Find CDF of quotient of uniform RVs

Question

Let $U_1,U_2 = \operatorname{Uniform}[0,1]$ and independent. Find the CDF of $Z=\frac{U_1}{U_2+1}$.

My attempt:

$P(Z\le z)=P(U_1\le (U_2+1)z, U_2+1> 0)+P(U_1\ge (U_2+1)z, U_2+1<0)$. The last term will be equal to $0$, since $U_2$ is never less than $-1$. This results in: $$ F_Z(z)=\int_0^1du_2\int_0^{(u_2+1)z}du_1=z\int_0^1(u_2+1)du_2=\frac32z.$$ This holds only if $0\le (u_2+1)z\le 1$ and since $2\ge u_2+1>0$, we get $0\le z\le \frac12.$

So, my answer would be $$ F_Z(z)=\begin{cases} 0 & \text{if }\quad z<0 \\ \frac32z&\text{if }\quad 0\le z\le\frac12\\1&\text{if }\quad z>\frac12\end{cases}.$$

However, in the solutions, there is an extra component: $2-\frac{1}{2z}-\frac{z}2$ if $\frac12\le z\le 1$. I don't see where this one comes from.

Thanks.

Answer 1

The actual bound for the variable $U_1$ is $min((u_2+1)z , 1)$, since $U_1$ is always less than $1$. Hence you need to compute

$$\int_0^1 \int_0^{min((u_2+1)z , 1)} du_1du_2 = \int_0^1 min((u_2+1)z , 1) du_2$$ We see that $(u_2+1)z \leq 1$ exactly when $u_2 \leq \frac1z -1$, hence for $z\in (1/2,1)$ we get, that $$\int_0^1 min((u_2+1)z , 1) du_2=\int_0^{1/z-1}(u_2+1)zdu_2+\int_{1/z-1}^1du_2 = 2-\frac{1}{2z}-\frac{z}{2}$$

Answer 2

Get a sheet of paper and plot the lines. You properly defined the problem but you messed up with the bounds.

We have $0 \leq z \leq 1$, and we plot the straight line $$y=z(x+1)$$ because the cdf is $$F_Z(z)=P(U_1 \leq z(U_2+1))$$ The probability is the intersection between the square $S$ : $[0,1] \times [0,1]$ and the area under the straight line $Y$ of equation $y=z(x+1)$

The line $Y$ intercepts the line of equation $x=1$ at the point $(1,2z)$. This point is within $S$ if $0 \leq z \leq \frac{1}{2}$. The ared under $Y$ intersected with $S$ is a trapezoid whose small base is $z$, the bigger one is $2z$ and the height is one. Therefore the probability is $$1*\frac{z+2z}{2}=\frac{3}{2}z$$

If $z >\frac{1}{2}$, the point $(1,2z)$ is outside $S$. The line $Y$ will intercept the line $y=1$ at the point $(\frac{1}{z}-1,1)$. The area under $Y$ intersected with $S$ is the area of $S$ minus the area of the triangle with points $(0,z)$, $(0,1)$ and $(\frac{1}{z}-1,1)$, and its area is $$\frac{(1-z)\left(\frac{1}{z}-1\right)}{2}$$

So the probability is $$1-\frac{(1-z)\left(\frac{1}{z}-1\right)}{2}=2-\frac{1}{2z}-\frac{z}{2}$$