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Let X = $(X_1, \dots, X_n)$ be a sample from Weibull distribution $W(\alpha, \beta)$ with fixed and known $\alpha$. Find MLE of parametric function $g(\beta) = \beta^{\alpha}$. Check if bias is equal to $0$. Show it is consistent and asymptotically normal.

Weibull's density is: $$ f(x, \alpha, \beta) = \alpha \beta^{-\alpha}x^{\alpha-1}e^{-(\frac{x}{\beta})^\alpha} = $$ so $$ l(p) = \alpha^n\beta^{-n\alpha}(x_1 \cdot \dots \cdot x_n)^{\alpha -1}e^{-\frac{1}{\beta^{\alpha}}(x_1 \cdot \dots \cdot x_n)^\alpha} \\ L(p) = n\ln \alpha - n \alpha \ln \beta + (\alpha -1)(\ln x_1 + \dots + \ln x_n)-(\frac{x_1}{\beta})^\alpha - \dots - (\frac{x_n}{\beta})^\alpha \\ (L(p))' = \frac{-n \alpha}{\beta} + \alpha(\frac{x_1}{\beta})^{\alpha-1}\frac{x_1}{\beta^2}+\dots + \alpha(\frac{x_n}{\beta})^{\alpha-1}\frac{x_n}{\beta^2}$$ so $$ \frac{n \alpha}{\beta} = \alpha \frac{\frac{x_1^\alpha}{\beta^{\alpha-1}}+\dots+\frac{x_n^\alpha}{\beta^{\alpha-1}}}{\beta^2} \\ \beta n = \frac{x_1^\alpha + \dots + x_n^\alpha}{\beta^{\alpha-1}} $$ and finally $$ \beta^\alpha = \frac{x_1^\alpha + \dots + x_n^\alpha}{n} $$

so the MLE of g is $\frac{x_1^\alpha + \dots + x_n^\alpha}{n}$.

How can I proceed with calculating bias, consistency or the asymptotical normality?

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    $\begingroup$ Please add self-study tag if this is homework. Search for invariance property of MLE. What is $\theta$ here? $\endgroup$ – StubbornAtom Mar 28 '20 at 15:02
  • $\begingroup$ Well, by $\mathbb{E}(\hat{\theta}) = \theta$ I meant the (un)biasedness of the estimator. I need to check if bias is equal to 0. $\endgroup$ – Никита Васильев Mar 28 '20 at 15:19
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    $\begingroup$ What is the MLE of $\beta$? Add it in your post. $\endgroup$ – StubbornAtom Mar 28 '20 at 15:36
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    $\begingroup$ Here's a hint: look at $x$ in the exponential term in your first and second lines - the pdf and the likelihood function. See how the terms differ? That's a start. $\endgroup$ – jbowman Mar 28 '20 at 16:41
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    $\begingroup$ Bias is pretty straightforward at this point; derive the distribution of $x^a$ using a standard change-of-variable approach and find the expectation (should be almost immediately obvious.) You can apply the CLT to the sample mean of $x^a$ to deduce the asymptotic normality. If the estimator is unbiased, it's consistent too. $\endgroup$ – jbowman Mar 28 '20 at 20:12
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Combining the comments into an answer:

Density of $X$ is $$f_X(x)=\alpha\beta^{-\alpha}x^{\alpha-1}e^{-(x/\beta)^{\alpha}}\mathbf1_{x>0}\quad;\,\alpha,\beta>0$$

So likelihood function for known $\alpha$ given the sample $x_1,\ldots,x_n$ is

$$L(\beta)\propto \beta^{-n\alpha}\exp\left\{-\frac1{\beta^\alpha}\sum_{i=1}^n x_i^\alpha\right\}\mathbf1_{x_1,\ldots,x_n>0}\quad,\,\beta>0$$

Log-likelihood is $$\ell(\beta)=\text{constant }-n\alpha\ln\beta-\frac1{\beta^\alpha}\sum_{i=1}^n x_i^\alpha$$

And $$\ell'(\beta)=-\frac{n\alpha}\beta+\frac{\alpha}{\beta^{\alpha+1}}\sum_{i=1}^n x_i^\alpha$$

MLE of $\beta^\alpha$ is indeed $$\widehat{\beta^\alpha}(X_1,\ldots,X_n)=\frac1n\sum_{i=1}^n X_i^\alpha$$

By a change of variables $Y=X^\alpha$, you can see that density of $Y$ is

$$f_Y(y)=f_X(y^{1/\alpha})\left|\frac{dx}{dy}\right|=\frac1{\beta^\alpha}e^{-y/\beta^\alpha}\mathbf1_{y>0}\quad;\,\alpha,\beta>0$$

So $Y$ is Exponential with mean $\beta^\alpha$.

  • For verifying consistency of MLE, use (weak) law of large numbers.

  • For verifying asymptotic normality of MLE, appeal to classical CLT.

As expected in 'regular' cases, MLE is consistent and asymptotically normal.

A brief discussion of consistency and asymptotic normality of MLE can be found here.

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