1

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 :)

nicefella
  • 1,069
  • 1
    The joint density function of $X$ and $Y$ is indeed the product of the individual densities. But your suggestion about the pdf of $W$ is not correct. You are asked for the probability that $W\le w$. Express that as an integral involving the joint density. If you run into difficulties, a more detailed start can be suggested. – André Nicolas Mar 27 '13 at 22:54
  • Thank you! So to find the pdf of W, I should get the P(W < w) = P(X/Y < w). However I do not know the next step. What can i replace w with? I know that I need to take the double integral of the joint probability density function but I am confused as to how I can go forward from P(X/Y < w)? Thank you :) Hope I am on the right track! – nicefella Mar 27 '13 at 23:14
  • 1
    Everything is positive. Rewrite as $X\le Yw$. Now look at line $x=yw$, or more familiarly $y=x/w$. Draw that line. We want the probability of being above that line. That's the integral of the joint density over a certain infinite "triangle." – André Nicolas Mar 27 '13 at 23:22
  • Excellent! I understand now, thanks again! :) – nicefella Mar 27 '13 at 23:42
  • 1
    You are welcome. Just in case there were more questions, I had typed an answer. But it is always best if possibly with some help, you do things yourself. Note the importance of the basic geometry. – André Nicolas Mar 28 '13 at 00:00

1 Answers1

1

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 $W$.

Community
  • 1
André Nicolas
  • 507,029