I just have a small technical question. I am in the midst of solving a problem where I have gotten 2 different exponential probability density functions that are as follows:
pdf #1: 3e^(-3x)
pdf #2: 5e^(-5y)
The question then asks of me to find the cumulative distribution function and the probability density function of W = X/Y. Note: the variables X & Y are independent.
Here is where I am confused:
To find the cumulative distribution function, all I would have to do is take the integral of (3e^(-3x) * 5e^(-5y)) to get the cdf? I believe I can multiply the pdfs since the two variables are independent!
To find the pdf of W, I am not entirely sure but again, since the variables are independent can I not just have pdf W = (pdf X)/(pdf Y)?
Help is greatly appreciated! Thanks 🙂
Best Answer
By independence, the joint density function of $X$ and $Y$ is $(3e^{-3x})(5e^{-5y})$ when $x$ and $y$ are positive, and $0$ elsewhere. Let $W=\dfrac{X}{Y}$. We want to find the cumulative distribution function $F_W(w)$ of $W$.
If $w\le 0$, then $F_W(w)=0$. Now let $w$ be positive. For a long time we will think of $w$ as being fixed, like $w=1.7$.
We have $\dfrac{X}{Y}\le w$ if and only if $Y\ge \dfrac{X}{w}$.
Draw the line $y=\dfrac{x}{w}$. Then $\dfrac{X}{Y}\le w$ if and only if $(X,Y)$ is above or on that line. Let $T$ be the first-quadrant region which is above or on that line. Then $$\Pr(W\le w)=\iint_T (3e^{-3x})(5e^{-5y})\,dy\,dx.$$
Now we just have an integration problem. It is marginally easier to integrate first with respect to $y$, from $x/w$ to $\infty$.
We get $3e^{-3x}e^{-5x/w}$.
Finally, integrate with respect to $x$ (Edit: this originally said y, i believe it should be x?), from $0$ to $\infty$. The integration is straightforward, we are just integrating $3\exp(-(3+5/w)x)$, and a substitution does it.
Now that we have the cdf $F_W(w)$, differentiate it with respect to $w$ to get the density function $f_W(w)$ of $W$.