S3.1 - Combinations of random variables

Learning objectives

S3.1.1Linear combinations of Normal variablesDistribution of linear combinations If X ∼ N(μ, σ 2) and Y∼ N(μ, σ 2) independently, then x x y y of independent Normal random aX± bY ∼ N(aμ ± bμ, a2σ 2 + b2σ 2). variables. x y x y No proofs required.

A linear combination of independent Normal random variables is also Normal. If $X\sim N(\mu_x,\sigma_x^2),\qquad Y\sim N(\mu_y,\sigma_y^2)$ independently, then its mean follows the signs in the combination, while each variance contribution uses the square of its coefficient.

$aX\pm bY\sim N\left(a\mu_x\pm b\mu_y,\ a^2\sigma_x^2+b^2\sigma_y^2\right)$

Combination	Mean	Variance
$aX+bY$	$a\mu_x+b\mu_y$	$a^2\sigma_x^2+b^2\sigma_y^2$
$aX-bY$	$a\mu_x-b\mu_y$	$a^2\sigma_x^2+b^2\sigma_y^2$

Subtraction changes the centre because it reverses Y's contribution to the value. It does not subtract uncertainty: deviations in either variable create spread in the combination, so independent variance contributions add after scaling by squared coefficients.

For independent copies, add one mean and one variance contribution for each copy. If male load $M\sim N(80,100)$ and female load $W\sim N(69,25)$ , then the load of six men and three women is $T=M_1+\cdots+M_6+W_1+\cdots+W_3\sim N(687,675).$ Therefore $P(T>700)=P\left(Z>\frac{700-687}{\sqrt{675}}\right)=P(Z>0.500\ldots)\approx0.3085.$

Convert comparisons into one variable before standardising. If $A\sim N(20,9)$ and $B\sim N(8,4)$ independently, then $D=A-2B\sim N(4,25),$ so $P(A>2B)=P(D>0)=P\left(Z>\frac{0-4}{5}\right)=P(Z>-0.8)=0.7881.$

Check both Normality and independence before using the rule. Keep the second parameter of N(mean, variance) as a variance, take its square root only when standardising, and never subtract variances for a difference. No proof of the combination rule is required.

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