# Question about probability of a random variable with uniform distribution

Why is it that when a random variable has a uniform distribution the following statement it's true? $\Pr \left ( X_{1} \leqslant c \right ) = c$

The question arised when I was doing this exercise where there were a sequence of random variables $X_1, X_2,\ldots, X_n$ iid with common distribution $U(0,\theta)$. And another random variable $M$ defined as follows: $M : =\max\{ X_1, X_2,\ldots, X_n \}$ . It was asked then to evaluate $\Pr\left( M\leq \frac{\theta }{2} \right )$ The answer for the probability of each $X_{i}$ was $\theta/2$ .

But , as I learned from textbooks the probability would be the area under the function or geometrically $\frac{x - a}{b - a}$ which would give $1/2$ . I don't know what else to do, was this a typo? I've been the whole day trying to figure this out but now I'm exhausted and considering an accidental error. Thanks!

• For the statement $\Pr(X_1\le c)=c$, we would need to be $X_1$ to be $U(0,1)$. [The statement would then hold for all $c\in [0,1]$. Oct 12 '14 at 2:07

The cumulative distribution function $F_X(x) = \operatorname{P}(X\leq x)$ for the continuous uniform distribution with support $x \in [a,b]$
$$F(x)= \begin{cases} 0 & \text{for }x < a \\[8pt] \frac{x-a}{b-a} & \text{for }a \le x < b \\[8pt] 1 & \text{for }x \ge b \end{cases}$$
If, in your case, $a=0$ and $b=1$ and take $x=c$ you can see why your statement is true.