# Showing Independence for X and frac(X + Y)

Suppose that we have independent samples $X_1, X_2 \sim \text{unif}[0, 1)^d$. I'm asked to show that $Y_1 = X_1$ and $Y_2 = X_1 + X_2 - \lfloor X_1 + X_2 \rfloor$ are also independent uniform samples from $[0, 1)^d$.

I'm able to show that $Y_2$ is uniform on $[0, 1)^d$ by using a convolution for $X_1 + X_2$ and then reasoning about $\lfloor X_1 + X_2 \rfloor$ to obtain the PDF.

However, I'm not sure how to formally show that $Y_1, Y_2$ are independent. This post indicates that identical distribution does not imply independence. I also don't think the joint PDF is calculable given only the marginals without assuming independence?

Any advice on what approach to use to show independence that uses the fact that $X_1, X_2$ are independent?

• If $X_1$ and $X_2$ are independent, it should not be too difficult to show that given $X_1 = x_1$ you have $X_1 + X_2$ conditionally uniformly distributed on $[x_1,x_1+1)$ and so you have $X_1 + X_2 - \lfloor X_1+X_2\rfloor$ conditionally uniformly distributed on $[0,1)$. Since this latter conditional distribution does not depend on $x_1$, you get independence of $Y_1$ and $Y_2$ – Henry Oct 2 '17 at 17:34
• Thanks! I had something written down to the effect that the final PDF doesn't depend on $X_1$ but wanted to know if there was a more formal treatment (since this was just my intuition). – Aaron Zou Oct 2 '17 at 19:49

This is a collection of $d$ identical statements, one for each coordinate, so it suffices to examine the case $d=1$.
Let $0 \le y \lt 1$. Define the random variable $$X_2^{(y)} = \left\{\matrix{X_2+1 & X_1 \le y \\ X_2&\text{otherwise.}}\right.$$ $(X_1, X_2^{(y)})$ has a uniform distribution on a region consisting of the unit square $[0,1)^2$ from which a triangle bounded by $(0,0), (y,0), (0,y)$ (yellow in the figure) has been removed and translated one unit upwards to the triangle bounded by $(0,1), (y,1), (0,1+y)$.
Writing $[z] = z-\lfloor z\rfloor$ for the fractional part of any number $z$,
$$Y_2 \le y \iff [X_1+X_2] \le y \iff 1 \le X_1 + X_2^{(y)} \le 1_y.$$
The probability of $X_1 + X_2^{(y)}$ within this region is concentrated in the parallelogram with vertices $(0,1), (1,0), (1,y), (0,1+y)$ (the blue region in the figure). If, simultaneously, $X_1\le x$ (for $0 \le x \lt 1$), the probability is the area of this parallelogram to the left of $X_1=x$. That region, too, is a parallelogram: its height is still $y$ but its base is only $x$, whence its area is $$\Pr(Y_1 \le x,\ Y_2 \le y) = xy = \Pr(Y_1 \le x)\Pr(Y_2 \le y).$$
We have seen how, throughout the support of $(Y_1,Y_2)$, its distribution function factors into the product of the marginal distributions. Consequently $(Y_1,Y_2)$ is independent.