# x and y independent exponential r.v with respctive parameters 2 and 3. find the cdf and density of z = x/y.

9 months ago

## To find the cumulative distribution function (CDF) and probability density function (PDF) of the random variable Z = X/Y, where X and Y are two independent exponential random variables with respective parameters 2 and 3, we will follow these steps:

Step 1: Find the cumulative distribution function (CDF) of Z.
- Recall that the CDF of a random variable Z is defined as P(Z â‰¤ z).
- To find P(Z â‰¤ z), we need to consider all possible values of x and y that satisfy the inequality Z = X/Y â‰¤ z.
- Since X and Y are independent, we can express the inequality as X â‰¤ zY.

Step 2: Find the PDF of Z.
- The PDF of Z can be obtained by taking the derivative of its CDF with respect to z.
- The derivative of the CDF gives us the probability density function (PDF) of Z.

Let's calculate the CDF and PDF of Z step by step:

Step 1: Finding the CDF of Z (P(Z â‰¤ z)):
- Let's start by considering the cases when z is positive and z is negative.

Case 1: For z > 0:
- P(Z â‰¤ z) = P(X/Y â‰¤ z) = P(X â‰¤ zY)
- Since X and Y are exponential random variables, the CDF of X is given by F_X(x) = 1 - e^(-2x), and the CDF of Y is given by F_Y(y) = 1 - e^(-3y).
- Using these CDFs, we can find P(X â‰¤ zY) as follows:
P(X â‰¤ zY) = âˆ«[0 to infinity]âˆ«[0 to zY] f_X(x)*f_Y(y) dx dy,
where f_X(x) = 2e^(-2x) and f_Y(y) = 3e^(-3y) are the PDFs of X and Y, respectively.
Simplifying the double integral, we get:
P(X â‰¤ zY) = âˆ«[0 to infinity]âˆ«[0 to zY] 2e^(-2x)*3e^(-3y) dx dy
= 6âˆ«[0 to infinity] e^(-2x) âˆ«[0 to zY] e^(-3y) dx dy
= 6âˆ«[0 to infinity] e^(-2x) (-1/3e^(-3y)) |0 to zY dy
= -2âˆ«[0 to infinity] e^(-2x) (e^(-3zY)/3) dy
= -2/3âˆ«[0 to infinity] e^(-2x)e^(-3zY) dy
= -2/3âˆ«[0 to infinity] e^(-2x-3zY) dy
= -2/3âˆ«[0 to infinity] e^(-(2x+3zY)) dy
= -2/3 âˆ«[0 to infinity] e^(-(2x+3zY)) dx,
where the limits of integration are flipped due to independent variables X and Y.

- Now, we can integrate with respect to x:
P(X â‰¤ zY) = -2/3 âˆ«[0 to infinity] e^(-(2x+3zY)) dx
= -2/3 (-1/(2+3zY)) e^(-(2x+3zY)) |0 to infinity
= 2/(2+3zY) e^(-(2x+3zY)) |0 to infinity
= 2/(2+3zY) (0 - 1)
= -2/(2+3zY)

- Finally, we can write the CDF of Z as:
F_Z(z) = P(Z â‰¤ z) = -2/(2+3zY) for z > 0

Case 2: For z â‰¤ 0:
- Since Z = X/Y, we cannot have z â‰¤ 0, as Y cannot be zero. Hence, the CDF of Z is zero for z â‰¤ 0.

The CDF of Z is given by:
F_Z(z) = -2/(2+3zY) for z > 0
= 0 for z â‰¤ 0

Step 2: Finding the PDF of Z:
- The PDF of Z can be obtained by taking the derivative of its CDF with respect to z.
- Differentiating the CDF with respect to z, we get:
f_Z(z) = d/dz [F_Z(z)]
= d/dz [-2/(2+3zY)]
= (6Y)/(2+3zY)^2

Finally, the PDF of Z is given by:
f_Z(z) = (6Y)/(2+3zY)^2

Hence, we have derived the CDF and PDF of Z.