Let $\:X_{i}=(\theta/2)t_{i}^2+\epsilon_{i}, i=1,...,n $ where $\epsilon_{i}$ are independent normal random variables with mean 0 and known variance $\sigma^2$. Using a pivot based on the MLE $(2\sum_{i}^{n}(t_{i}^2*X_{i}))/\sum_{i}^n t_{i}^4 $ of $\theta $ find a fixed length level $(1-\alpha)$ confidence interval for $\theta$:
I am confused about how to get the MLE. My thought is to use $(MLE- \theta)/\sigma$ to get the confidence interval, but I do not know how to proceed the question.
Assuming $X_i=\frac{\theta}{2}t_i^2+\varepsilon_i$ where $\varepsilon_i$'s are i.i.d $N(0,\sigma^2)$ and $t_i,\sigma$ are fixed for all $i=1,2,\ldots,n$.
In other words, $X_i$'s are independent $N\left(\frac{\theta}{2}t_i^2,\sigma^2\right)$ variables, $i=1,2,\ldots,n$.
The MLE of $\theta$ is given by $$\hat\theta=\frac{2\sum_{i=1}^n X_i t_i^2}{\sum_{i=1}^n t_i^4}$$
Now,
$$2 X_it_i^2 \sim N\left(\theta t_i^4,4\sigma^2t_i^4\right)\,,\,\text{ independently for all }i$$
This implies $$2\sum_{i=1}^n X_it_i^2=\hat\theta\sum_{i=1}^n t_i^4\sim N\left(\theta \sum_{i=1}^n t_i^4,4\sigma^2\sum_{i=1}^nt_i^4\right)$$
Standardizing this variable, you have a pivot
$$\frac{\sum_{i=1}^n t_i^4(\hat\theta-\theta)}{2\sigma\sqrt{\sum_{i=1}^n t_i^4}}=\frac{\sqrt{\sum_{i=1}^n t_i^4}}{2\sigma}(\hat\theta-\theta)\sim N(0,1)$$
You can now derive a confidence interval.