The Exact Confidence Interval for an MLE of a Gamma Distribution  Above here is the information I've been given for one of my seminar questions, so far I have calculated the fisher information and from there I computed the asymptotic distribution for $$\hat{\lambda}$$ is:

$$\lambda_n = N\left(\lambda,\frac{1}{nI(\lambda)}\right) = N\left(\lambda, \frac{\lambda^2}{dn}\right)$$

After that I derived the 90% confidence interval for λ as:

$$T_{1,2}=\hat{\lambda} \pm \frac{z_{\alpha/2}}{\sqrt{nI(\lambda)}}= \frac{dn}{x} \pm 1.64 \frac{\lambda}{\sqrt{nd}}$$

And from here I need to find the exact 90% confidence interval but this is where I'm stuck. Can anyone provide any assistance?

• You cannot find the exact CI from an asymptotic CI; you have to start from scratch. You are given a hint on what to do in the last line of your question. That is pretty much the answer. – StubbornAtom Mar 29 at 6:10
• So how exactly would I do this, would I need to calculate the likelihood of this new function and go from there? @StubbornAtom – king Mar 29 at 10:20
• Can you verify that $2\lambda\sum X_i$ has a chi-square distribution? If you can, then this is your pivotal quantity from which the CI follows. – StubbornAtom Mar 29 at 10:25