# I Integral limits

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1. Apr 21, 2016

### S_David

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

2. Apr 21, 2016

### andrewkirk

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 $x_2=0$.

3. Apr 21, 2016

### S_David

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

4. Apr 21, 2016

### andrewkirk

Given that additional info, and the discussed correction to the lower integration limit, what you've written is correct, but it is better practice to use upper case for random variables, to distinguish them from quantiles of those variables, ie:

$$F_X(x)=\int_{x_3=0}^{\infty}\int_{x_2=0}^{\infty}\text{Pr}\left[X_1\leq\frac{x_2\,x}{G \gamma_Q}(\frac{x_3 \gamma_P}{G}+1)\right]f_{X_2}(x_2)f_{X_3}(x_3)\,dx_2\,dx_3$$