Further Probability

Variance of a linear transformation (single variable, always):

$\text{Var}(aX + b) = a^2 \text{Var}(X).$

The $+b$ shifts but does not stretch.

Variance of a sum — INDEPENDENCE REQUIRED:

$\text{Var}(aX + bY) = a^2 \text{Var}(X) + b^2 \text{Var}(Y).$

Note the squared coefficients and that subtraction still adds variances: $\text{Var}(X - Y) = \text{Var}(X) + \text{Var}(Y)$.

Worked example. $X$ has $E(X) = 5, \text{Var}(X) = 4$. $Y$ has $E(Y) = 8, \text{Var}(Y) = 9$. $X, Y$ independent. Find $E(2X - 3Y + 1)$ and $\text{Var}(2X - 3Y + 1)$.

$E = 2(5) - 3(8) + 1 = 10 - 24 + 1 = -13$.

$\text{Var} = 4(4) + 9(9) = 16 + 81 = 97$. (constant $+1$ contributes 0)

• Mean: linearity ALWAYS holds.
• Variance: squared coefficients AND independence.
• $\text{Var}(X - Y) = \text{Var}(X) + \text{Var}(Y)$ (PLUS!).

$aX + bY \sim N(a\mu_X + b\mu_Y, a^2\sigma_X^2 + b^2\sigma_Y^2).$

Worked example. $X \sim N(60, 25)$ (mass of bag), $Y \sim N(58, 16)$. Independent. Find $P(X > Y)$.

Let $D = X - Y$. Then $D \sim N(60 - 58, 25 + 16) = N(2, 41)$, so $\sigma_D = \sqrt{41} \approx 6.403$.

$P(X > Y) = P(D > 0) =$ normCdf$(0, 10^{99}, 2, 6.403) \approx 0.624$.

• Normal sums are normal — no need for limit theorems.
• Variances ADD, never subtract.
• State the resulting distribution before using GDC.