Affine images of normal vectors
Theorem taught in Functions of random variables and limit theorems, part of Probability
Affine images of normal vectors
| Type | Theorem |
| Taught in | Functions of random variables and limit theorems |
| Part | Probability |
| Worked examples | 3 |
If X_i\sim\operatorname{N}(\mu_i,\sigma_i^2) are independent and a,b_1,\ldots,b_n are constants, then
a+\sum_{i=1}^n b_i X_i \sim \operatorname{N}\Bigl(a+\sum_{i=1}^n b_i\mu_i,\;\sum_{i=1}^n b_i^2\sigma_i^2\Bigr).
The same conclusion holds for a jointly normal vector and any affine image, with the output variance \mathbf{b}^\top\Sigma\mathbf{b}.
Examples
Ball bearings and a sieve
Worked example from Functions of random variables and limit theorems
Ball bearings have diameters X\sim\operatorname{N}(1,0.01) cm. They are dropped through a sieve whose holes have diameters Y\sim\operatorname{N}(1.1,0.04) cm, independent of the balls.
Question. What is the probability that a ball falls through a hole?
Solution. The ball falls through when Y>X. Because X and Y are independent normals, Y-X is normal with mean 1.1-1=0.1 and variance 0.04+0.01=0.05. So
\mathbb{P}(Y-X>0)=\mathbb{P}\Bigl(Z>\frac{-0.1}{\sqrt{0.05}}\Bigr)=\Phi(0.447)=0.67.
The variances add even though the variable is a difference.
Partial sums of standard normals
Question page: Partial sums of standard normals
Let X_1,X_2,\ldots be independent and standard normal, and set Y_n=X_1+\cdots+X_n, with Y_1=X_1.
Describe a typical path of X_1,\ldots,X_n and of Y_1,\ldots,Y_n for large n.
Find \mathbb{E} Y_n and \operatorname{Var}(Y_n).
Find \operatorname{Cov}(Y_m,Y_n) for m\le n.
The X_i fluctuate about 0 and do not wander. The partial sums Y_n form a random walk: the path is continuous only in the discrete-time sense of successive additions, and it drifts away from 0 on the scale \sqrt{n}.
\mathbb{E} Y_n=0 and \operatorname{Var}(Y_n)=n.
For m\le n, Y_n=Y_m+X_{m+1}+\cdots+X_n. The extra sum is independent of Y_m and has mean 0, so \operatorname{Cov}(Y_m,Y_n)=\operatorname{Var}(Y_m)=m.
Thickness of a circuit board
Question page: Thickness of a circuit board
Boards should have thickness between 0.146 cm and 0.154 cm. A machine produces thickness X\sim\operatorname{N}(0.151, 0.003^2).
Find the probability that one board meets the specification. Use \Phi(1)\approx 0.8413 and \Phi(1.67)\approx 0.9525.
Let \overline X be the mean thickness of 25 independent boards from this machine. Find the probability that the batch mean meets the same specification.
- Standardise the limits 0.146 and 0.154 with mean 0.151 and standard deviation 0.003:
\mathbb{P}(0.146\le X\le 0.154)=\Phi(1)-\bigl(1-\Phi(1.667)\bigr)\approx 0.8413-(1-0.9525)=0.794.
- The batch mean is \operatorname{N}(0.151, 0.003^2/25). The standardised limits are -8.33 and 5. Both are so far into the tails that the probability is 1 for any ordinary purpose.
Used in
- Taught in Functions of random variables and limit theorems, where the derivation and examples are.
- All 21 questions on Functions of random variables and limit theorems
See also
Multivariate normal distribution · Transformation rule · Normal linear model · Law of large numbers · Central limit theorem