Distribution Function f(x)= .5e^|x|, find EX and Var(x)

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The discussion focuses on finding the expected value (EX) and variance (Var(X)) of a continuous random variable with the corrected density function f(x) = 0.5e^{-|x|}. It clarifies that the original function provided was not a valid density function due to improper integration limits. The expected value can be determined using symmetry, eliminating the need for complex integration. The variance calculation involves integrating by parts to evaluate the integral of x^2e^{-|x|}. The conversation emphasizes the importance of correctly defining the density function for accurate statistical analysis.
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Let X be a continuous random variable with density function

f(x)= .5e^|x|

for x range R. Find EX and Var(x)

help please!
 
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Check your problem again. That is NOT a density function. \int_{-\infty}^\infty f(x)dx is not even defined, much less being 1. Did you mean f(x)= 0.5e^{-|x|}?
 
haha yeah its f(x)= 0.5e^{-|x|}

sorry about that.
 
Then E(x)= \int_{-\infty}^{\infty}xe^{-|x|}dx which you should be able to get by "symmetry" without needing to do the integral.

And then Var(X)= \int_{-\infty}^{\infty}x^2e^{-|x|}dx= 2\int_0^\infty x^2e^{x}dx which you can do integrating by parts (twice).
 
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