Notes: ECON 141 / Econometrics

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These notes are for Spring 2022’s Economics 141: Econometrics.

Midterm 1



\(\mathbb{E}[X] = \int xf_x(x) dx\) \(\mathbb{E}[aX + b] = a\mathbb{E}[X] + b\) \(Var(X) := \mathbb{E}[X^2] - \mathbb{E}[X]^2\) \(Cov(X, Y) := \mathbb{E}[XY] - \mathbb{E}[X] \mathbb{E}[Y]\) \(\mathbb{E}[XY] = \int \int xy f_{x, y} (x, y) dx dy\) \(\mathbb{P}(Y = y | X = x) = \frac{\mathbb{P}(Y = y, X = x)}{\mathbb{P}(X = x)}\) \(Var(X + Y) = Var(X) + Var(Y) + 2Cov(X, Y)\)

\[X \sim \mathcal{N}(\mu, \sigma^2) \implies f_X(x) = \frac{1}{\sqrt{2\pi} \sigma} \text{exp}(-\frac{(x - \mu)^2}{2\sigma^2})\]

For independent variables, $Cov(X, Y) = 0$. But for $Y = X^2$ and $X \sim U[-1, 1]$, we have $Cov(X, Y) = 0$ despite dependence.


Suppose we use $\hat{\mu}$ as an estimator for r.v. $X$. If $\mathbb{E}{\hat{\mu}} = \mathbb{E}[X]$, then $\hat{\mu}$ is an unbiased estimator.

The Law of Large Numbers dictates that for $\hat{\mu}n = \frac{1}{n} \sum{i=1}^n x_i$ with $n$ samples $x_i$, we have:

\[\mathbb{P}(\hat{\mu}_n - \mu) > \epsilon) \to 0\]

So the average $\hat{\mu}_n$ is a consistent estimator of $\mu$, i.e. it becomes asymptotically close. The variance is $Var(\hat{\mu}_n) = \frac{Var(X)}{n} = \sigma^2_x / n$, which approaches $0$ as $n \to \infty$.

The z-score is:

\[z = \frac{\bar{x} - \mu}{\sigma_x / \sqrt{n}}\]

According to the Central Limit Theorem, as $n \to \infty$, the distriubution of the z-score is $\mathcal{L}(u) \sim \mathcal{N}(0, 1)$.

The difference between the estimators $\hat{\mu} = \bar{x}$ and $\check{\mu} = x_1$ is different weights, $w_1 = (1, 0, …, 0)$ and $w_{\bar{x}} = (\frac{1}{n}, \frac{1}{n}, …, \frac{1}{n})$. It turns out that $\hat{\mu} = \bar{x}$ is the most efficient estimator among weigted averages, as measured by variance.

Notice that since we use the samples to estimate $\bar{x}$, there is one fewer degree of freedom among the demeaned samples $x_i - \bar{x}$ (since knowing $n-1$ of them tells you the value of the $n$th), so we compute:

\[X_x^2 = \frac{1}{n-1} \sum_{i=1}^n (x_i - \bar{x})^2\]

Intuitively, if we have fewer samples, then the mean of our sample will be a biased estimator of the actual mean, underestimating the variance.

Hypothesis Testing

Suppose our significance level is 0.5. The acceptance region is collection of $\bar{x}$s such that $\bar{x} \in [\mu - 1.96 * \sigma_x, \mu + 1.96 * \sigma_x]$. For be correct, we should accept true hypotheses $(A, T)$ and reject false hypotheses $(R, F)$, or else we have a type 1 error $(R, T)$ or type 2 error $(A, F)$. The power of the test $\alpha$ is derived from $\mathbb{P}(\text{type 1 error}) \leq \alpha$.

We set $t = \sqrt{n} (\frac{\bar{x} - \mu_0}{\hat{\sigma}_x})$. As $n \to \infty$, $\hat{\sigma}_x \to \sigma_x$, and $t \sim N(0, 1)$. For the one-sided case, the required confidence level is usually $z(x) > 1.65$.


Our estimator is $Y_i = \beta_0 + \beta_1 X_i + u_i$, with goal of minimizing the sum of squared residuals $u_i$:

\[\min_{b_0, b_1} \sum_{i=1}^n [Y_i - (b_0 + b_1 X_x)]^2\]

Deriving with respect to both $b_0$ and $b_1$, we have $0 = \frac{1}{n} \sum_{i=1}^n \hat{u}i$, which means the sample average of residuals must be 0, and $\frac{1}{n} \sum{i=1}^n X_i \hat{u}_i$, which means that the coveraince between residuals and $X_i$s must be 0. It yields:

\[\hat{\beta}_1 = \frac{s_{XY}}{s_X^2}\] \[\hat{\beta_0} = \bar{Y} - \hat{B}_1 \bar{X}\]

Not ehtat residuals and errors are different, since the residuals $\hat{u_i} := Y_i - (\hat{\beta}_0 + \hat{\beta}_1 X_i)$ are estimators of the true errors $u_i := Y_i - (\beta_0 + \beta_1 X_i)$.

The SSR or sum of squared residuals is how much of the data can’t be explained by linear regression. The ESS or explained sum of squares is how much can be explained. The TSS or total sum of squares is the total spread of the data.

\[TSS =\sum_{i=1}^n (Y_i - \bar{Y})^2 = \sum_{i=1}^n \hat{u_i}^2 + \hat{\beta_1}^2 \sum_{i=1}^n (X_i - \bar{X})^2 = ESS + SSR\]

Finally, $R^2 = \frac{ESS}{TSS}$ and represents the proportion of the variation in $Y$ explained by the variation in $X$.

Homoskedastic errors have $Var(u_i) = \sigma_u^2$. Under heteroskedasticity, $\hat{\beta}_1$ will be consistent, unbiased, and asymptotically normal but not efficent. We use a different formula for standard error which usually produces higher values.

Omitted variable bias occurs when we have a omitted variable $Z$ and the coefficient $\beta_2 \neq 0$ and $\hat{sigma}{XZ} \neq 0$, resulting in a bias of $\hat{B_1} = \beta_1 + \beta_2 \frac{\hat{\sigma}{XZ}}{\hat{\sigma}X^2} + \frac{\hat{\sigma}{Xu}}{\hat{\sigma}_X^2}$.