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Suppose I have the following cdf of a random variable $X$.

$ F(x) = \begin{cases} 0 & \text{if } x < 0; \\ x/2 & \text{if } 0 \leq x < 1; \\ 2/3 & \text{if } 1 \leq x < 2; \\ 11/12 & \text{if } 2 \leq x < 3; \\ 1 & \text{if } x \geq 3.\end{cases} $

First, what is $\mathbb{P}(X<3)$? Is it $\frac{11}{12}$? Second, what is $\mathbb{P}(X=1)$? Is it $\frac{2}{3} - \frac{1}{2} = \frac{1}{6}$?

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    CDFs are right-continuous functions. Thus, $F(a) = P\{X \leq a\}$ is the limiting value of $F(x)$ as $x \to a$ from the right while P\{X < a\} is the limiting value of $F(x)$ s $x \to a$ from the left. The limits are different if there is$a$jump discontinuity at $a$, and the difference (if any) is the probability that $X = a$.2012-09-12

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As $F(x) = \mathbb{P}(X \leqslant x)$ is right-continuous, i.e. $F(x) = \lim\limits_{\epsilon \to 0^+} F(x+\epsilon)$, we have: $ \mathbb{P}(X=x) = F(x) - \lim\limits_{\epsilon \to 0^+} F(x-\epsilon) $ and $ \mathbb{P}(X

Numerical values you computed are correct: $ \mathbb{P}(X < 3) = \lim_{\epsilon \to 0^+} F(3-\epsilon) = \lim_{\epsilon \to 0^+} \frac{11}{12} = \frac{11}{12} $ $ \mathbb{P}(X = 1) = F(1) - \lim_{\epsilon \to 0^+} F(1-\epsilon) = \frac{2}{3} - \lim_{\epsilon \to 0^+} \frac{1-\epsilon}{2} = \frac{1}{6} $