I have been reading Davidson's Econometric theory and methods and I cannot understand one particular paragraph about the distribution of a test statistic.In the paragraph below he says that under the alternative hypothesis if the true $\beta$ is equal to $\beta_1$ then $\hat b= \beta_1+\hat\gamma$. Then he defines $z=(\beta_1+\hat\gamma-\beta_0)/(\sigma^2/\nu)$ and finds the mean and variance as given by 4.04. Why is he doing that? I understand how he gets the mean and variance but I don't get why is he doing that because at the end we dont get a standard normal variable but a variable with mean $\lambda$ and variance 1.enter image description here

I have been stuck here for a while so any help would be highly appreciated.

  • $\begingroup$ Without (4.01) not much can be said about what you posted. $\endgroup$
    – MathUser
    Commented May 16, 2016 at 7:59
  • $\begingroup$ @Ronaldo777, Are you sure that z is as define here? I've edited you post, but the original doesn't seem to be correct either... $\endgroup$ Commented May 16, 2016 at 15:54

2 Answers 2


Reading the book, $z=\frac{\hat\beta-\beta_0}{(\sigma^2/n)^{(1/2)}}$

The distribution of $z$ under $H_0$ will determine the critical values for a given level of significance $\alpha$, i.e. it will determine the probability of Type I errors (prob of rejecting the null, when null is true).

The distribution of $z$ under $H_1$ (4.04) will define the probability of Type II errors(not rejecting null when alternative is the true hypothesis).

The closer - the averages are close - the distributions of z under the null and the alternative are, the higher is the probability of Type II errors. In the picture below, we have several pdf for normal distributions. In blue, the distribution has zero mean; in red mean 1; and in yellow mean 2.

enter image description here

At the usual critical values of $1.96$ and $-1.96$ (for an $\alpha=0.05$), we can see a vertical line. When we're on the blue curve, i.e. under the null, we can calculate the prob of type I error, as the area to the right of the rightmost vertical line, to the left of the leftmost vertical line and below the blue line.

To calculate the type II error, we check in the interval between the vertical lines(not rejecting $H_0$ area), under the red or yellow curve, depending on the values that $\beta$ really takes. Notice that the area below such a curve will increase the closer the curve is to the one under the null distribution. In this case, the prob of type II error is greatest is when the alternative is the true model and the mean of the distribution is 1.

By deducing the distribution of z under the alternative, the authors want to give an idea of the type of rejection region one will get, when doing inference in this setting.


The model is

$$y_t = \beta +u_t,\;\; u_t\sim N(0, \sigma^2),\; t=1,...,n$$

and the sample is independent. The estimator is

$$\hat \beta = \frac1n\sum_{t=1}^n y_t = \frac1n\sum_{t=1}^n (\beta +u_t) = \beta + \frac1n\sum_{t=1}^n u_t$$

If $\beta = \beta_1$ (where $\beta_1$ is some value different than the $\beta_0$ we set as the null hypothesis), then the authors set $\frac1n\sum_{t=1}^n u_t \equiv \hat \gamma$ and so they write that under the alternative we have

$$\hat \beta = \beta_1 +\hat \gamma$$

Since the $u'$ are i.i.d normal their sum/average is also normal. Then the distribution of the estimator is

$$\hat \beta \sim N(\beta, \sigma^2/n)$$

Assume we set as our null hypothesis that $H_0:\beta = \beta_0$. and the alternative $H_1:\beta \neq \beta_0$.

Then we form the statistic (which is a function of the estimator, not the estimator itself)

$$z = \frac{\hat \beta - \beta_0}{\sigma/ \sqrt{n}}$$

The distribution of this statistic is (before specifying any hypothesis to test)

$$z \sim N\left(\frac{\beta - \beta_0}{\sigma /\sqrt{n}},1\right)$$

Assume that we pose as our null hypothesis that $H_0: \beta = \beta_0$. Then if the null hypothesis is true we get

$$z|_{H_0} \sim N\left(0,1\right)$$

If the alternative is true then we substitute for $\hat \beta$ to get

$$z|_{H_1} = \frac{\beta_1 +\hat \gamma - \beta_0}{\sigma /\sqrt{n}} = \frac{\beta_1 - \beta_0}{\sigma \sqrt{n}}+\frac{\hat \gamma }{\sigma /\sqrt{n}}$$

The first term is a constant, the second term is a standard normal r.v. (remember what $\hat \gamma$ stands for). So the distribution of the statistic under the alternative is

$$z|_{H_1} \sim N\left(\frac{\beta_1 - \beta_0}{\sigma /\sqrt{n}},1\right)$$

and the authors write

$$\lambda \equiv \frac{\beta_1 - \beta_0}{\sigma /\sqrt{n}}$$


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