2

Let $X_1,X_2$ be independent $\text{Uniform}(0,1)$ random variables. Define $U_1 = X_1 - \lfloor X_1 \rfloor$ and $U_2 = X_1 + X_2 - \lfloor X_1 + X_2 \rfloor$ where $\lfloor a \rfloor$ is the largest integer less or equal to $a \in \mathbb{R}$. We can take as given that $U_1$ and $U_2$ are also $\text{Uniform}(0,1)$.

We want to show that $U_1$ and $U_2$ are independent. We can do so by arguing in cases that $P(U_1 \leq u_1, U_2 \leq u_2) = P(U_1 \leq u_1) P(U_2 \leq u_2) = u_1 u_2$ for (1) $u_1 > u_2$ and (2) $u_1 \leq u_2$. I'm struggling with arguing two lines in case (2). The solution given is as follows: \begin{align} P(U_1 \leq u_1, U_2 \leq u_2) &= P(X_1 - \lfloor X_1 \rfloor \leq u_1, X_1 + X_2 - \lfloor X_1 + X_2 \rfloor \leq u_2) \\\\ &= P\left(X_1 - \lfloor X_1 \rfloor \leq u_1, 0 \leq X_1 + X_2 \leq u_2\right) + P(X_1 - \lfloor X_1 \rfloor \leq u_1, 1 \leq X_1 + X_2 \leq 1 + u_2) \\\\ &= E\left[P(X_1 - \lfloor X_1 \rfloor \leq u_1, 0 \leq X_1 + X_2 \leq u_2 | X_1 = x)\right] \\\\ &\ \ \ \ + E\left[P(X_1 - \lfloor X_1 \rfloor \leq u_1, 1 \leq X_1 + X_2 \leq 1 + u_2 | X_2 = y)\right] \\\\ &= \int_0^{u_1} P(0 \leq X_1 + X_2 \leq u_2 | X_1 = x)\ dx + \int_0^{u_1} P(1 \leq X_1 + X_2 \leq 1 + u_2 | X_2 = y)\ dy \\\\ &= \int_0^{u_1} P(X_2 \leq u_2 - x)\ dx + \int_0^{u_1} P(1 - y \leq X_2 \leq 1)\ dy \\\\ &= \int_0^{u_1} u_2 - x\ dx + \int_0^{u_1} y\ dy \\\\ &= u_1 u_2 \end{align} as desired. Between the third and fourth lines, I'm unclear why it must be the case that $$ E\left[P(X_1 - \lfloor X_1 \rfloor \leq u_1, 1 \leq X_1 + X_2 \leq 1 + u_2 | X_2 = y)\right] = \int_0^{u_1} P(1 \leq X_1 + X_2 \leq 1 + u_2 | X_2 = y)\ dy. $$ "Pulling" the indicator $I(X_1 \leq u_1)$ conditioned on $X_1 = x$ makes sense for the first expectation, but it doesn't seem justified conditioned on $X_2 = y$. That is, $X_1$ and $X_1 + X_2$ aren't conditionally independent given $X_2 = y$. Additionally, assuming this holds, I'm struggling with arguing that $$\int_0^{u_1} P(1 \leq X_1 + X_2 \leq 1 + u_2 | X_2 = y)\ dy = \int_0^{u_1} P(1 - y \leq X_2 \leq 1)\ dy$$ between the fourth and fifth lines. It's possible that there's a typo in the solution where the second expectation in the third line should actually be conditioned on $X_1 = x$, say. While this would rectify my first issue, I'm still unclear why the equality holds between the fourth and fifth lines.

Joe Robert
  • 23

1 Answers1

1

Yes, there is a typo between the third and fourth lines. It is supposed to be $$ E\left[P(X_1 - \lfloor X_1 \rfloor \leq u_1, 1 \leq X_1 + X_2 \leq 1 + u_2 | X_1 = y)\right] = \int_0^{u_1} P(1 \leq X_1 + X_2 \leq 1 + u_2 | X_1 = y)\ dy. $$ As for your second question we have $y\leq u_1$ on this interval which by the intial assumption implies $y\leq u_2\longrightarrow 1\leq 1+u_2-y$.

So the right side of the inequality $P(1-y\leq X_2\leq 1+u_2-y)$ vanishes to $P(1-y\leq X_2\leq 1)$ because $X_2\leq 1$.

Ryan
  • 1,666