# CDF of 2 variables diverges

Suppose you have 2 i.i.d. drawn variables X,Y $$\sim\mathcal{U}(0,1)$$ , I'm trying to calculate the CDF of $$Z = X + (1-X)Y\in[0,1]$$. I get stuck around the CDF calculation.

$$F_Z(z) = P(Z \leq z) = P(Y \leq \frac{z-X}{1-X})$$

I'm trying to find this by integrating out $$f_{XY}$$. This, however, leads to divergence:

\begin{align} \int\limits_0^1\int\limits_0^\frac{z-x}{1-x} \frac{z-x}{1-x} \, \textrm{d}x \textrm{d}y &= \int\limits_0^1\min \left\{ \frac{z-x}{1-x}, 1\right\} \,\textrm{d}x \\ &= \int\limits_0^1 \frac{x-z}{x-1}\,\textrm{d}x \end{align}

How come this is divergent? Intuitively it seems like it should converge properly, but I think I am missing something ...

Update: a random simulation show a similar finding of divergence (as I decrease the bin size, the right bar goes up) so I guess my question changes to: what does it mean for a CDF to be divergent? Is it not proper then?

Update 2: despite the density keyword in the above code, the y-axis in the above plot is incorrect as described here.

• Are you accounting for the minimum in your final line? Commented Jul 11, 2020 at 0:03
• I think so, the minimum is $\frac{z-x}{1-x}$ when $\frac{z-x}{1-x}\leq 1$ and $1$ otherwise. The first condition translates to $z \leq 1$ which is always true due to $z\in[0,1]$.
– ciri
Commented Jul 11, 2020 at 1:07
• You've plotted the PDF (Probability Density Function), not the CDF (Cumulative Distribution Function). The CDF is $P(Z\le z)$ by definition, so its values are always in the interval $[0,1].$ In R, you can simulate this using x <- runif(10000, 0, 1);y <- runif(10000, 0, 1);z <- x + (1-x)*y;cdf <- ecdf(z);plot(cdf). Commented Jul 14, 2020 at 22:43
• Thanks @r.e.s. , yes it seems like I didn't use mpl correctly here. I was looking at the behavior of the right-most bar which seemed to increase in size and therefore indicated to me that the CDF had a singular point at the edge. As it turns out this is due to my incorrect interpretation of the y-axis (the bars are plotted such that the total area=1 so decreasing bar width increases y axis). Thanks for pointing it out though, I've updated my answer to reflect your comment.
– ciri
Commented Jul 14, 2020 at 22:46

I am not encountering an issue with divergence. Here's how I proceeded; define random variables $$U$$ and $$V$$ by $$U:=X$$ and $$V:=X+(1-X)Y$$. It's easy to see that $$(U,V)$$ has joint density $$f_{UV}$$ where $$\\f_{UV}(u,v)=f_{XY}(u,\frac{v-u}{1-u})\left\lvert\frac{\partial(x,y)}{\partial(u,v)}\right\lvert=\frac{1}{1-u}$$ whenever $$(u,v)\in\Omega$$ and $$f_{UV}(u,v)=0$$ elsewhere. Here, $$\Omega:=\{(u,v)\in(0,1)^2|u. We can obtain the density for $$f_V$$ simply by "integrating away" the $$u$$ variable in the joint density. Doing so gives us $$f_V(v)=\ln\big(\frac{1}{1-v}\big)$$ for $$v\in(0,1)$$ and $$f_V(v)=0$$ elsewhere. From basic integration in Calculus we get that $$F_V(v)=(1-v)\ln(1-v)+v$$ for $$v\in(0,1)$$, $$F_V(v)=0$$ for $$v\leq0$$, and $$F_V(v)=1$$ for $$v\geq1$$. This is the CDF you're looking for.

• Thank you very much Matthew, the final result looks correct. I'm wondering how you get to $f_{UV}(u,v)=\frac{1}{1-u}$ though. If I calculate $f_{XY}(u,\frac{v-u}{1-u})=v$ and $|\frac{\partial (x,y)}{\partial (u,v)}|=\frac{1}{1-u}$ I obtain $f_{UV}(u,v)=\frac{v}{1-u}$.
– ciri
Commented Jul 14, 2020 at 19:37
• If $X$ and $Y$ are i.i.d. uniform random variables on $[0,1]$ then the joint density of $(X,Y)$ factors as $f_{XY}(x,y)=f_{X}(x)f_{Y}(y)$. This means we can say $f_{XY}(x,y)=1$ on the unit square $[0,1]^2$ and $f_{XY}(x,y)=0$ elsewhere.
– user801306
Commented Jul 14, 2020 at 21:52
• Clear, thank you!
– ciri
Commented Jul 14, 2020 at 21:54

You can actually proceed with your initial method to arrive at this answer as well. Notice for $$z\in[0,1]$$ fixed we have that

$$P\big(Y\leq\frac{z-X}{1-X}\big)=\int_{ -\infty}^{\infty}P\big(Y\leq\frac{z-x}{1-x}\lvert X=x\big)f_X(x)\,dx=\int_{0}^{z}F_{Y|X}(\frac{z-x}{1-x}|x)f_X(x)\,dx$$

Independence suggests that $$F_{Y|X}(y|x)=F_Y(y)$$ so the last integral simplifies to $$\int_{0}^{z}\big(\frac{z-x}{1-x}\big)\,dx=(1-z)\ln(1-z)+z$$. I was going to show you this method initially but I took the opportunity to show you how to find the density of a random variable using transformations because I think it's cool af :)

• Thanks for this! I actually do need to calculate more complex ones with up to 6 variables so I'm already in the process of getting used to it now, learning as I go. I think for this particular question the key was the integral domain.
– ciri
Commented Jul 14, 2020 at 22:44