Absolute moment existence

Question

I have given

$$F_X(x) = \frac{1}{1+e^{-x}} ; \quad x \geq0$$

I have to calculate $k$, s.t. $E[|X|^k]<\infty$

I did calculate the density by $f_X(x) = \frac{dF}{dx} =\frac{e^{-x}}{(1+e^{-x})^2} $ . Using this to calculate the absolute moments, I found it to be very cumbersome so I think that this might not be the best way to solve this. Does anyone have thoughts?

Best

Answer 1

$\int |x|^{k} \frac {e^{-x}} {1+e^{-x}}dx \geq \int_{-\infty}^{-1} |x|^{k} \frac {e^{-x}} {1+e^{-x}}dx =\int_1^{\infty} x^{k} \frac {e^{x}} {1+e^{x}}dx \geq \frac e {1+e} \int_1^{\infty} x^{k}dx=\infty$ for any $k \geq 0$. Now let $k<0$. Then $\int |x|^{k} \frac {e^{-x}} {1+e^{-x}}dx \geq \int_{0}^{1} x^{k}\frac {e^{-x}} {1+e^{-x}}dx \geq \frac1 {2e} \int_0^{1}x^{k}dx =\infty$ if $k+1 <0$ or $k <-1$.

Finally let $-1\leq k <0$. I will let you use a similar argument by considering the integral from $-\infty$ to $-1$ to see that then integral is divergent in this case too. Hence there is no $k$ for which $E|X|^{k} <\infty$.

Answer 2

Your density is not correct. Differentiation of the CDF gives $$f_X(x) = \frac{dF_X}{dx} = -(1 + e^{-x})^{-2} (-e^{-x}) = \frac{e^{-x}}{(1+e^{-x})^2} = \frac{e^{-x}}{e^{-x}(e^{x/2} + e^{-x/2})^2} = \frac{1}{4 \cosh^2 \frac{x}{2}}.$$

If $k$ is a positive integer, it is not difficult to show $\operatorname{E}[|X|^k] < \infty$. I leave this as an exercise; in particular, one might wish to consider the moment-generating function of $|X|$.