Mean and Variance

**dynas7y** · March 31st 2010, 08:45 AM

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.

**Anonymous1** · March 31st 2010, 10:53 AM

Well you could use the $MGF$ but I guess that is defined by integration...

**Anonymous1** · March 31st 2010, 10:58 AM

Oh wait! I remember how to do this:

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

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

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

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

Which simplifies to

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

Where the last equality holds by an identity

**dynas7y** · March 31st 2010, 11:58 AM

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

**Anonymous1** · March 31st 2010, 12:32 PM

Start like this

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

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

Um... gimme a sec.

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