So I have this question:
The measurement errors made by a measuring device operating under two different conditions, $i=1$ and $i=2$ are Normally distributed with mean $0$ and variance $\sigma_i^2$ (Assuming independence)
a) Based on a simple sample of measurements made under condition $1,$ the measurement errors $x_1, x_2,\ldots,x_n$ were recorded. Similarly under condition 2, measurement errors $y_1, y_2, \ldots, y_m$ were recorded. Write down the combined likelihood in terms of $\sigma_1$ and $\sigma_2$. If $\sigma_1=\sigma_2$, write down the combined likelihood in terms of $\sigma_1$
This is what I have written so far but I am not sure if it is correct
$$L(\sigma_1,\sigma_2)=\prod_{i=1}^n\frac{1}{\sigma_1\sqrt{2\pi}}\exp\left(-\frac{x_i^2}{2\sigma_1^2}\right)\prod_{j=1}^m\frac{1}{\sigma_{2} \sqrt{2\pi}} \exp\left(-\frac{y_{j}^2}{2\sigma_{2}^2}\right)$$
And I am not too sure of how to go on from here
b)Derive the test statistics for a maximum likelihood ratio test of $H_0$: $\sigma_1 = \sigma_2$ versus $H_1$: $\sigma_1\neq\sigma_2$
The MLE for $\sigma_1$ and $\sigma_2$ over the entire parameter space are $$\sqrt\frac{\sum_{i=1}^n {x_i}^2}{n}$$ and $$\sqrt\frac{\sum_{j=1}^m {y_j}^2}{m}$$
For this part of the question I worked out the maximum likelihood estimator (MLE) for $\sigma_1$ as $$\hat\sigma_1=\sqrt{\frac{\sum_{i=1}^n {x_i}^2 + \sum_{j=1}^m {y_j}^2}{n+m}}$$ I then substituted everything into the formula and got $$\lambda=\frac{\sqrt{\frac{\sum_{i=1}^n {x_i}^2 + \sum_{j=1}^m {y_j}^2}{n+m}}}{\sqrt\frac{\sum_{i=1}^n {x_i}^2}{n} + \sqrt\frac{\sum_{j=1}^m {y_j}^2}{m}}$$
But I am not sure how to simplify it further so as to derive the test statistics from it