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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!

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  • $\begingroup$ 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 $c\in [0,1]$. $\endgroup$
    – paw88789
    Oct 12 '14 at 2:07
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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.

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