Continuous Random Variable and Density Function

**clipperdude21** · November 1st 2007, 08:37 PM

Let X be a continuous random variable with density function f(x)=.5e^-abs(x) for x in all real numbers. Find EX and var(y)

I got 0 for both values and im not sure if i did this right.... Can anyone help?

Thanks!!!

**kalagota** · November 1st 2007, 09:07 PM

Originally Posted by clipperdude21

Let X be a continuous random variable with density function f(x)=.5e^-abs(x) for x in all real numbers. Find EX and var(y)

I got 0 for both values and im not sure if i did this right.... Can anyone help?

Thanks!!!

$f(x) = \frac{1}{2}e^{-|x|}$
so,
$E[X] = \int_R xf(x)dx = \int_{-\infty}^0 \frac{1}{2}xe^{x}dx + \int_{0}^{\infty} \frac{1}{2}xe^{-x}dx$

if this is what you did, probably you are right..

**clipperdude21** · November 1st 2007, 10:09 PM

thats what i set it up as but somehow im second guessing my calculus i integrated it and got .5E^x(x-1) and -.5e^-x(x+1) for the other and i didnt know if i did that right...

**kalagota** · November 1st 2007, 11:04 PM

Originally Posted by clipperdude21

thats what i set it up as but somehow im second guessing my calculus i integrated it and got .5E^x(x-1) and -.5e^-x(x+1) for the other and i didnt know if i did that right...

can you show your complete solution?

**kalagota** · November 2nd 2007, 05:28 AM

Originally Posted by CaptainBlack

If that is what you did it does integrate to 0, but is wrong, you should have had:

$E[X] = \int_R xf(x)dx = \int_{-\infty}^0 \frac{1}{2}xe^{-x}dx + \int_{0}^{\infty} \frac{1}{2}xe^{-x}dx$

which does not integrate up to 0.

RonL

notice that there is an absolute value sign there and by definition,
$|x| = \left\{ {\begin{array}{cc} x&x \geq 0\\ -x&x < 0\\ \end{array}}\right. $

or i just had a wrong interpretation?

**CaptainBlack** · November 2nd 2007, 05:50 AM

Originally Posted by kalagota

notice that there is an absolute value sign there and by definition,
$|x| = \left\{ {\begin{array}{cc} x&x \geq 0\\ -x&x < 0\\ \end{array}}\right. $

or i just had a wrong interpretation?

No ignore my post I had misread the integrals

RonL

**kalagota** · November 2nd 2007, 05:57 AM

Originally Posted by CaptainBlack

No ignore my post I had misread the integrals

RonL

ok.. Ü

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