Compute $E(X)$ if $F_X(x)=e^{-\lambda x^{-\delta}}$

Question

I am trying to calculate the expected value for this cdf:

$F_X(x)=e^{-\lambda x^{-\delta}} , x \ge 0, \lambda >0 ,\delta >0.$

which gives the pdf as:

$f_X(x) = \lambda \delta x^{\delta -1} e^{-\lambda x^{-\delta}}$

I am trying to calculate the expectation using:

$E\{x\} = \int_0^ \infty \lambda \delta x^{\delta} e^{-\lambda x^{-\delta}} dx $

but the calculation of above expectation gives me infinity as expected value!I am confused wether I am wrong in calculation or there is any trick to get rid of infinity or the infinity makes sense as the expectation.

Answer 1

With your chosen parametrization, $$\begin{align*} \operatorname{E}[X] &= \int_{x=0}^\infty x f_X(x) \, dx \\ &= \int_{x=0}^\infty \lambda \delta x^{-\delta} \exp(-\lambda x^{-\delta}) \, dx, \quad u = \lambda x^{-\delta}, \quad du = -\lambda \delta x^{-\delta - 1} \, dx, \quad x = (u/\lambda)^{-1/\delta},\\ &= \int_{u=0}^\infty (u/\lambda)^{-1/\delta} e^{-u} \, du \\ &= \lambda^{1/\delta} \int_{u=0}^\infty u^{(1 - 1/\delta) - 1} e^{-u} \, du \\ &= \lambda^{1/\delta} \Gamma(1 - 1/\delta), \end{align*}$$ where we have used the definition of the gamma function $$\Gamma(z) = \int_{t=0}^\infty t^{z-1} e^{-t} \, dt.$$ Indeed, it is easy to see that the $k^{\rm th}$ raw moment under your parametrization is $$\operatorname{E}[X^k] = \lambda^{k/\lambda} \Gamma(1-k/\lambda).$$

Answer 2

You made a mistake in the sign of the derivative! A $-$ sign should precede $\delta$ in the power of $x$