Linear combinations of random variables
Introduced| English | Meaning |
|---|---|
| variance/ˈveərɪəns/ | A measure of spread based on the squared deviations of values from their mean. |
| linear combination/ˈlɪnɪə ˌkɒmbɪˈneɪʃn/ | linear combination |
| random variable/ˈrændəm ˈveərɪəbl/ | A numerical quantity whose value depends on the outcome of a random process. |
| linear transformation/ˈlɪnɪə trænsfɔːˈmeɪʃn/ | A mapping that preserves vector addition and multiplication by scalars. |
| independent/ˌɪndɪˈpendənt/ | Describing quantities, events or choices that do not depend on one another in the relationship being considered. |
The mathematics of combining
- If you know the mean and variance 方差 of two separate processes, what happens when you add them together?
- Linear combinations 线性组合 let you predict the behaviour of combined random variables 随机变量 — essential for quality control, finance, and engineering.
Linear combination lab
E(aX + b) = aE(X) + b
Change a scaling factor and see how the expected value scales.
Linear transformations 线性变换 of a single variable
- For a single variable changed by a linear rule:
Worked example. $E(X) = 5$, $\text{Var}(X) = 4$. Let $Y = 3X + 2$. $E(Y) = 3(5) + 2 = 17$. $\text{Var}(Y) = 3^2(4) = 36$.
The constant $b$ vanishes from variance. Adding a constant shifts the mean but doesn't change the spread. $\text{Var}(X + 5) = \text{Var}(X)$, not $\text{Var}(X) + 25$.
The Poisson distribution Po(3); its mean and variance both equal lambda
X has mean 5. What is E(3X − 1)?
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E(3X − 1) = 3E(X) − 1 = 3×5 − 1 = 14.
X has variance 4. What is Var(3X − 1)?
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Var(3X − 1) = 3²×Var(X) = 9×4 = 36 (the −1 has no effect).
In Var(aX + b), the constant b:
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Adding a constant shifts the data but does not change its spread, so b drops out.
Two independent 独立 variables
- Note: variances always add (even for $X - Y$), because $(-1)^2 = 1$.
X and Y are independent. Var(X) = 4, Var(Y) = 9. Var(X + Y) = ?
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Var(X + Y) = Var(X) + Var(Y) = 4 + 9 = 13 (variances add for independent variables).
X and Y are independent. Var(X) = 4, Var(Y) = 9. Var(X − Y) = ?
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Var(X − Y) = Var(X) + (−1)²Var(Y) = 4 + 9 = 13. Variances always add.
Distributions under linear combinations
- If $X$ is normal, so is $aX + b$; the sum of independent normals is normal.
- The sum of independent Poissons is Poisson: if $X \sim \text{Po}(\lambda_1)$ and $Y \sim \text{Po}(\lambda_2)$, then $X + Y \sim \text{Po}(\lambda_1 + \lambda_2)$.
If X ~ Po(3) and Y ~ Po(5) are independent, then X + Y ~ Po(8).
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The sum of independent Poisson variables is Poisson with parameter λ₁ + λ₂ = 3 + 5 = 8.
Worked example — combining normals
- $X \sim N(10, 4)$ and $Y \sim N(20, 9)$, independent. Find the distribution of $X + Y$.
- $E(X + Y) = 10 + 20 = 30$. $\text{Var}(X + Y) = 4 + 9 = 13$.
- So $X + Y \sim N(30, 13)$.
- Linear combinations of normal distributions are normal; sums of independent Poisson distributions are Poisson.
You've got it
- $E(aX+b) = aE(X) + b$; $\text{Var}(aX+b) = a^2\,\text{Var}(X)$ (the $b$ drops, $a$ is squared)
- independent: $\text{Var}(aX+bY) = a^2\text{Var}(X) + b^2\text{Var}(Y)$
- normal stays normal under a linear rule; sums of independent Poissons are Poisson
Write from an expectation of 7 and a variance of 12
- E(X) = 4 and Var(X) = 3. Y = 2X - 1.
- Find E(Y) and Var(Y).
E(X) is 4 and Var(X) is 3. For Y = 2X - 1, what are E(Y) and Var(Y)?
8 minus 1 is 7. 4 times 3 is 12.
What is Var(Y)?
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The coefficient 2 is squared, so the variance is multiplied by 4. Other options — 36 (the lesson variance); 3 (the original variance left unscaled).