Been trying to solve this problem for some time now. Any help would be appreciated.

X and Y are independent R.V's with distribution exp(a) each.

I'm asked to find the pdf of Z with:


I proceeded as the following:

$W=1+Y$ with $f_W(w)=f_Y(w-1)=ae^{-a{(w-1)}}$

As I got that I did my normal routine of setting $$Z=X/W$$ and $$U=W$$

so, $f_{Z,U}(z,u)=f_X(zu)f_W(u).|u|$

Following that, I have $f_Z(z)=\int_1^{\infty}f_{Z,U}(z,u)du$


I've got: $$f_Z(z)=\frac{a^2(e^{-(za)})(1-za+a)}{(za+a)^2}$$ which is a bit far from the answer that is :$$f_Z(z)=\frac{z+1+a}{a(z+1)^2}e^{-z/a}$$

Thank you

  • 1
    $\begingroup$ In order not to commit errors, I tend to use cdf instead of the pdf. So you would be interested by the probability $\mathbb P\left[ \frac{X}{1+Y}\leq z \right]$ for any $z>0$, then you can use the law of total probability to get the result and differentiate with respect to $z$ if possible to get the pdf. $\endgroup$ – P. Quinton Jan 26 at 8:32
  • $\begingroup$ how would you use that property on this problem? $\endgroup$ – Mahamad A. Kanouté Jan 26 at 8:52

Let's compute the cdf of $Z$, for any $z>0$ \begin{align*} F_Z(z) &= \mathbb P\left[ \frac{X}{1+Y} \leq z \right]\\ &= \mathbb P\left[ X \leq z (1+Y) \right]\\ &= \int_0^\infty \mathbb P\left[ X \leq z (1+Y)\middle| Y=y \right] f_Y(y) dy\\ &= \int_0^\infty \mathbb P\left[ X \leq z (1+y)\right] f_Y(y) dy\\ &= \int_0^\infty (1-\exp(-a z (1+y))) a\exp(-ay) dy\\ &= \int_0^\infty a\exp(-ay) dy - \exp(-az)\int_0^\infty a\exp(-a(1+z)y) dy\\ &= 1 -\frac{\exp(-az)}{1+z} \end{align*}

differentiating with respect to $z$ gives you the pdf : \begin{align*} f_Z(z) &= \frac{d}{dz} \left[ 1 -\frac{\exp(-az)}{1+z} \right]\\ &=\exp(-az)\frac{a+az+1}{(1+z)^2} \end{align*}

  • $\begingroup$ Quiet frankly I see why my professor says to always condition on something. Thank you for showing me this method ! $\endgroup$ – Mahamad A. Kanouté Jan 26 at 11:00
  • $\begingroup$ Would it be reasonable to use the same technique for example if we have Z=X+Y instead of drawing the region and integrating ? $\endgroup$ – Mahamad A. Kanouté Jan 26 at 12:50
  • $\begingroup$ Yes it would, actually it is the same but using the knowledge of the cdf of the exponential. $\endgroup$ – P. Quinton Jan 26 at 13:22

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