Let $Y = X − [X]$, where $X\sim U(0,\theta)$. Show that $Y \sim U (0, 1)$

Question

Let $X \sim U(0,\theta)$, where $\theta$ is a positive integer. Let $Y = X − [X]$, where $[x]$ is the largest integer $≤ x$. Show that $Y \sim U (0, 1)$

Clearly, the support of $Y$ is $S_Y = [0,1]$.

In order to show this, I want to be able to prove that for $y\in [0,1]$

$F_Y(y) = y$

But for $y\in [0,1]$

$$F_Y(y) = Pr(X \leq y + [X]) = \sum_{i \; = \;0}^\theta Pr(X \leq y + i, \; [X] = i) =\frac1\theta \sum_{i \; = \;0}^\theta \int_{i}^{y+i} dx =\frac1\theta \sum_{i \; = \;0}^\theta y = \frac{\theta +1}\theta y \neq y$$

Is this question wrong or am I solving it incorrectly?

Answer 1

Hint: Your integral cannot extend up to $y+i$. For example when $i=\theta$ the interval $(i,y+i)$ is completely outside $(0,\theta)$ so the density function is $0$ there.

Answer 2

If $X\sim U(0;\theta)$ the distribution cannot change with a linear transformation. Then the resulting distribution of Y must be still uniform.

The only issue you have to proof is the Y-support, but it is enough to observe that Y is exactly the decimal part of X, that is $Y \in [0;1]$ as you already noted