Lesson M01.L01: Descriptive Statistics & Summation Notation
| Field | Value |
|---|---|
| Module | M01 — Probability & Statistics Review |
| Level | Graduate (ECMT5001) |
| Duration | ~60 minutes |
| Learning Objective | Master summation algebra and compute sample descriptive statistics (mean, variance, covariance, correlation) as the building blocks for all subsequent regression analysis. |
| Provenance | Wooldridge Appendix A (Summation Notation); ECMT1020 Weeks 1–3 |
Explanation
1. Summation Notation
Summation notation is the language of econometrics. It compresses long arithmetic into a concise form, and every formula in this course — OLS estimators, standard errors, hypothesis tests — is built from it.
The notation \(\sum_{i=1}^{n} x_i\) reads: "sum \(x_i\) as \(i\) runs from 1 to \(n\)." Explicitly:
Linearity Rules
For a constant \(c\) and variables \(x_i\), \(y_i\):
Warning: \(\sum x_i y_i \neq \left(\sum x_i\right)\left(\sum y_i\right)\) in general. Products do not split.
Double Summation
A double sum iterates over two indices. For an \(n \times m\) array \(x_{ij}\):
This simply sums every element. Order of summation is interchangeable when all sums are finite.
2. Sample Mean
The sample mean (average) of observations \(x_1, x_2, \ldots, x_n\) is:
The sample mean is the single number that minimises the sum of squared deviations — a property directly inherited by OLS estimators.
Key property: The deviations from the mean always sum to zero:
3. Sample Variance & Standard Deviation
The sample variance measures the average squared deviation from the mean:
Why divide by \(n-1\) instead of \(n\)? Dividing by \(n\) would give a biased estimate of the population variance \(\sigma^2\). Because we used the sample to estimate \(\bar{x}\) first (consuming one degree of freedom), the deviations \(x_i - \bar{x}\) carry only \(n-1\) pieces of independent information. Dividing by \(n-1\) makes \(s^2\) an unbiased estimator: \(E(s^2) = \sigma^2\).
A computationally equivalent formula:
The sample standard deviation is \(s = \sqrt{s^2}\), expressed in the same units as \(x\).
Z-scores (Standardisation)
A z-score transforms \(x_i\) into units of standard deviations from the mean:
After standardising, the new variable has mean 0 and standard deviation 1.
4. Sample Covariance
Covariance measures how two variables move together:
- \(s_{xy} > 0\): \(x\) and \(y\) tend to move in the same direction.
- \(s_{xy} < 0\): they tend to move in opposite directions.
- \(s_{xy} = 0\): no linear association.
The population covariance uses the same idea but weights by probabilities:
Covariance rules (used repeatedly in OLS derivations):
| Rule | Formula |
|---|---|
| Symmetry | \(\text{Cov}(X,Y) = \text{Cov}(Y,X)\) |
| Constant | \(\text{Cov}(X, c) = 0\) |
| Scalar | \(\text{Cov}(aX, bY) = ab\,\text{Cov}(X,Y)\) |
| Additivity | \(\text{Cov}(X+Z, Y) = \text{Cov}(X,Y) + \text{Cov}(Z,Y)\) |
5. Sample Correlation
Covariance depends on units (e.g. dollars vs. cents), making comparisons awkward. Correlation rescales it to be unit-free:
Key properties: - \(-1 \leq r_{xy} \leq 1\) always. - \(r = 1\): perfect positive linear relationship; \(r = -1\): perfect negative. - \(r = 0\): no linear relationship (could still be a nonlinear one).
Worked Example
Australian Wage Data (Small Sample)
Suppose we have hourly wages (AUD) and years of education for five randomly selected Australian workers:
| Worker \(i\) | Wage \(w_i\) | Education \(e_i\) |
|---|---|---|
| 1 | 25 | 10 |
| 2 | 35 | 12 |
| 3 | 45 | 14 |
| 4 | 30 | 11 |
| 5 | 55 | 16 |
Step 1 — Sample means:
Step 2 — Sample variance of wages:
| \(i\) | \(w_i - \bar{w}\) | \((w_i - \bar{w})^2\) |
|---|---|---|
| 1 | \(-13\) | \(169\) |
| 2 | \(-3\) | \(9\) |
| 3 | \(7\) | \(49\) |
| 4 | \(-8\) | \(64\) |
| 5 | \(17\) | \(289\) |
| Sum | \(0\) | \(580\) |
Step 3 — Sample covariance:
| \(i\) | \(w_i - \bar{w}\) | \(e_i - \bar{e}\) | Product |
|---|---|---|---|
| 1 | \(-13\) | \(-2.6\) | \(33.8\) |
| 2 | \(-3\) | \(-0.6\) | \(1.8\) |
| 3 | \(7\) | \(1.4\) | \(9.8\) |
| 4 | \(-8\) | \(-1.6\) | \(12.8\) |
| 5 | \(17\) | \(3.4\) | \(57.8\) |
| Sum | \(116.0\) |
Step 4 — Variance of education (similar calculation gives \(s_e^2 = 5.3\), \(s_e \approx 2.302\)).
Step 5 — Correlation:
Hmm — that exceeds 1, which is impossible! Let us recompute \(s_e^2\):
Deviations: \(-2.6,\,-0.6,\,1.4,\,-1.6,\,3.4\); squares: \(6.76,\,0.36,\,1.96,\,2.56,\,11.56\); sum \(= 23.2\); \(s_e^2 = 23.2/4 = 5.8\); \(s_e \approx 2.408\).
The correlation is essentially 1.00, indicating an almost perfect positive linear relationship between wages and years of education in this (deliberately constructed) small dataset — workers with more education earn proportionally more.
Common Misconception
"A high correlation between \(X\) and \(Y\) means \(X\) causes \(Y\)."
Correlation measures linear association, not causation. Ice cream sales and drowning deaths are positively correlated — both rise in summer. The confounder is warm weather. In Australian economics, house prices in Sydney correlate strongly with GDP growth, but neither directly causes the other; both are driven by shared macroeconomic cycles.
Establishing causation requires additional assumptions — randomised experiments, natural experiments, or structural econometric models (the central theme of the rest of this course). Correlation is simply the first diagnostic step.
Practice Prompts
Prompt 1
Given the following 4 observations of GDP growth rates (%) for Australia: 2, 4, 3, 7. (a) Compute the sample mean \(\bar{x}\). (b) Compute the sample variance \(s^2\). (c) Compute the z-score for the observation \(x = 7\).
Answer
**(a) Sample mean:** $$\bar{x} = \frac{2+4+3+7}{4} = \frac{16}{4} = 4\%$$ **(b) Sample variance:** | $x_i$ | $x_i - \bar{x}$ | $(x_i - \bar{x})^2$ | |:---:|:---:|:---:| | 2 | $-2$ | $4$ | | 4 | $0$ | $0$ | | 3 | $-1$ | $1$ | | 7 | $3$ | $9$ | | **Sum** | $0$ | $14$ | $$s^2 = \frac{14}{3} \approx 4.67 \quad (\%)^2$$ **(c) Z-score for $x=7$:** $s = \sqrt{4.67} \approx 2.16$ $$z = \frac{7 - 4}{2.16} \approx 1.39$$ Interpretation: 7% growth is about 1.4 standard deviations above the mean.Prompt 2
Prove algebraically that \(\sum_{i=1}^{n}(x_i - \bar{x}) = 0\).
Answer
$$\sum_{i=1}^{n}(x_i - \bar{x}) = \sum_{i=1}^{n} x_i - \sum_{i=1}^{n}\bar{x} = \sum_{i=1}^{n} x_i - n\bar{x}$$ By definition, $\bar{x} = \frac{1}{n}\sum x_i$, so $n\bar{x} = \sum x_i$. Therefore: $$\sum_{i=1}^{n} x_i - n\bar{x} = \sum_{i=1}^{n} x_i - \sum_{i=1}^{n} x_i = 0 \qquad \blacksquare$$Prompt 3
You have income \(Y_i\) (AUD thousands) and consumption \(C_i\) (AUD thousands) for three households:
| \(i\) | \(Y_i\) | \(C_i\) |
|---|---|---|
| 1 | 50 | 40 |
| 2 | 70 | 58 |
| 3 | 90 | 70 |
Compute the sample covariance \(s_{YC}\) and the sample correlation \(r_{YC}\). Interpret the result.
Answer
$\bar{Y} = 70$, $\bar{C} = 56$. | $i$ | $Y_i - \bar{Y}$ | $C_i - \bar{C}$ | Product | $(Y_i-\bar{Y})^2$ | $(C_i-\bar{C})^2$ | |:---:|:---:|:---:|:---:|:---:|:---:| | 1 | $-20$ | $-16$ | $320$ | $400$ | $256$ | | 2 | $0$ | $2$ | $0$ | $0$ | $4$ | | 3 | $20$ | $14$ | $280$ | $400$ | $196$ | | **Sum** | | | $600$ | $800$ | $456$ | $$s_{YC} = \frac{600}{2} = 300$$ $$r_{YC} = \frac{300}{\sqrt{800} \times \sqrt{456}} = \frac{300}{\sqrt{364800}} \approx \frac{300}{604.0} \approx 0.497$$ **Interpretation:** There is a moderate positive correlation ($r \approx 0.50$) between income and consumption — higher-income households spend more, but the relationship is not perfectly linear. This is consistent with Keynesian consumption theory.Further Resources
- Wooldridge, Introductory Econometrics (7th ed.): Appendix A (Summation Notation), Appendix B.1 (Random Variables)
- Khan Academy — Statistics and Probability: khanacademy.org/math/statistics-probability — "Summarizing quantitative data" module
- R Reference:
mean(),var(),cov(),cor()in base R;psych::describe()for full descriptive summaries - ECMT1020 Notes (USYD): Weeks 1–3 — Random variables, sample estimators, and correlation