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In a class we looked at this example:

Let $X_1,...,X_n\sim U(0,\theta)$. Then the maximum likelihood function is

$\mathcal{L}(\theta) = \begin{cases} \dfrac{1}{\theta^{n}} & \text{if } \text{max}\{X_1,\dotsc,X_n\}\leq\theta \\ 0 & \text{otherwise} \end{cases}$

Only the fact that the maximum likelihood estimator $\hat{\theta}=\text{max}\{X_1,\dotsc,X_n\}$ is asymptotically unbiased and consistent was mentioned but I'm curious about why this is true. Could anyone help me to see how to prove this?

Thanks.

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We want to obtain the distribution of the estimator $\hat{\theta} = max_{i = 1}^n{X_i}$:

$$P(max_{i=1}^n{X_i} \leq x) = \prod_{i=1}^nP(X_i \leq x) = P(X_1 \leq x)^n = (\frac{x - 0}{\theta - 0})^n = \frac{x^n}{\theta^n} I_{[0, \theta]}(x)$$

Which means that, differentiating the function with respect to $x$, we obtain:

$$P(max_{i_1}^n{X_i} = x) = \frac{nx^{n-1}}{\theta^n} I_{[0, \theta]}(x)$$

Which is exactly what we wanted, thereby:

$$f(x) = \frac{n x^{n-1}}{\theta^n} I_{[0, \theta]}(x)$$

Hence:

$$E[\hat{\theta}] = \int_{-\infty}^{+\infty} x f(x) dx = \int_{-\infty}^{+\infty} x \frac{n x^{n-1}}{\theta^n} I_{[0, \theta]}(x) dx = \int_0^\theta \frac{n}{\theta^n} x^n dx = \frac{n}{\theta^n} \frac{\theta^{n+1}}{n+1} = \frac{n}{n+1} \theta$$

Now, lets see what happens with

$$V[\hat{\theta}] = E[\hat{\theta}^2] - E[\theta]^2$$

And

$$E[\hat{\theta}^2] = \int_{-\infty}^{+\infty} x^2 f(x) dx = \int_{-\infty}^{+\infty} x^2 \frac{n x^{n-1}}{\theta^n} I_{[0, \theta]}(x) dx = \int_0^\theta \frac{n}{\theta^n} x^{n+1} dx = \frac{n}{\theta^n} \frac{\theta^{n+2}}{n+2} = \frac{n}{n+2} \theta^2$$

Thus

$$V[\hat{\theta}] = \frac{n}{n+2} \theta^2 - (\frac{n}{n+1})^2(\theta)^2 = \frac{n}{(n+2)(n+1)^2} \theta^2$$

So, we have that $E[\hat{\theta}] \to \theta$ as $n \to +\infty$, meaning that it is asymptotically unbiased. Also, $V[\hat{\theta}] \to 0$ as $n \to +\infty$, which proves that it is consistent.

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  • $\begingroup$ Thanks a lot for your answer. There is just one thing that isn't clear to me: Where did $f(x)$ come from? And what does $I_{[0,\theta]}(x)$ mean? $\endgroup$ – Cristopher May 26 '15 at 19:33
  • $\begingroup$ I'll extend my answer in a sec. However, $I_{[0, \theta]}(x) = "x \in [0, \theta]"$, meaning, it is 1 if x is in that inverval, or 0 otherwise $\endgroup$ – Misguided May 26 '15 at 19:52
  • $\begingroup$ Thanks! :) I understand it now. $\endgroup$ – Cristopher May 26 '15 at 21:38

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