Linear combinations of independent Normal variables

What you’ll learn

• Combine means and variances with correct coefficients.
• Distinguish repeated independent observations from a multiple of one observation.
• Calculate a probability for a sum or difference.

Understand the idea

A linear combination of independent Normal variables is Normal. Its mean follows the signs of the coefficients, while its variance uses the squares of the coefficients. Thus independent variation accumulates even in a difference: subtracting a random quantity does not subtract its uncertainty.

Independence is essential for the simple variance-addition rule. If two measurements share a source of variation, a covariance term may be needed and the independent formula is not justified. State the assumption when combining separately sampled quantities.

The sum of two independent copies $X_1+X_2$ differs from $2X$, which doubles the same outcome. Both have mean $2\mu$, but their variances are $2\sigma^2$ and $4\sigma^2$ respectively. Once the combined Normal distribution is identified, use ordinary standardisation.

Key formulae

$$E(aX+bY+c)=a\mu_X+b\mu_Y+c$$

$$\operatorname{Var}(aX+bY+c)=a^2\sigma_X^2+b^2\sigma_Y^2\quad(X,Y\text{ independent})$$

$$aX+bY+c\sim N\!\left(a\mu_X+b\mu_Y+c,a^2\sigma_X^2+b^2\sigma_Y^2\right)$$

Worked example

Independent variables satisfy $X\sim N(20,4)$ and $Y\sim N(12,9)$. Find $P(2X-Y>33)$.

1. Let $D=2X-Y$. Its mean is $2(20)-12=28$.
2. Its variance is $2^2(4)+(-1)^2(9)=25$, so its standard deviation is 5.
3. Therefore $D\sim N(28,25)$ and the boundary standardises to $(33-28)/5=1$.
4. The probability is $1-\Phi(1)=0.1587$ approximately.

Answer: $0.1587$ approximately.