= Solution
For the mean parameter $\theta=1/\lambda>0$, the <log-likelihood> is $\ell(\theta)=-n\log\theta-\sum_iY_i/\theta$. Its derivative is $n(\bar Y-\theta)/\theta^2$, positive below $\bar Y$ and negative above it. Therefore $\boxed{\widehat\theta=\bar Y}$ is the unique <maximum-likelihood estimator>.
The <exponential distribution> has mean $\theta$ and <variance> $\theta^2$. The <central limit theorem> gives the pivot
$$
T_n=\sqrt n\,\frac{\bar Y-\theta}{\theta}\ \Rightarrow\ N(0,1).
$$
Let $z=\Phi^{-1}(1-\alpha/2)$ for fixed $0<\alpha<1$. For $n>z^2$, the event $-z<T_n<z$ is exactly
$$
\frac{\bar Y}{1+z/\sqrt n}<\theta<\frac{\bar Y}{1-z/\sqrt n}.
$$
Both 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 $n\leq z^2$, the displayed finite positive upper endpoint is not valid, although the large-sample claim is unaffected.
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