# Find the pdf of Z=X/(1+Y)

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:

$$Z=\frac{X}{1+Y}$$

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

$$f_Z(z)=\int_1^{\infty}a^2ue^{-u(za+a)}e^adu=a^2e^a\int_1^{\infty}ue^{-u(za+a)}du$$

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

• 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. – P. Quinton Jan 26 at 8:32
• how would you use that property on this problem? – 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*}