# Weibull's MLE consistency and asymptotic normality

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?

• Please add self-study tag if this is homework. Search for invariance property of MLE. What is $\theta$ here? – StubbornAtom Mar 28 '20 at 15:02
• 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. – Никита Васильев Mar 28 '20 at 15:19
• What is the MLE of $\beta$? Add it in your post. – StubbornAtom Mar 28 '20 at 15:36
• 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. – jbowman Mar 28 '20 at 16:41
• 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. – jbowman Mar 28 '20 at 20:12

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.