- #1

- 1,367

- 61

## Main Question or Discussion Point

Hi.

I have the following expression

[tex]X=\frac{G\frac{x_1}{x_2}\gamma_Q}{\frac{1}{G}x_3 \gamma_P+1}[/tex]

where ##x_i## is an exponential random variable with mean 1. All other parameters are nonzero positive constants. Basically I want to find the probability density function (pdf) of ##X##. So I started with cumulative distribution function (cdf) as

[tex]F_X(x)=\text{Pr}\left[\frac{G\frac{x_1}{x_2}\gamma_Q}{\frac{1}{G}x_3 \gamma_P+1}\leq x\right][/tex]

which I evaluated it as

[tex]F_X(x)=\int_{x_3=0}^{\infty}\int_{x_3=0}^{\infty}\text{Pr}\left[x_1\leq\frac{x_2\,x}{G \gamma_Q}(\frac{1}{G}x_3 \gamma_P+1)\right]f_{X_2}(x_2)f_{X_3}(x_3)\,dx_2\,dx_3[/tex]

where ##f_{X_i}(x_i)## is the pdf of the random variable ##x_i##. Is what I did correct?

Thanks

I have the following expression

[tex]X=\frac{G\frac{x_1}{x_2}\gamma_Q}{\frac{1}{G}x_3 \gamma_P+1}[/tex]

where ##x_i## is an exponential random variable with mean 1. All other parameters are nonzero positive constants. Basically I want to find the probability density function (pdf) of ##X##. So I started with cumulative distribution function (cdf) as

[tex]F_X(x)=\text{Pr}\left[\frac{G\frac{x_1}{x_2}\gamma_Q}{\frac{1}{G}x_3 \gamma_P+1}\leq x\right][/tex]

which I evaluated it as

[tex]F_X(x)=\int_{x_3=0}^{\infty}\int_{x_3=0}^{\infty}\text{Pr}\left[x_1\leq\frac{x_2\,x}{G \gamma_Q}(\frac{1}{G}x_3 \gamma_P+1)\right]f_{X_2}(x_2)f_{X_3}(x_3)\,dx_2\,dx_3[/tex]

where ##f_{X_i}(x_i)## is the pdf of the random variable ##x_i##. Is what I did correct?

Thanks