{-2X+4Y-3}^2=(-2)^2\sigma_X^2+(4)^2\sigma_Y^2$$

$$\sigma_Z^2=\sigma_{-2X+4Y-3}^2=(-2)^2(5)+(4)^2(3)=68$$

If $X$ and $Y$ are not independent then

$$\sigma_{aX+bY+c}^2=a^2 \sigma_X^2+b^2\sigma_Y^2+2ab\sigma_{XY}$$

$$\sigma_Z^2=\sigma_{-2X+4Y-3}^2=(-2)^2(5)+(4)^2(3)+2(-2)(4)(1)=52$$