Week 3.2: Variance & Standard Deviation
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# Week 3.2: Variance & Standard Deviation > **Prerequisites:** Expected Value ([Week 3.1: Expected Value](/notes/01-foundation-bsma1004-stats-2-week03-08-expectation)) > **Cross-links:** BSMA3012 (Linear Stat Models) — residual variance > **Core question:** How spread out is a random variable around its mean? * * *...

Week 3.2: Variance & Standard Deviation
Prerequisites: Expected Value (Week 3.1: Expected Value) Cross-links: BSMA3012 (Linear Stat Models) — residual variance Core question: How spread out is a random variable around its mean?
1. Intuition: Beyond the Average
Expected value tells us the centre, but two RVs can have the same mean yet behave very differently:
- X≡10 (constant) → mean 10, no spread
- Y∈{9,11} equally → mean 10, small spread
- Z∈{0,20} equally → mean 10, large spread We need a measure of dispersion or spread. The variance quantifies how far values typically fall from the mean.
2. Formal Definition
>Var(X)=E[(X−μ)2],where μ=E[X].>Definition (Variance & Standard Deviation) The variance of X is:
Var(X)=E[X2]−(E[X])2.The standard deviation is SD(X)=Var(X). Alternative formula (easier for computation):
Proof:
3. Properties
- Var(c)=0 for a constant c
- Var(aX+b)=a2Var(X) (scaling changes variance, shifting does not)
- SD(aX+b)=∣a∣SD(X)
- If X and Y are independent: Var(X+Y)=Var(X)+Var(Y)
4. Variance of Common Distributions
| Distribution | PMF | E[X] | Var(X) |
|---|---|---|---|
| Bernoulli( p ) | px(1−p)1−x | p | p(1−p) |
| Binomial( n,p ) | (xn)px(1−p)n−x | np | np(1−p) |
| Geometric( p ) | (1−p)x−1p | 1/p | (1−p)/p2 |
| Poisson( λ ) | e−λλx/x! | λ | λ |
| Uniform {1,…,n} | 1/n | (n+1)/2 | (n2−1)/12 |
5. Standardised Random Variables
Z=σX−μDefinition X is standardised if E[X]=0 and Var(X)=1. For any X with finite mean and variance:
is standardised. This Z-score expresses values in "standard deviation units".
6. Practice Questions
Q1 (Easy)
X has PMF: fX(−2)=0.1, fX(0)=0.6, fX(2)=0.3. Find Var(X).
Full Solutionμ=(−2)(0.1)+0(0.6)+2(0.3)=−0.2+0+0.6=0.4.E[X2]=4(0.1)+0(0.6)+4(0.3)=0.4+0+1.2=1.6.Var(X)=1.6−(0.4)2=1.6−0.16=1.44.
Q2 (Medium)
X∼Binomial(100,0.3). Find E[X] and Var(X).
Full SolutionE[X]=np=100×0.3=30. Var(X)=np(1−p)=100×0.3×0.7=21. SD(X)=21≈4.58.
Q3 (Hard)
X∼Poisson(5). Find E[X2] and Var(X).
Full SolutionFor Poisson(λ): μ=λ, Var(X)=λ=5.Var(X)=E[X2]−μ2⟹E[X2]=Var(X)+μ2=5+25=30.
Next topic: Week 3.3: Covariance & Correlation — Covariance and correlation coefficient. Join Discord PreviousWeek 3.1: Expected ValueNextWeek 3.3: Covariance & Correlation