How to show $\frac{1}{\bar{X}}$ is a consistent estimator of $\theta$? Let $X_1,...,X_n$ be iid where $X_i \sim \text{exponential}(\theta)$ 
I am trying to show that $\frac{1}{\bar{X}}$ is a consistent estimator of $\theta$. 
As I understand it, it is sufficient to show that  $$\lim_{n\rightarrow \infty} \text{Bias}_\theta(\hat{\theta},\theta) = 0$$
$$\lim_{n\rightarrow \infty} \text{Var}_\theta(\hat{\theta}) = 0$$
I already showed that $\lim_{n\rightarrow \infty} \text{Bias}_\theta(\hat{\theta},\theta) = 0$
I calculated the variance to be $$\frac{\theta^2}{(n-1)(n-2)}-\left(\frac{n\theta}{(n-1)}\right)^2$$
But the limit of this is $-\theta^2$ which isn't zero.
Did I make a mistake? If not, what is the intuition behind why the MLE isn't consistent? 
 A: Your variance isn't correct.  You are missing a factor of $n^2$ in the numerator of the first term.
If $\bar X^{-1}$ is your estimator for $\theta$ for the sample $$X_1, \ldots, X_n \sim \operatorname{Exponential}(\theta)$$ with common PDF $$f_X(x) = \theta e^{-\theta x}, \quad x > 0, \quad \theta > 0,$$ then the PDF of the sum $$Y = \sum_{i=1}^n X_i = n\bar X \sim \operatorname{Gamma}(n, \theta)$$ with PDF $$f_Y(y) = \frac{\theta^n y^{n-1} e^{-\theta y}}{\Gamma(n)}, \quad n > 0, \theta > 0, y > 0,$$ hence the distribution of $\bar X$ is $\bar X \sim \operatorname{Gamma}(n, n\theta)$ with PDF $$f_{\bar X}(x) = \frac{(n\theta)^n x^{n-1} e^{-n\theta x}}{\Gamma(n)}, \quad n > 0, \theta > 0, x > 0.$$  The distribution of $W = \bar X^{-1}$ is therefore inverse gamma:  $$f_W(w) = \frac{1}{w^2} \cdot \frac{(n\theta)^n w^{-(n-1)} e^{-n\theta / w}}{\Gamma(n)} = \frac{(n\theta)^n e^{-n\theta /w}}{w^{n+1} \Gamma(n)}.$$  If we accept that the integral of $f_W(w)$ for $w > 0$ is $1$ for any legitimate choice of $n, \theta$, then we can compute the $k^{\rm th}$ raw moments as follows:  $$\begin{align*} \operatorname{E}[W^k] &= \int_{w=0}^\infty w^k f_W(w) \, dw \\ &= \frac{(n\theta)^n}{\Gamma(n)} \int_{w=0}^\infty \frac{e^{-n\theta/w}}{w^{n-k+1}} \, dw \\ &= \frac{(n\theta)^k \Gamma(n-k)}{\Gamma(n)} \int_{w=0}^\infty \frac{(n\theta)^{n-k} e^{-n\theta/w}}{w^{n-k+1} \Gamma(n-k)} \, dw \\ &=  \frac{(n\theta)^k \Gamma(n-k)}{\Gamma(n)}, \end{align*} $$ since the last integral is for an inverse gamma PDF with parameters $n-k$ and $n\theta$, which equals $1$.  Therefore, for $k = 1$ we recover the expectation $$\operatorname{E}[\bar X^{-1}] = \frac{n\theta}{n-1}, \quad n > 1,$$ and for $k = 2$ we get $$\operatorname{E}[(\bar X^{-1})^2] = \frac{n^2 \theta^2}{(n-1)(n-2)}, \quad n > 2.$$  Therefore the variance is $$\operatorname{Var}[\bar X^{-1}] = \frac{n^2 \theta^2}{(n-1)^2 (n-2)},$$ and the asymptotic variance as $n \to \infty$ is simply $0$, as desired.
