[Date Prev][Date Next][Thread Prev][Thread Next][Date Index][Thread Index]

From |
Jeph Herrin <junk@spandrel.net> |

To |
statalist@hsphsun2.harvard.edu |

Subject |
Re: st: mean of a distribution |

Date |
Sun, 18 Oct 2009 20:31:06 -0400 |

I think you are confusing the random variable X and the values in its domain, x. It's true taht E(X) = integral xf(x) but you have to integrate over the real line - it's an integral, not just an anti-derivative. In this case, you find the anti derivative (I think using integration by parts) and plug in the end points of the reals (plus & minus infinity), and the answer is 1/lambda. hope this helps, Jeph carol white wrote:

Hi, How to calculate the mean of the distribution of a random variable? Take the exponential distribution with the probability density function f(x)=lambda.exp(-lambda.x) where lambda is a constant and x is a random variable. The mean of this distribution is the reciprocal of lambda. If the mean is the expected value of x, which for a continuous random variable E(x) = Integral (x.f(x))dx, how could E(x) be the reciprocal of lambda? Regards, Carol * * For searches and help try: * http://www.stata.com/help.cgi?search * http://www.stata.com/support/statalist/faq * http://www.ats.ucla.edu/stat/stata/

* * For searches and help try: * http://www.stata.com/help.cgi?search * http://www.stata.com/support/statalist/faq * http://www.ats.ucla.edu/stat/stata/

**References**:**st: mean of a distribution***From:*carol white <wht_crl@yahoo.com>

- Prev by Date:
**Re: st: RE: automatic coding so that legends assume the label variables** - Next by Date:
**st: scatter line and reshape question** - Previous by thread:
**st: RE: mean of a distribution** - Next by thread:
**Re: st: RE: mean of a distribution** - Index(es):

© Copyright 1996–2015 StataCorp LP | Terms of use | Privacy | Contact us | What's new | Site index |