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≡10X \equiv 10 (constant) → mean 1010, no spread
  • Y∈{9,11}Y \in \{9,11\} equally → mean 1010, small spread
  • Z∈{0,20}Z \in \{0,20\} equally → mean 1010, large spread We need a measure of dispersion or spread. The variance quantifies how far values typically fall from the mean.

2. Formal Definition

Definition (Variance & Standard Deviation) The variance of XX is:
>Var⁡(X)=E[(X−μ)2],where μ=E[X].>> \operatorname{Var}(X) = \mathbb{E}\left[(X - \mu)^2\right], \quad \text{where } \mu = \mathbb{E}[X]. >
The standard deviation is SD⁡(X)=Var⁡(X)\operatorname{SD}(X) = \sqrt{\operatorname{Var}(X)}. Alternative formula (easier for computation):
Var⁡(X)=E[X2]−(E[X])2.\operatorname{Var}(X) = \mathbb{E}[X^2] - (\mathbb{E}[X])^2.
Proof:
Var⁡(X)=E[(X−μ)2]=E[X2−2μX+μ2]=E[X2]−2μE[X]+μ2=E[X2]−2μ2+μ2=E[X2]−μ2.\begin{aligned} \operatorname{Var}(X) &= \mathbb{E}[(X-\mu)^2] = \mathbb{E}[X^2 - 2\mu X + \mu^2] \\ &= \mathbb{E}[X^2] - 2\mu\mathbb{E}[X] + \mu^2 = \mathbb{E}[X^2] - 2\mu^2 + \mu^2 = \mathbb{E}[X^2] - \mu^2. \end{aligned}

3. Properties

  1. Var⁡(c)=0\operatorname{Var}(c) = 0 for a constant cc
  2. Var⁡(aX+b)=a2Var⁡(X)\operatorname{Var}(aX + b) = a^2 \operatorname{Var}(X) (scaling changes variance, shifting does not)
  3. SD⁡(aX+b)=∣a∣SD⁡(X)\operatorname{SD}(aX + b) = |a| \operatorname{SD}(X)
  4. If XX and YY are independent: Var⁡(X+Y)=Var⁡(X)+Var⁡(Y)\operatorname{Var}(X+Y) = \operatorname{Var}(X) + \operatorname{Var}(Y)

4. Variance of Common Distributions

DistributionPMFE[X]\mathbb{E}[X]Var⁡(X)\operatorname{Var}(X)
Bernoulli( pp )px(1−p)1−xp^x(1-p)^{1-x}ppp(1−p)p(1-p)
Binomial( n,pn,p )(nx)px(1−p)n−x\binom{n}{x}p^x(1-p)^{n-x}npnpnp(1−p)np(1-p)
Geometric( pp )(1−p)x−1p(1-p)^{x-1}p1/p1/p(1−p)/p2(1-p)/p^2
Poisson( λ\lambda )e−λλx/x!e^{-\lambda}\lambda^x/x!λ\lambdaλ\lambda
Uniform {1,…,n}\{1,\dots,n\}1/n1/n(n+1)/2(n+1)/2(n2−1)/12(n^2-1)/12

5. Standardised Random Variables

Definition XX is standardised if E[X]=0\mathbb{E}[X] = 0 and Var⁡(X)=1\operatorname{Var}(X) = 1. For any XX with finite mean and variance:
Z=X−μσZ = \frac{X - \mu}{\sigma}
is standardised. This Z-score expresses values in "standard deviation units".

6. Practice Questions

Q1 (Easy)

XX has PMF: fX(−2)=0.1f_X(-2)=0.1, fX(0)=0.6f_X(0)=0.6, fX(2)=0.3f_X(2)=0.3. Find Var⁡(X)\operatorname{Var}(X).
Full Solution
μ=(−2)(0.1)+0(0.6)+2(0.3)=−0.2+0+0.6=0.4\mu = (-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\mathbb{E}[X^2] = 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\operatorname{Var}(X) = 1.6 - (0.4)^2 = 1.6 - 0.16 = 1.44.

Q2 (Medium)

X∼Binomial(100,0.3)X \sim \text{Binomial}(100, 0.3). Find E[X]\mathbb{E}[X] and Var⁡(X)\operatorname{Var}(X).
Full Solution
E[X]=np=100×0.3=30\mathbb{E}[X] = np = 100 \times 0.3 = 30. Var⁡(X)=np(1−p)=100×0.3×0.7=21\operatorname{Var}(X) = np(1-p) = 100 \times 0.3 \times 0.7 = 21. SD⁡(X)=21≈4.58\operatorname{SD}(X) = \sqrt{21} \approx 4.58.

Q3 (Hard)

X∼Poisson(5)X \sim \text{Poisson}(5). Find E[X2]\mathbb{E}[X^2] and Var⁡(X)\operatorname{Var}(X).
Full Solution
For Poisson(λ\lambda): μ=λ\mu = \lambda, Var⁡(X)=λ=5\operatorname{Var}(X) = \lambda = 5.
Var⁡(X)=E[X2]−μ2  ⟹  E[X2]=Var⁡(X)+μ2=5+25=30\operatorname{Var}(X) = \mathbb{E}[X^2] - \mu^2 \implies \mathbb{E}[X^2] = \operatorname{Var}(X) + \mu^2 = 5 + 25 = 30.

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