Expected value of an estimator of scale parameter of the Weibull distribution

Question

We consider $n$ iid random variables $X_1,...,X_n$ sampled from a Weibull distribution:

$$\mathscr{W}(\lambda,k) \qquad \lambda,k > 0 \qquad pdf: f(x) = \frac{kx^{k-1}}{\lambda^k}e^{-(x/\lambda)^k} _{\Bbb{R_+^*}}(x)$$

Here $k$ is known and we want to estimate $\lambda$. With maximum likelihood estimation we find:

$$\hat{\lambda}_n = \left(\frac{\sum X_i^k}{n}\right)^{1/k}$$

I would like to compute the expected value $E_{\lambda}(\hat{\lambda}_n)$ in order to compute the estimator bias but it is very difficult due to the power $1/k$ in the expression. My question is: is it possible to compute this expected value ? And if not, can we prove it ? I heard about integral convergence so maybe we can show that the integral formula of the expected value of $\hat{\lambda}_n$ doesn't converge, but I really don't know how to do this...

Answer 1

Let $X \sim \operatorname{Weibull}(\lambda, k)$ with the parametrization you chose, and let $Y = X^k$. Then $$f_Y(y) = f_X(y^{1/k}) \left|\frac{d}{dy}[y^{1/k}]\right| = \frac{ky^{1-1/k}}{\lambda^k} e^{-(y^{1/k}/\lambda)^k} \frac{1}{k} y^{1/k-1} = \lambda^{-k} e^{-\lambda^{-k} y}, \quad y > 0. \tag{1}$$ This shows that $$Y \sim \operatorname{Exponential}(\lambda^{-k}), \tag{2}$$ where $\lambda^{-k}$ is a rate parameter. Consequently, $$S = \sum_{i=1}^n X_i^k = \sum_{i=1}^n Y_i \sim \operatorname{Gamma}(n, \lambda^{-k}), \tag{3}$$ with density $$f_S(s) = \frac{s^{n-1} e^{-s/\lambda^k}}{\lambda^{nk} \Gamma(n)}, \quad s > 0 \tag{4}$$ where the parametrization is by shape and rate. So our estimator $\hat \lambda = (S/n)^{1/k}$ has density $$\begin{align} f_{\hat \lambda}(r) &= f_S(n r^k) \left|\frac{d}{dr}[nr^k]\right| \\ &= \frac{(nr^k)^{n-1} e^{-(nr^k)/\lambda^k}}{\lambda^{nk} \Gamma(n)} nkr^{k-1} \\ &= \frac{k n^n (r/\lambda)^{nk} e^{-n(r/\lambda)^k}}{r \Gamma(n)}. \tag{5}\\ \end{align}$$

Integration of this expression for the expectation yields $$\operatorname{E}[\hat \lambda] = \int_{r=0}^\infty r f_{\hat \lambda}(r) \, dr = \frac{\lambda \Gamma(n + \frac{1}{k})}{n^{1/k} \Gamma(n)}. \tag{6}$$ I leave the details of computation of this integral to you as an exercise. The density $(5)$ is called a generalized gamma distribution.