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Week 6.2: Sampling Distributions

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Now · 1. Intuition: The Distribution of a Statistic

Week 6.2: Sampling Distributions

Prerequisites: CLT (Week 5.2: Law of Large Numbers & Central Limit Theorem), Normal Distribution (Week 4.2: Common Continuous Distributions) Cross-links: BSMA3012 (Linear Stat Models) — inference Core question: What distributions describe sample statistics before we observe data?

1. Intuition: The Distribution of a Statistic

A statistic is any function of the sample (e.g., Xˉ\bar{X}, S2S^2, sample median). Its sampling distribution is the probability distribution of that statistic over repeated sampling.
Key idea: Before we draw the sample, Xˉ\bar{X} is a random variable with its own distribution. After we draw, it's a number.

2. Sampling Distribution of the Sample Mean

If X1,,Xni.i.d. (μ,σ2)X_1,\dots,X_n \sim \text{i.i.d. } (\mu, \sigma^2):
  1. Exact (normal data): If XiN(μ,σ2)X_i \sim \mathcal{N}(\mu, \sigma^2), then XˉN(μ,σ2/n)\bar{X} \sim \mathcal{N}(\mu, \sigma^2/n).
  2. Approximate (CLT): For large nn, XˉN(μ,σ2/n)\bar{X} \approx \mathcal{N}(\mu, \sigma^2/n). Standard error: SE(Xˉ)=σ/n\text{SE}(\bar{X}) = \sigma / \sqrt{n}.

3. Sampling Distribution of the Sample Variance

Definition The sample variance is S2=1n1i=1n(XiXˉ)2S^2 = \frac{1}{n-1} \sum_{i=1}^n (X_i - \bar{X})^2. For XiN(μ,σ2)X_i \sim \mathcal{N}(\mu, \sigma^2) i.i.d.:
(n1)S2σ2χn12,\frac{(n-1)S^2}{\sigma^2} \sim \chi^2_{n-1},
where χn12\chi^2_{n-1} is the chi-squared distribution with n1n-1 degrees of freedom.

4. The Chi-Squared Distribution

Definition (χk2\chi^2_k) If Z1,,ZkN(0,1)Z_1,\dots,Z_k \sim \mathcal{N}(0,1) i.i.d., then i=1kZi2χk2\sum_{i=1}^k Z_i^2 \sim \chi^2_k. Properties:
  • E[χk2]=k\mathbb{E}[\chi^2_k] = k
  • Var(χk2)=2k\operatorname{Var}(\chi^2_k) = 2k
  • As kk \to \infty, χk2N(k,2k)\chi^2_k \approx \mathcal{N}(k, 2k) PDF:
f(x)=12k/2Γ(k/2)xk/21ex/2,x>0.f(x) = \frac{1}{2^{k/2} \Gamma(k/2)} x^{k/2-1} e^{-x/2}, \quad x > 0.

5. The t-Distribution

Definition (tkt_k) If ZN(0,1)Z \sim \mathcal{N}(0,1) and Vχk2V \sim \chi^2_k are independent, then:
>T=ZV/ktk.>> T = \frac{Z}{\sqrt{V/k}} \sim t_k. >
Properties:
  • Symmetric around 0, heavier tails than normal
  • As kk \to \infty, tkN(0,1)t_k \to \mathcal{N}(0,1)
  • Var(T)=k/(k2)\operatorname{Var}(T) = k/(k-2) for k>2k > 2 Critical use: When σ\sigma is unknown, replace with SS:
XˉμS/ntn1.\frac{\bar{X} - \mu}{S/\sqrt{n}} \sim t_{n-1}.

6. The F-Distribution

Definition (Fd1,d2F_{d_1,d_2}) If V1χd12V_1 \sim \chi^2_{d_1} and V2χd22V_2 \sim \chi^2_{d_2} are independent:
>F=V1/d1V2/d2Fd1,d2.>> F = \frac{V_1/d_1}{V_2/d_2} \sim F_{d_1,d_2}. >
Use: Comparing two variances: S12/S22Fn11,n21S_1^2/S_2^2 \sim F_{n_1-1, n_2-1} under equal variances.

7. Summary Table

StatisticDistributionWhen
Xˉμσ/n\frac{\bar{X} - \mu}{\sigma/\sqrt{n}}N(0,1)\mathcal{N}(0,1)Known σ\sigma , normal data or large nn
XˉμS/n\frac{\bar{X} - \mu}{S/\sqrt{n}}tn1t_{n-1}Unknown σ\sigma , normal data
(n1)S2σ2\frac{(n-1)S^2}{\sigma^2}χn12\chi^2_{n-1}Normal data
S12/σ12S22/σ22\frac{S_1^2/\sigma_1^2}{S_2^2/\sigma_2^2}Fn11,n21F_{n_1-1,n_2-1}Two normal samples

8. Practice Questions

Q1 (Easy)

If XiN(10,4)X_i \sim \mathcal{N}(10, 4) i.i.d. for n=25n=25, what is the distribution of Xˉ\bar{X}?
Full Solution
XˉN(10,4/25)=N(10,0.16)\bar{X} \sim \mathcal{N}(10, 4/25) = \mathcal{N}(10, 0.16).

Q2 (Medium)

For n=20n=20 normal samples, xˉ=15\bar{x}=15, s=3s=3. Find a distribution for Xˉμs/n\frac{\bar{X}-\mu}{s/\sqrt{n}}.
Full Solution
XˉμS/nt19\frac{\bar{X} - \mu}{S/\sqrt{n}} \sim t_{19} (t-distribution with 19 df).

Q3 (Hard)

If S2S^2 is the sample variance from n=10n=10 normal samples, find E[S2]\mathbb{E}[S^2] and Var(S2)\operatorname{Var}(S^2).
Full Solution
9S2σ2χ92\frac{9S^2}{\sigma^2} \sim \chi^2_9.
E[9S2/σ2]=9    E[S2]=σ2\mathbb{E}[9S^2/\sigma^2] = 9 \implies \mathbb{E}[S^2] = \sigma^2 (unbiased). ✓
Var(9S2/σ2)=18    81Var(S2)/σ4=18    Var(S2)=18σ481=2σ49\operatorname{Var}(9S^2/\sigma^2) = 18 \implies 81 \cdot \operatorname{Var}(S^2)/\sigma^4 = 18 \implies \operatorname{Var}(S^2) = \frac{18\sigma^4}{81} = \frac{2\sigma^4}{9}.

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