If X and Y are independent exponential random variables with respective parameters λ1 and λ2, how do I find the distribution of Z = X/Y ?

The distribution of Z = X/Y is  λ₁ / (λ₁+λ₂).

The probability distribution of random variables is described using the cumulative distribution function. The probability for a discrete, continuous, or mixed variable may be described using it. The cumulative probability for a random variable is calculated by adding the probability density function.

Given:

X and Y are independent exponential random variables with respective parameters λ₁ and λ₂.

To find the distribution of Z = X/Y:

First, we have

[tex]f_x_y[/tex](x,y) = λ₁λ₂[tex]e^{-\lambda_1x[/tex][tex]e^{-\lambda_2y[/tex]

First, we find the cumulative distribution function (CDF) for Z = X/Y.

Derivative of Z, f(z) and put  a = x/y

[tex]F_z[/tex](a)  = P (X/Y ≤ a)

        = P (X ≤ aY)

        = [tex]\int\limits^{\infty}_ {x} \ \int\limits^{ay}_0 {\lambda_1\lambda_2e^{-\lambda_1x}e^{-\lambda_2y}\ dx dy[/tex]

       [tex]= \int\limits^{\infty}_ {0} {\lambda_1\lambda_2e^{-\lambda_2y}\ dy [ -1/{\lambda_1}{e^{-\lambda_1x}]\limits^{ay}_ {0}[/tex]

       [tex]= \lambda_2\int\limits^{\infty}_ {0} {e^{-\lambda_2y} - e^{-y(\lambda_2 + \lambda_1a)} \ dy[/tex]

        [tex]= \lambda_2[ [{1/{\lambda_2}+{\lambda_1a]-{e^{-\lambda_2y} + e^{-y(\lambda_2 + \lambda_1a)}]\limits^{\infty}_ {0}[/tex]

        [tex]= -[({\lambda_2}/{\lambda_2 + \lambda_1) - 1][/tex]

        [tex]= -[({\lambda_2}/{\lambda_2 + \lambda_1a) - 1][/tex]

       [tex]= [({\lambda_1a}/{\lambda_2 + \lambda_1a)][/tex]

So, P(X<Y) = P (X/Y <1)

                 = λ₁ / (λ₁+λ₂)

Therefore, distribution is λ₁ / (λ₁+λ₂).

To learn more about the cumulative distribution function;

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