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Hi is the following a true statement.

I have a random variable $P(X)$, and I need to find $P(1\leq X <\frac{3}{2})$.

So I was thinking whether the following statement is true:

$P(1\leq X <\frac{3}{2})=P(X=1)+P(1<X<\frac{3}{2})$.

Jakob Jul Elben
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    Yes, that is definitely true. That is because the event ${1 \leq X < \frac{3}{2}}$ is a union of two disjoint events ${ X = 1 }$ and ${ 1 < X < \frac{3}{2}}$ (and the probability law $P$ is additive). – Sangchul Lee Oct 08 '16 at 12:19
  • So if I have the following cumulative distributions function $F(X)=\begin{cases}0, x\leq 0\ \frac{x}{3}, 0<x\leq 1\ \frac{2x-1}{3},1<x\leq 2\ 1, x>2 \end{cases}$ means that. $P(1<X\leq \frac{3}{2})=P(1<X< \frac{3}{2})+P(X=1)=P(1<X< \frac{3}{2})=F(\frac{3}{2})-F(1)$ – Jakob Jul Elben Oct 08 '16 at 12:24
  • Notice that your CDF defines a continuous distribution and consequently $P(X = 1) = 0$. (Or simply, $P(X = 1) = F(1) - F(1^-) = \frac{1}{3} - \frac{1}{3} = 0$.) So, Yes. – Sangchul Lee Oct 08 '16 at 12:28

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I suppose you meant "random variable $X$".

Yes, that is true, because probability (measure) is a measure, hence it must satisfy countable aditivity - in symbols (without exact details), let $\{A_i\}_{i = 1}^{\infty}$ be the set of pairwise disjoint measurable sets and $P$ be the probability measure. Then $P(\bigcup_{i = 1}^{\infty} A_i) = \sum_{i = 1}^{\infty} P(A_i)$.

This also applies to finite disjoint union, because $\bigcup_{i = 1}^{n}A_i = A_1 \cup A_2 \cup \cdots \cup A_n \cup \emptyset \cup \emptyset \cup \cdots$.

And this applies to your situation, because $\{1 \leq X < 3/2 \} = \{X = 1\} \cup \{1 < X < 3/2\}$ and $\{X = 1 \} \cap \{1 < X < 3/2\} = \emptyset$.

Pan Miroslav
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