integrate e^(-|x|) but with a twist

G

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Hi
This is one of my homework problems from my book. The answer is 1/2, but I don't know why.

If the sample space is (wierd C) = {c: -infinity < c < infinity} and if C is a subset of (wierd C) is a set for which the integral C e^(-|x|) dx exists, show that this set function is not a probablity set function. What constant do we multiply the integrand by to make it a probablity set function?

I keep getting the answer of 0. I use the limits (- infinity, infinity) and it keeps coming up 0. So the answer in the back of the book says I would multiply the integrand by 1/2, I would just keep getting 0. So I'm really confused.

Thanks in advance for your help on this
Take care,
Beckie
 
Well the integral from minus infinite to infinity of e<SUP>-|x|</SUP> is 2.
If e<SUP>-|x|</SUP> is multiplied by (1/2) we get a pdf.
 
Thank you.

I understand now because if you do one of the integrals from 0 to infinity and the other integral from -infinity to 0, you get 1 + 1 = 2 and then 1/2 * that would be 1.

I see.

Thanks.
 
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