 Solution

For the mean parameter $\theta =1/\lambda >0$, the log-likelihood is $\ell (\theta )=-n\log \theta -{\sum }_i{Y}_i/\theta$. Its derivative is $n(\bar{Y}-\theta )/{\theta }^2$, positive below $\bar{Y}$ and negative above it. Therefore $\boxed{\hat{\theta }=\bar{Y}}$ is the unique maximum-likelihood estimator.

The exponential distribution has mean $\theta$ and variance ${\theta }^2$. The central limit theorem gives the pivotLet $z={\Phi }^{-1}(1-\alpha /2)$ for fixed $0<\alpha <1$. For $n>z^2$, the event $-z<T_n<z$ is exactlyBoth denominators are then positive, so the algebra neither reverses the ordering nor changes the parameter domain. Taking probabilities and using continuity of the limiting normal distribution proves the exponential-mean confidence interval by pivot inversion has coverage probability tending to $\boxed{1-\alpha }$. This is inversion of the pivot with the true mean in its denominator; it is not the ordinary symmetric Wald interval centered at $\bar{Y}$. For a small sample with $n\le z^2$, the displayed finite positive upper endpoint is not valid, although the large-sample claim is unaffected.