THe Laplace random variable has a PDF that is a double exponential

[nbalderaz]

THe Laplace random variable has a PDF that is a double exponential, fT(t)=ae^(-|t|/2) for all values of t and a, a constant to be determined.


A) Find a
(Answer 1/4)

B)Find the expected value of T, given T is greater than or equal to -1.
(Answer 1.31)

 

[HallsofIvy]

THe Laplace random variable has a PDF that is a double exponential, fT(t)=ae^(-|t|/2) for all values of t and a, a constant to be determined.


A) Find a
(Answer 1/4)

The total "cumulative probability" must be 1 and since this pdf is symmetric, we must have $\int_0^{\infinity}ae^\frac{-t}{2}dt= 1/2$. Do that integration and solve for a.

B)Find the expected value of T, given T is greater than or equal to -1.
(Answer 1.31)

Knowing that "T is greater than or equal to -1" tells us that the distribution must be "normalized" so that the total integral from -1 to infinity must now be 1. Put the value of a you found in (A) in the pdf and integrate it from -1 to infinity (again, using symmetry, that is the same as $\frac{1}{2}+ \int_0^1 ae^\frac{-t}{2}dt$). Divide the original pdf by that. Using "A" for "a" in that modified pdf, the expected value of T is
$A\int_{-1}^{\infinity}te^\frac{-t}{2}dt$. You can integrate that using "integration by parts".