Convolution of Uniform Distribution and Square of Uniform Distribution

Question

I am trying to find the CDF of $Z=X+Y$ whereby $X$ and $Y$ are random variables. Given that the CDF of Z is:

$$F_Z(z)=\int F_X\left(z-y\right)f_Y(y)dy$$

Given that $X$ is uniform distribution over $[0,1]$ with:

$$f_X \left(x \right)=\frac{1}{\sigma }\left [ H\left ( x \right ) -H\left ( x-\sigma \right )\right ]$$ $$F_X \left(x \right)=\frac{x}{\sigma }\left [ H\left ( x \right ) -H\left ( x-\sigma \right )\right ]$$

And $Y$ is square of uniform distribution over $[0,1]$ with:

$$f_Y \left(x \right)=\frac{log\left ( \frac{\sigma ^{2}}{x} \right )}{\sigma ^{2}}$$ $$F_Y \left(x \right)=\frac{x\left ( 1+2log\left ( \sigma \right ) - log\left ( x \right )\right )}{\sigma ^{2}}$$

For $\sigma=1$, how I started to solve is by partitioning the problem to $0<z<1$ and $1<z<2$. I used the integration below with its limits to obtain the solution:

$$\int_0^z (z-y) \log \left(\frac{1}{y}\right) \, dy=\frac{1}{4} z^2 (3-2 \log (z))$$ $$\int_{z-1}^1 (z-y) \log \left(\frac{1}{y}\right) \, dy=\frac{1}{4} \left(2 \left(z^2-1\right) \log (z-1)-3 (z-2) z\right)$$

Checking against numerical results, the first solution I obtained is correct. The second is not because it gives me the following graph for $1<z<2$:

Where am I wrong in the second integral? Is it the limits? Or is it something else? Thanks

Answer 1

If $V\sim U[0,1]$ then $Y:=V^2$ has: $$\begin{eqnarray*} &i)&f_{Y}(y)=\frac{{\bf 1}_{\{0\leqslant y\leqslant 1\}}}{2\sqrt{y}}\\ &ii)&F_{Y}(y)={\bf 1}_{\{y > 1\}} + {\bf 1}_{\{0\leqslant y\leqslant 1\}}\sqrt{y} \end{eqnarray*} $$

This is in contrast with your pdf $f_Y(y)=\log (1/y)$.

In addition, assuming that $X$ and $Y$ are independent, we have $$ \begin{eqnarray*} F_{Z}(z) &=& {\bf 1}_{\{z \geqslant 2\}} + {\bf 1}_{\{0\leqslant z <2\}}\int_{-\infty}^{\infty}F_{X}(z-y)f_{Y}(y)\text dy \\ &=& {\bf 1}_{\{z \geqslant 2\}} + {\bf 1}_{\{0\leqslant z <2\}}\int_{-\infty}^{\infty}\frac{z-y}{2\sqrt{y}}{\bf 1}_{\{0\vee z-1\ \leqslant y\leqslant\ z \wedge 1\}}\text dy\\ &=& {\bf 1}_{\{z \geqslant 2\}} + {\bf 1}_{\{0\leqslant z <1\}}\int_{0}^{z}\frac{z-y}{2\sqrt{y}}\text dy + {\bf 1}_{\{1\leqslant z <2\}}\Big(\int_{z-1}^{1}\frac{z-y}{2\sqrt{y}}\text dy + \int_{0}^{z-1}\frac{1}{2\sqrt{y}}\text dy\Big). \end{eqnarray*} $$

Hence,

$$ \begin{eqnarray*} F_{Z}(z) &=& \ \ {\bf 1}_{\{z \geqslant 2\}} + {\bf 1}_{\{0\leqslant z < 1\}}\frac{2}{3}z^{3/2} \\ && + {\bf 1}_{\{1\leqslant z < 2\}}\Big(z-\frac{1}{3}-z(z-1)^{1/2}+\frac{1}{3}(z-1)^{3/2}+(z-1)^{1/2}\Big)\,. \end{eqnarray*} $$