# Mean and Variance

• Mar 31st 2010, 08:45 AM
dynas7y
Mean and Variance
Given the pdf and cdf of a continuous random variable, say X, describe how you would find the mean and variance of X without using integration.
• Mar 31st 2010, 10:53 AM
Anonymous1
Well you could use the $\displaystyle MGF$ but I guess that is defined by integration...
• Mar 31st 2010, 10:58 AM
Anonymous1
Oh wait! I remember how to do this:

Here is an example:
$\displaystyle X$~$\displaystyle exp(\lambda)$

Differentiating the density identity (ofcourse I'm assuming we can swap the order of differentiation and integration).

$\displaystyle \frac{d}{d\lambda} \int_0^{\infty}\lambda e^{-\lambda x}dx = 1$

$\displaystyle \Rightarrow \int_0^{\infty} (-\lambda x +1) e^{-\lambda x}dx = 0$

Which simplifies to

$\displaystyle \int_0^{\infty} \lambda x e^{-\lambda x} = \int_0^{\infty} xe^{-\lambda x} = \frac{1}{\lambda} = E[X]$

Where the last equality holds by an identity
• Mar 31st 2010, 11:58 AM
dynas7y
Thank you for the response, but I'm not quite following. Can you use the following as an example and find the mean using the method you described using no integration:

pdf: fx(X) = ((x-5)^3)/4 for 5 <= x <= 7 and 0 elsewhere
cdf: Fx(X) = ((x-5)^4)/16 FOR 5 <= x <= 7

Thanks
• Mar 31st 2010, 12:32 PM
Anonymous1
Start like this

$\displaystyle \frac{d}{dx}\int_5^7 \frac{x^3 - 15x^2 + 75x - 125}{4} dx = \frac{d}{dx}1$

$\displaystyle \Rightarrow \int_5^7 \frac{3x^2-30x+75}{4} = 0$

Um... gimme a sec.