# Find Maximum-Likelihood-Estimator (MLE) for $\alpha$

Consider the following PDF:

$$w_{\alpha,\beta}(x):=\alpha \beta x^{\beta-1}e^{-\alpha x^{\beta}} \mathbf{1}_{(0,\infty)}(x)$$

This is the Weibull distribution often used in material science. Assume we know $$\beta$$ and we want to estimate $$\alpha$$. Let $$X_1,\ldots X_n$$ be i.i.d weibull-distributed.

1. Find the MLE $$\hat{\alpha}$$ for the parameter $$\alpha$$.
2. Find a $$c \in \mathbb R$$ such that $$c \cdot \alpha$$ is an unbiased estimator.

Question: The result I am getting for the MLE doesn't look correct but I don't know what I am doing wrong. For Part 2, do I just have to show that $$\operatorname{E}(\hat{\alpha}-\alpha)=0$$?

My attempt:

Step 1: Write down the ML function:

$$L(\alpha)=\prod_{i=1}^n \alpha \beta x_i^{\beta-1}e^{-\alpha x_i^{\beta}}$$

Step 2: Take the natural log:

\begin{aligned}\ln(L(\alpha))&=\sum_{i=1}^n \ln\left(\alpha \beta x_i^{\beta-1}e^{-\alpha x_i^{\beta}} \right) \\[5pt] &=\sum_{i=1}^n \ln\left(\alpha\beta x_i^{\beta-1}\right)+\ln \left( e^{-\alpha x_i^{\beta}}\right) \\[5pt] &=\sum_{i=1}^n\ln\left(\alpha \right)+\ln(\beta)+(\beta-1)\ln\left(x_i \right)-\alpha x_i^{\beta} \end{aligned}

Step 3: Differentiate and set equal to zero:

\begin{aligned}&\frac{\partial }{\partial \alpha}\ln(L(\alpha))=\sum_{i=1}^n \frac{1}{\alpha}-x_i^{\beta}=0 \\[5pt] &\iff \sum_{i=1}^n\frac{1}{\alpha}=\sum_{i=1}^n x_i^{\beta} \\[5pt] & \iff \frac{n}{\alpha}=\sum_{i=1}^nx_i^{\beta} \iff \alpha=\frac{n}{\sum_{i=1}^nx_i^{\beta}}\end{aligned}

For part 2 I was thinking of setting $$c=\alpha \bar{X}$$, where $$\bar{X}$$ is the average of the $$x_i^{\beta}$$'s. Then it would follow that: $$E\left[\alpha \cdot \frac{1}{n} \cdot \sum_{i=1}^n x_i^{\beta} \cdot \frac{n}{\sum_{i=1}^n x_i^{\beta}}\right]=\alpha \\ \iff \alpha =\alpha$$

But I am not sure if am allowed to set $$c$$ equal to that.

• Looks correct to me except your last line is missing an exponent. You get $1/\alpha$ being the sample average of the $x_i^\beta$s. Jul 14, 2021 at 19:53
• @A rural reader Thanks for your comment. I edited the last line. But what is my estimator$\hat{\alpha}$ then? Isn't that what I am supposed to find and what I need for part 2?
– qmd
Jul 14, 2021 at 20:03
• That $\alpha$, the critical point, is the MLE $\hat\alpha$. I imagine this estimator is a biased estimator of $\alpha$, so the next step is to find a constant for each $n$ for which $\operatorname{E}[c\hat\alpha] = \alpha$. Jul 14, 2021 at 20:10
• @Aruralreader So in my case: $$\hat{\alpha}=\frac{n}{\sum_{i=1}^n x_i^{\beta}}$$ and I need to show that $$E[c \cdot \frac{n}{\sum_{i=1}^n x_i^{\beta}}]= \alpha$$ for all $n$? Is there some standard strategy I can use to approach this?
– qmd
Jul 14, 2021 at 20:21
• @Aruralreader I was maybe thinking of setting $c=\alpha \bar{X}$, where $\bar{X}$ is the average of the $x_i^{\beta}$'s. Then it would follow that: $$E\left[\alpha \cdot \frac{1}{n} \cdot \sum_{i=1}^n x_i^{\beta} \cdot \frac{n}{\sum_{i=1}^n x_i^{\beta}}\right]=\alpha \\ \iff \alpha =\alpha$$
– qmd
Jul 14, 2021 at 20:47

Hint to get you going: $$X^\beta$$ is distributed exponentially with parameter $$\alpha$$ [use the "transformation technique" for this].

Transformation technique: $$Y=X^\beta\Rightarrow s(y)=x=y^{1/\beta}\Rightarrow \frac{ds(y)}{dy}=\frac 1\beta y^{\frac1\beta-1}$$

$$g(y)=\alpha\beta (y^{1/\beta})^{\beta-1}e^{-\alpha(y^{1/\beta})^\beta}\cdot \frac 1\beta y^{\frac1\beta-1}=\alpha e^{-\alpha y}$$

Thus $$\sum_{i=1}^n X_i^\beta$$ is Gamma(n, $$\alpha$$). Then you would try to calculate the expectation of $$\hat\alpha$$ and wrt to the gamma pdf, and then unbias it by multiplying with a $$c\in\mathbb R$$. Let $$Z=\sum_{i=1}^n X_i^\beta$$.

$$\begin{split}E(\hat\alpha)&=\int_0^\infty\frac n{z}\cdot\frac{\alpha^n}{(n-1)!}z^{n-1}e^{-\alpha z}dz\\ &=\alpha\frac n{n-1}\int_0^\infty \frac{\alpha^{n-1}}{(n-2)!}z^{n-2}e^{-\alpha z}dz\\ &=\alpha \frac{n}{n-1}\end{split}$$

I get: $$c=\frac{n-1}{n}$$

$$E\left(c\hat\alpha - \alpha\right)=0$$

• Thanks for your answer! I was just about to write something when you edited it and most of my questions were answered by your edit. Thanks! One question that I have is that i thought the (lowercase) $x_i$'s that I used in my calculation for the MLE are not random variables but just "numbers" but in your case you seem to be using them as random variables. Am I confusing something here?
– qmd
Jul 14, 2021 at 21:59
• @qmd i think the MLE is a random variable as $\frac{n}{\sum_{i=1}^n X_i}$ has a distribution. sometimes they are called maximum likelihood estimator versus maximum likelihood estimate (where the latter is $\frac{n}{\sum_{i=1}^n x_i}$ with the observed values plugged in)
– Vons
Jul 14, 2021 at 22:09
• Thanks! That makes sense. Thank you very much for your help :)
– qmd
Jul 14, 2021 at 22:25
• you are welcome :)
– Vons
Jul 14, 2021 at 23:38