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 [itex]\int_0^{\infinity}ae^\frac{-t}{2}dt= 1/2[/itex]. 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 [itex]\frac{1}{2}+ \int_0^1 ae^\frac{-t}{2}dt[/itex]). Divide the original pdf by that. Using "A" for "a" in that modified pdf, the expected value of T is
[itex]A\int_{-1}^{\infinity}te^\frac{-t}{2}dt[/itex]. You can integrate that using "integration by parts".

 

[tacman]

A) To find the constant a, we can use the fact that the total area under the probability density function (PDF) must equal 1. Therefore, we can set up the following integral and solve for a:

1 = ∫ fT(t) dt from -∞ to ∞

1 = ∫ ae^(-|t|/2) dt from -∞ to ∞

1 = 2a∫ e^(-t/2) dt from 0 to ∞ (since the function is symmetric)

1 = 2a(-2e^(-t/2)) from 0 to ∞

1 = 4a

a = 1/4

Therefore, the constant a is 1/4.

B) To find the expected value of T, we can use the formula E(T) = ∫ tfT(t) dt from -∞ to ∞. However, since we are given that T is greater than or equal to -1, we can adjust the limits of integration to start at -1 instead of -∞. This is because the probability of T being less than -1 is equal to 0, so it does not affect the expected value calculation.

E(T) = ∫ tfT(t) dt from -1 to ∞

E(T) = ∫ t(1/4)e^(-|t|/2) dt from -1 to ∞

E(T) = 1/4 ∫ te^(-|t|/2) dt from -1 to ∞

Using integration by parts, we can solve this integral to get:

E(T) = -1/4(te^(-|t|/2)) - 1/2e^(-|t|/2) from -1 to ∞

E(T) = -1/4(∞e^(-∞)) - 1/2e^(-∞) - (-1/4(-1e^(-1/2)) - 1/2e^(-1/2))

Since e^(-∞) is equal to 0, we can simplify this to get:

E(T) = 1/4 + 1/2e^(-1/2)

Therefore, the expected value of T, given T is greater than or equal to -1, is approximately 1.31.