1
$\begingroup$

I have

$$X_1 \dots X_n \sim f_\theta(x) = \begin{cases} \exp(\theta-x) & x\geq\theta\\ 0& otherwise \end{cases}$$

And I have the estimator $\hat\theta_n = X_{(1)}=\min\{X_1 \dots X_n\}$

I have found the CDF and PDF of the estimator

$$F_{\hat\theta_n}(x) = 1-e^{n(\theta-x)}$$ $$f_{\hat\theta_n}(x) = ne^{n(\theta-x)}$$

Now I want to test consistency so for $\epsilon >0$

$$\lim_{n\to\infty}\Pr(|\hat\theta_n - \theta| < \epsilon) = 1$$

Then we have $$ \Pr(-\epsilon < \hat\theta_n - \theta < \epsilon)$$ $$ \Pr(\theta-\epsilon < \hat\theta_n < \theta + \epsilon)$$

$= 1-e^{-n\epsilon}-1+e^{n\epsilon}$

Which goes to $+\infty$ as $n\to\infty$. Did I do everything correctly?

What do I conclude from this? Is the estimator consistent or not? Is this the same as the probability going to $1$?

EDIT: I believe I have found my error. The CDF should be

$$F_{\hat\theta_n}(x) = 1-e^{n(\theta-x)} 1_{x \geq \theta}$$

Then you get the probability: $1-e^{-n\epsilon}$ which goes to $1$ as expected.

$\endgroup$
1
  • 2
    $\begingroup$ The fact is, the probability that $\hat{\theta}_n$ is less than $\theta$ is $0$. $\endgroup$
    – Zhanxiong
    Commented Oct 20, 2017 at 2:40

2 Answers 2

2
$\begingroup$

It may be more convenient to work with $P(|\hat{\theta}_n - \theta| \geq \varepsilon)$ (the reason will be clear as the computation flows): \begin{align*} & P(|\hat{\theta}_n - \theta| \geq \varepsilon) \\ = & P(|X_{(1)} - \theta| \geq \varepsilon) = P(X_{(1)} - \theta \geq \varepsilon) \\ = & P(X_{(1)} \geq \theta + \varepsilon) \\ = & P(X_1 \geq \theta + \varepsilon, \ldots, X_n \geq \theta + \varepsilon) \\ = & P(X_1 \geq \theta + \varepsilon)^n \quad \text{by the i.i.d. assumption} \\ = & \left(\int_{\theta + \varepsilon}^\infty e^{\theta - x} dx \right)^n \\ = & e^{-n\varepsilon} \to 0 \end{align*} as $n \to \infty$. This shows that $\hat{\theta}_n$ is a consistent estimate of $\theta$.

$\endgroup$
0
$\begingroup$

I think you have to write $\lim_{n\to\infty}\Pr(|\hat\theta_n - \theta| < \epsilon) =\lim_{n\to\infty}\Pr(\theta < \hat\theta_n < \theta + \epsilon)$ since $\theta $ is the minimum value of $\hat{\theta}$.

Then you can show that \begin{align*} \lim_{n\to\infty}\Pr(\theta < \hat\theta_n < \theta + \epsilon) &= lim_{n\to\infty}\left \{F_{\hat{\theta}}(\theta+\epsilon)-F_{\hat{\theta}}(\theta)\right \} \\&=lim_{n\to\infty}\left \{[1-e^{n[\theta-(\theta+\epsilon)]}]-[1-e^{n(\theta-\theta)}]\right \}\\&=lim_{n\to\infty}\left \{1-e^{-n\epsilon}-1+1\right\}\\&=1 \end{align*}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.