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$X_1,X_2, \cdots , X_n$ be an iid sample from an exponential distribution with unknown parameter $\theta$

I need to show that $lim _{n \to \infty}$ Pr$(| (1/ \bar{X_n} ) - \theta | \ge \epsilon) = 0$. I am basically testing the consistency of the estimator $(1/ \bar{X_n} )$.

I know that Law of Large numbers $lim _{n \to \infty}$ Pr$(| \bar{X_n} - (1/\theta) | \ge \epsilon) = 0$ can be used but I am not able to figure out how?

Also I am also not able to use Markov's inequality $ Pr(| (1/ \bar{X_n} ) - \theta | \ge \epsilon) \le \cfrac{E[| (1/ \bar{X_n} ) - \theta |]}{\epsilon}$ (If at all it is to be used).

But I want to learn to use both of the above methods (LLN and Markov's inequality) to solve the above question. Any Help/hint would be appreciated.

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Here are some quick hints:

I suppose you know that $E(1/\bar X_n) = n\theta/(n-1),$ so $T_n = 1/\bar X_n$ is a biased estimator of $\theta$. Of course, $n/(n-1) \rightarrow 1$ with increasing $n$, so it is asymptotically unbiased.

The LLN shows directly that $\bar X_n$ converges in distribution (or probability) to $\mu = 1/\theta.$ It seems possible, but a little messy, to deal with reciprocals here. (I have dim memories of working through something similar once, but not recently, so no guarantee.)

Perhaps an approach similar to yours with Markov's inequality is easier to undertstand: $$ 0 \leq P\{|(1/\bar X_n) - \theta| \ge \epsilon\} \le E[(1/\bar X_n) - \theta)]^2/\epsilon^2 = e_n.$$ Then show that $e_n \rightarrow 0$. The numerator of $e_n$ is the variance of $T_n =1/\bar X_n$ plus its squared bias and $T_n$ has an inverse gamma distribution with a known variance (e.g., Wikipedia). [Note, this may be a little round about because convergence in squared mean implies convergence in probability, but that is not what you asked.]

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