Suppose that you observe an IID random sample \(\{X_t\}_{t=1}^n\) of size \(n=5\) from \(\mathsf{Unif}\left[1,\alpha\right]\), where \(\alpha\in\mathbb{R}\) and \(\alpha>1\).

Below you will find an observed sample of size \(n=5\):

`## [1] 2.41 2.27 1.28 1.75 1.90`

Answer the following questions given the information above:

- Provide two moment conditions which will allow you to estimate \(\alpha\).
- Provide two estimates of \(\alpha\) given the observed sample and the moment conditions you provided.

- Two possible moment conditions which will allow estimation of \(\alpha\) are:

\[ \begin{eqnarray} \mathbb{E}\left(X_t-\frac{1+\alpha}{2}\right) &=& 0 \\ \mathbb{E}\left(X_t^2-\frac{\alpha^2+\alpha+1}{3}\right) &=& 0 \end{eqnarray} \]

To show that these two moment conditions hold, you have to calculate

\[ \mathbb{E}\left(X_t\right)=\int_1^{\alpha} t \cdot \frac{1}{\alpha-1} \,\,dt \qquad \mathbb{E}\left(X_t^2\right)=\int_1^{\alpha} t^2 \cdot \frac{1}{\alpha-1} \,\,dt \]

- An estimator based on the moment condition \[\mathbb{E}\left(X_t-\frac{1+\alpha}{2}\right)=0\] involves solving for \(\widehat{\alpha}\) in the equation

\[\frac{1}{n}\sum_{t=1}^n X_t - \frac{1+\widehat{\alpha}}{2}=0.\]

An estimate of \(\alpha\) based on the data is \(2*(1.922)-1=2.844\).

- An estimator based on the moment condition \[\mathbb{E}\left(X_t^2-\frac{\alpha^2+\alpha+1}{3}\right)=0\] involves solving for \(\widetilde{\alpha}\) in the equation \[\frac{1}{n}\sum_{t=1}^n X_t^2-\frac{\widetilde{\alpha}^2+\widetilde{\alpha}+1}{3}=0.\] Here we need solve a quadratic equation in \(\widetilde{\alpha}\), which is \(\widetilde{\alpha}^2+\widetilde{\alpha}+1-3*(3.85438)=0\). The two roots of the quadratic equation based on the data are \(2.788\) and \(-3.788\). Choose \(\widetilde{\alpha}=2.788\).

An acceptable solution is to calculate the variance.

- But this may take more time, especially when calculating an estimate.

Another issue is that it is

**incorrect**to write \[\mathbb{E}\left(X_t\right) = \frac{1}{n}\sum_{t=1}^n X_t\]One thing to note is that it is very likely that \(\widehat{\alpha}\neq \widetilde{\alpha}\).

Try writing down a GMM criterion function which exploits both moments. You can use a weighting matrix equal to the identity matrix.

In the next slide, you will see a very simple R program to implement one version of GMM.

The data used in the quiz was from \(\mathsf{Unif}\left[1,3\right]\). In this context, \(\beta^o = 3\).

```
# Create GMM criterion function
# Weighting matrix is identity matrix
gmm.obj <- function(alpha)
{
# Draw random numbers from uniform
x <- runif(5, 1, 3)
# Setup the GMM objective function
return((sum(x)-(1+alpha)/2)^2+(sum(x^2)-(alpha^2+alpha+1)/3)^2)
}
# Command to find minimizers/maximizers
# Valid for one-dimensional problems only
# Minimize is the default
# c(1, 100) is the interval [1,100]. This is to tell
# optimize where to search for a minimizer.
optimize(gmm.obj, c(1,100))
```

```
## $minimum
## [1] 8.46
##
## $objective
## [1] 102
```

Try adjusting the interval \([1,100]\) to \([1,4]\) or \([1,5]\), or some other interval. Check how sensitive the results are to the choice of where to look for a minimizer.

Try increasing the number of observations from \(n=5\) to \(n=200\) for example.

Adjust the code to use a weighting matrix of the form \[\left(\begin{array}{cc} 1 & 0\\ 0 & 4 \end{array}\right)\]

What do you notice? Try adjusting the interval where you look for a minimizer.