Originally Posted by
meymathis 1. F(x) = 1 - 1/x (x greater than or equal to 1)
You can do this in two ways. You can calculate the PDF, f, and then integrate $\displaystyle xf(x)$. The PDF is just the derivative of the CDF, if continuous.
$\displaystyle f(x) = 1/x^2$
$\displaystyle \int_1^\infty xf(x)dx = \int_1^\infty 1/x\,dx =\ln(x)|_1^\infty $which does not converge.
2. F(x) = 1 - exp(-x^2) (x greater than or equal to 1)
$\displaystyle f(x) = -2x\exp(-x^2)$ Mr F says: Small typo here. Should be 2x, not -2x .... The bigger problem is that it's not actually a valid cdf .... The support should be [0, +oo), NOT [1, +oo).
$\displaystyle \int_1^\infty xf(x)dx = \int_1^\infty -2x^2 \exp(-x^2)\,dx$
I guess I would try integration by parts with $\displaystyle u=x,\,dv=-2x \exp(-x^2)\,dx$
You will probably need to use the standard normal table in your stats book to do $\displaystyle \int_1^\infty \exp(-x^2)\,dx$ which should pop out after the integration by parts step. That integral is almost the area under the curve of the standard normal distribution, but it is missing a $\displaystyle 1/\sqrt{2\pi}$.