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Let $X_1,\ldots,X_n$ be an independent sample from random variable with pdf $$ f(x)= \frac 1 \theta e^{-x/\theta}, \qquad x ≥ 0 $$ How to check that the estimator $\theta(X) = \dfrac{X_1 + X_2} 2$ is consistent?

2 Answers 2

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Use definition of consistent estimator along with the fact that sum of exponentially distributed r.v.s has Gamma distribution.

In other words, you want to prove that $\mathbb{P}[\sum_{i=1}^{n}X_{i}\geq n\cdot\epsilon]\rightarrow0$ as $n\rightarrow\infty$. Since $\sum_{i=1}^{n}X_{i}$ has Gamma distribution with parameters $n$ and $\theta$, you want to show that $\Gamma(n,\frac{n\epsilon}{\theta})/\Gamma(n)$ tends to $1$ for $\epsilon>0$ and $\theta>0$.

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    $-1$ because you didn't mention that the proposed estimator, $(X_1+X_2)/2,$ is NOT consistent. $\qquad$ – 2017-01-31
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    @Michael Hardy +1 for explaining downvote. But, I believe MSE answers are not about getting the answer correct providing OP with all the details. I provided OP with a way to work with sum of exponentially distributed r.v.s and how to prove consistency of estimators. In fact how can an estimate be consistent if it throws away all but 2 observations. – 2017-01-31
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Note that you estimator does not depend on the sample size $n$, hence

$$ Var(\theta(X)) =\frac 1 4(\theta^2 + \theta^2)=\frac{\theta^2}{2}. $$ The variance (MSE, as this estimator unbiased) does not converges to $0$. Generally it is not sufficient condition to deduce that it is not consistent, but it gives the intuitive understanding that the probability mass does not concentrate at some point (parameter). More formally, for all $\epsilon > 0$,

$$\lim_{n\to \infty} P(| \theta(X) - \theta|\ge \epsilon) = \lim_{n\to\infty}\left( P(X_1+X_2\ge2(\theta+\epsilon))+ P(X_1+X_2\le 2(\theta - \epsilon))\right)$$

$$=P(X_1+X_2 \notin[2(\theta \pm \epsilon)]) = 1-\int_{2(\theta - \epsilon)}^{2(\theta + \epsilon)}f_{X_1+X_2}(x)dx \approx1.$$ In the last step you can use the fact $X_1+X_2 \sim \Gamma(2, 1/\theta)$, thus for small enaugh $\epsilon$, this integral will equal almost $0$.