Let 𝑋 =𝐸𝑥𝑝(𝜆). What is the pdf of 𝑌 = 𝑒xp(𝑋) ?

Question

Let 𝑋 =𝐸𝑥𝑝(𝜆). What is the pdf of 𝑌 = 𝑒xp(𝑋) ?

Answer 1

To find the probability density function (pdf) of Y = exp(X), where X is exponentially distributed with rate parameter 𝜆, we can use the method of transformations.

Step 1: Find the cumulative distribution function (CDF) of Y.
The CDF of Y, denoted as F_Y(y), is defined as P(Y ≤ y). We can express this in terms of X as follows:

F_Y(y) = P(Y ≤ y) = P(exp(X) ≤ y)

Step 2: Solve for X in terms of Y.
To do this, take the logarithm of both sides of the previous equation:

log(exp(X)) ≤ log(y)

X ≤ log(y)

Since X is exponentially distributed, its support is from 0 to infinity. Therefore, we have:

0 ≤ X ≤ log(y)

Step 3: Find the CDF of X.
The CDF of X, denoted as F_X(x), is given by:

F_X(x) = 1 - exp(-𝜆x) for x ≥ 0

Step 4: Calculate the pdf of Y.
The pdf of Y, denoted as f_Y(y), can be obtained by taking the derivative of the CDF of Y with respect to y:

f_Y(y) = d/dy [F_Y(y)]
= d/dy [P(X ≤ log(y))]
= d/dy [1 - exp(-𝜆log(y))]
= 𝜆 / y

Therefore, the pdf of Y is given by 𝑓𝑌(𝑦) = 𝜆 / y, where 𝑌 = 𝑒xp(𝑋).