understanding uniform distribution cdf equivalence

Question

I am confused about the following statement which I read some lecture notes. So the claim is: $Y$ is uniformly distributed in the interval $[0,\alpha]$ and the way the notes proves this is by showing that: $\Pr[Y\leq y]=y/\alpha$ for all $y$. Now, why is showing $\Pr[Y\leq y]=y/\alpha$, equivalent to showing $Y$ is uniform in the interval $[0,\alpha]$. I see if $Y$ is uniform then this statement is true, but not the converse, could someone clarify?

Answer 1

How are you mathematically defining the statement "$Y$ is uniformly distributed on $[0,a]$?"

If you are defining it as $Y$ has the density $$f_Y(y) = \frac{1}{a} \mathbb 1(0 \le y \le a),$$ then we can compute $$\begin{align*} f_Y(y) &= \lim_{\Delta y \to 0} \frac{\Pr[Y \le y + \Delta y] - \Pr[Y \le y]}{\Delta y} \\ &= \lim_{\Delta y \to 0} \frac{1}{\Delta y} \left(\frac{y + \Delta y}{a} - \frac{y}{a}\right)\mathbb 1 (0 \le y + \Delta y \le a) \\ &= \lim_{\Delta y \to 0} \frac{1}{a} \mathbb 1 (0 \le y + \Delta y \le a) \\ &= \frac{1}{a} \mathbb 1 (0 \le y \le a), \end{align*}$$ as claimed.

But if you define it in some other way, then you have to state this definition. For example, you can say that $Y$ is uniform if for any $0 \le y_1 < y_2 \le a$, $$\Pr[y_1 \le Y \le y_2] = \frac{y_2 - y_1}{a}.$$ This is also an acceptable definition. It means that the probability that $Y$ lies in some subinterval $[y_1, y_2]$ is proportional to the length of that subinterval, and that the constant of proportionality is independent of the choice of subinterval.