Consider a random sample $(L_1, R_1), . . . ,(L_n, R_n)$ of eye pressure measurements in the left and right eyes of n patients.
a) Suppose that $L_j$ and $R_j$ are independent and normally distributed with unknown mean µ and known variance $\sigma^2$, for j = 1, . . . , n. Obtain the likelihood function of the sample and derive the maximum likelihood estimator of µ. Construct a 100(1 − α)% confidence interval for µ.
b) Suppose now that for each j, we do not assume that Lj and Rj are independent. Instead we assume that Lj = Mj + Dj and Rj = Mj − Dj for independent Mj and Dj , where Mj ∼ N(µ, $\sigma_1^2$) and Dj ∼ N(0, $\sigma_1^2$), with $\sigma_1^2$ $\geq$ $\sigma_2^2$ and $\sigma_1^2 + \sigma_2^2 = \sigma^2$. Here µ is unknown but $\sigma_1^2$ and $\sigma_2^2$ are known. Find the distribution of $L_j$ and the distribution of $R_j$ , and find the maximum likelihood estimator for µ in terms of $L_j$ and $R_j$.
c) Assuming that $\sigma_1^2 = \sigma_2^2 = \sigma^2/2$, find a 100(1 − α)% confidence interval for µ.
d) Assuming now that $\sigma_1^2 > \sigma_2^2$, state qualitatively, without calculations, how this change will affect (i) the maximum likelihood estimator and (ii) the confidence interval.
So far I have attempted part (a) and obtained the maximum likelihood estimator $\hat{\mu}=\sum (L_j + R_j)/2n$ but I'm stuck on the later parts. Any help would be greatly appreciated!
Here's what I would do. To find your confidence interval in (a), first set $$\bar{X}:=\frac{1}{2n}\sum_{j=1}^{n}(L_j+R_j)$$ and observe how $\bar{X} \sim N(\mu,\frac{\sigma^2}{2n})$. This implies $\frac{\bar{X}-\,u}{\sigma/\sqrt{2n}} \sim N(0,1)$ and so $$-z_{\alpha/2}<\frac{\bar{X}-\,u}{\sigma/\sqrt{2n}}<z_{\alpha/2} \iff \mu\in\Big(\bar{X}-z_{\alpha/2}\frac{\sigma}{\sqrt{2n}},\bar{X}+z_{\alpha/2}\frac{\sigma}{\sqrt{2n}}\Big)$$ Replacing $\bar{X}$ with your MLE for $\mu$ should provide you with your desired confidence interval. For (b) its relatively easy to see how $L,R \sim N(\mu,\sigma^2)$. You can get the joint density for $(L,R)$, namely $f_{LR}$, by using the fact that $$f_{LR}(l,r)=f_{MD}\Bigg(\frac{l+r}{2},\frac{l-r}{2}\Bigg)\left|\frac{\partial(m,d)}{\partial(l,r)} \right|=\frac{1}{2}f_{M}\Bigg(\frac{l+r}{2}\Bigg)f_{D}\Bigg(\frac{l-r}{2}\Bigg)$$ Now that we have the joint density for $(L,R)$ we can compute the likelihood function and get the MLE for $\mu$. Let me know how these suggestions serve you.