Let $X_i$ be i.i.d. r.v. with $\sim Exp(\lambda)$ and for n to $\infty$ $$\sqrt{n} \cdot (Z_n - \theta) \overset{d}{\to} N(0,\sigma^2) $$
$\tilde{\lambda} = - ln(\bar{Y_n})$, where $Y_i = 1 \{X_i = 0\}$ -> Bernoulli
I am looking to find the asymptotic variance $V(\hat{\lambda})$
I can see that Y has the property of a Bernoulli $Y_i = 1_{(X_i=0)} = $ 1 if $(X_i = 0)$ and $= 0$ if $(X_i = 1)$
Applying the delta method (via Tayler expansion)
$$\sqrt{n}(g(Z_n)-g(\theta)) \overset{d}{\to} N(0, g'(\theta)^2 * \sigma^2)$$
$$\sqrt{n}(g(\bar{Y_n})-g(E(Y))) \overset{d}{\to} N(0, g'(E(Y))^2 * Var(Y))$$
Via the exponential pdf I calculate $P(X=0) = \lambda*e^{-\lambda*0} = \lambda$, which I will take as my $E[Y]$ and plug this term into the Bernoulli variance formula $p*(1-p)$: $$Var(Y) = \lambda*(1-\lambda) = \lambda-\lambda^2$$
Now for the term $g'(E[Y])^2$ I take the derivative of $g(E[Y]) = -ln(E[Y])$ and square it: $g'(E[Y])^2 = \frac{1}{E[Y]^2}$
With $E[Y] = \lambda$ calculated before, I get $g'(E[Y])^2 = \frac{1}{\lambda^2}$
Plugging both terms together I get $$V(\hat{\lambda}) = \frac{1}{\lambda^2} * (\lambda-\lambda^2) = \frac{1}{\lambda}-1$$
Since this solution is incorrect, where am I mistaken?