The likelihood function is given by \begin{align*} L(\beta,\sigma^2) &= \prod_{i=1}^n \frac{1}{\sqrt{2\pi \sigma^2}}\exp\left(-\frac{(y_i – \beta x_i)^2}{2\sigma^2}\right) \ &= \frac{1}{(2\pi \sigma^2)^{n/2}}\exp\left(-\frac{1}{2\sigma^2}\sum_{i=1}^n(y_i – \beta x_i)^2\right) \end{align*}

The score function is given by

$\frac{\partial \ln L}{\partial \beta}=\frac{1}{\sigma^2} \sum_{i=1}^n x_i\left(y_i-\beta x_i\right)$

The information matrix is given by

$I\left(\beta, \sigma^2\right)=-\frac{\partial^2 \ln L}{\partial \beta^2}=\frac{n}{\sigma^2}$

(b) To find the unrestricted maximum likelihood estimator, we need to maximize the likelihood function with respect to both $\beta$ and $\sigma^2$. Taking the partial derivatives of the log-likelihood function with respect to $\beta$ and $\sigma^2$, respectively, we obtain \begin{align*} \frac{\partial \ln L}{\partial \beta} &= \frac{1}{\sigma^2}\sum_{i=1}^n x_i(y_i – \beta x_i) = 0 \ \frac{\partial \ln L}{\partial \sigma^2} &= -\frac{n}{2\sigma^2} + \frac{1}{2\sigma^4}\sum_{i=1}^n(y_i – \beta x_i)^2 = 0 \end{align*}

Solving the first equation for $\beta$, we get

$\hat{\beta}=\frac{\sum_{i=1}^n x_i y_i}{\sum_{i=1}^n x_i^2}$

Substituting $\hat{\beta}$ into the second equation and solving for $\sigma^2$, we get

$\hat{\sigma}^2=\frac{1}{n} \sum_{i=1}^n\left(y_i-\hat{\beta} x_i\right)^2$

The Wald test for the null hypothesis $H_0: \beta=0$ is given by

$W=\frac{\hat{\beta}}{\sqrt{\widehat{\operatorname{Var}}(\hat{\beta})}} \sim \mathcal{N}(0,1)$

where $\widehat{\text{Var}}(\hat{\beta})$ is the estimated variance of $\hat{\beta}$, which is given by

$\widehat{\operatorname{Var}}(\hat{\beta})=\frac{\hat{\sigma}^2}{\sum_{i=1}^n x_i^2}$

To solve the restricted maximization problem, we need to maximize the restricted likelihood function under the null hypothesis $\beta=0$. The restricted likelihood function is obtained by substituting $\beta=0$ in the likelihood function:

$L\left(\sigma^2\right)=\prod_{i=1}^n \frac{1}{\sqrt{2 \pi \sigma^2}} \exp \left(-\frac{y_i^2}{2 \sigma^2}\right)=\frac{1}{\left(2 \pi \sigma^2\right)^{n / 2}} \exp \left(-\frac{1}{2 \sigma^2} \sum_{i=1}^n y_i^2\right)$

To maximize $L(\sigma^2)$ with respect to $\sigma^2$, we take the derivative with respect to $\sigma^2$ and set it equal to zero:

$\frac{\partial L\left(\sigma^2\right)}{\partial \sigma^2}=-\frac{n}{2 \sigma^2}+\frac{1}{2 \sigma^4} \sum_{i=1}^n y_i^2=0$

Solving for $\sigma^2$, we obtain:

$\hat{\sigma}^2=\frac{1}{n} \sum_{i=1}^n y_i^2$

This is the restricted maximum likelihood estimator of $\sigma^2$ under the null hypothesis $\beta=0$.

The Lagrange Multiplier test involves constructing a test statistic based on the difference between the unrestricted maximum likelihood estimator of $\sigma^2$ and the restricted maximum likelihood estimator of $\sigma^2$ under the null hypothesis $\beta=0$. The test statistic is given by:

$L M=\frac{\left(\hat{\sigma}{\text {unrestricted }}^2-\hat{\sigma}{\text {restricted }}^2\right) n}{\hat{\sigma}_{\text {uurestricted }}^2}$

where $\hat{\sigma}{\text{unrestricted}}^2=\frac{1}{n}\sum{i=1}^n (y_i-\hat{\beta}x_i)^2$ is the unrestricted maximum likelihood estimator of $\sigma^2$ and $\hat{\beta}$ is the maximum likelihood estimator of $\beta$.

Under the null hypothesis, the test statistic $LM$ follows a $\chi^2$ distribution with 1 degree of freedom. We reject the null hypothesis at level $\alpha$ if $LM>\chi_{1,\alpha}^2$, where $\chi_{1,\alpha}^2$ is the $(1-\alpha)$-quantile of the $\chi^2$ distribution with 1 degree of freedom.