# Thread: What is the mean of these two CDFs?

1. ## What is the mean of these two CDFs?

1. F(x) = 1 - 1/x (x greater than or equal to 1)

and

2. F(x) = 1 - exp(-x^2) (x greater than or equal to 1)

Many thanks.

2. 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 $xf(x)$. The PDF is just the derivative of the CDF, if continuous.

$f(x) = 1/x^2$
$\int_1^\infty xf(x)dx = \int_1^\infty 1/x\,dx =\ln(x)|_1^\infty$which does not converge.

2. F(x) = 1 - exp(-x^2) (x greater than or equal to 1)
$f(x) = -2x\exp(-x^2)$
$\int_1^\infty xf(x)dx = \int_1^\infty -2x^2 \exp(-x^2)\,dx$

I guess I would try integration by parts with $u=x,\,dv=-2x \exp(-x^2)\,dx$

You will probably need to use the standard normal table in your stats book to do $\int_1^\infty \exp(-x^2)\,dx$ 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 $1/\sqrt{2\pi}$.

3. 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 $xf(x)$. The PDF is just the derivative of the CDF, if continuous.

$f(x) = 1/x^2$
$\int_1^\infty xf(x)dx = \int_1^\infty 1/x\,dx =\ln(x)|_1^\infty$which does not converge.

2. F(x) = 1 - exp(-x^2) (x greater than or equal to 1)
$f(x) = -2x\exp(-x^2)$ 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).

$\int_1^\infty xf(x)dx = \int_1^\infty -2x^2 \exp(-x^2)\,dx$

I guess I would try integration by parts with $u=x,\,dv=-2x \exp(-x^2)\,dx$

You will probably need to use the standard normal table in your stats book to do $\int_1^\infty \exp(-x^2)\,dx$ 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 $1/\sqrt{2\pi}$.
..

4. Darn it. Details, details....

I'll have you know I checked that the first one was. And this was right after a long back and forth with someone about a CDF not really being a CDF. http://www.mathhelpforum.com/math-he...variables.html

Thanks Mr. F!

5. Thanks you two, that was extremely helpful.