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

In summary, the conversation is about finding the expected value and variance of a continuous random variable X with the density function f(x)=0.5e^{-|x|}. The expected value is given by E(x)=\int_{-\infty}^{\infty}xe^{-|x|}dx and the variance is Var(X)=\int_{-\infty}^{\infty}x^2e^{-|x|}dx=2\int_0^\infty x^2e^{x}dx. The person asking for help initially provided an incorrect density function, but it was later corrected to the correct one. The expected value can be obtained using symmetry, and the variance can be calculated by
  • #1
badgerbadger
7
0
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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  • #2
Check your problem again. That is NOT a density function. [itex]\int_{-\infty}^\infty f(x)dx[/itex] is not even defined, much less being 1. Did you mean [itex]f(x)= 0.5e^{-|x|}[/itex]?
 
  • #3
haha yeah its f(x)= 0.5e^{-|x|}

sorry about that.
 
  • #4
Then [itex]E(x)= \int_{-\infty}^{\infty}xe^{-|x|}dx[/itex] which you should be able to get by "symmetry" without needing to do the integral.

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

1. What is the formula for the distribution function f(x)=.5e^|x|?

The formula for the distribution function is f(x)=.5e^|x|.

2. How do you find the expected value (EX) of the distribution function f(x)=.5e^|x|?

To find the expected value, EX, of the distribution function f(x)=.5e^|x|, you need to integrate the function from -∞ to +∞ and multiply it by x. This can be written as ∫-∞+∞xf(x)dx.

3. What is the expected value (EX) for the distribution function f(x)=.5e^|x|?

Using the formula from question 2, the expected value (EX) for the distribution function f(x)=.5e^|x| is 0.

4. How do you find the variance (Var(x)) of the distribution function f(x)=.5e^|x|?

The variance (Var(x)) can be found by taking the integral of x^2f(x) from -∞ to +∞ and subtracting the square of the expected value (EX)^2. This can be written as ∫-∞+∞x^2f(x)dx - (EX)^2.

5. What is the variance (Var(x)) for the distribution function f(x)=.5e^|x|?

Using the formula from question 4, the variance (Var(x)) for the distribution function f(x)=.5e^|x| is 2.

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