# Integral limits

• I

## Main Question or Discussion Point

Hi.

I have the following expression

$$X=\frac{G\frac{x_1}{x_2}\gamma_Q}{\frac{1}{G}x_3 \gamma_P+1}$$

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

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

which I evaluated it as

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

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

Thanks

andrewkirk
Homework Helper
Gold Member
It is only correct if the three random variables are independent. If they are, it is valid to factorise the joint pdf $f_{X_1,X_2,X_3}(x_1,x_2,x_3)$ into $f_{X_1}(x_1)f_{X_2}(x_2)f_{X_3}(x_3)$ as you have done. If they are not independent, that factorisation is incorrect.

Also, one of your two lower limits for integration needs to be $x_2=0$.

It is only correct if the three random variables are independent. If they are, it is valid to factorise the joint pdf $f_{X_1,X_2,X_3}(x_1,x_2,x_3)$ into $f_{X_1}(x_1)f_{X_2}(x_2)f_{X_3}(x_3)$ as you have done. If they are not independent, that factorisation is incorrect.

Also, one of your two lower limits for integration needs to be $x_2=0$.
Yes, they are independent and identically distributed random variables. Right, the first integral is over all values of $x_2$, it's a typo. Thanks

andrewkirk
Yes, they are independent and identically distributed random variables. Right, the first integral is over all values of $x_2$, it's a typo. Thanks
$$F_X(x)=\int_{x_3=0}^{\infty}\int_{x_2=0}^{\infty}\text{Pr}\left[X_1\leq\frac{x_2\,x}{G \gamma_Q}(\frac{x_3 \gamma_P}{G}+1)\right]f_{X_2}(x_2)f_{X_3}(x_3)\,dx_2\,dx_3$$