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Could someone explain why this probablity ISN'T 0? Despite being a CDF?

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$P(X = 1/2) \neq 0$? Shouldn't it be $P(X = 1/2) = F(1/2) - F(1/2) = 0$

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In gerneral from cumulative F(x), P(X=c) is F(c) minus the limit of F(x) as x -> c from the left. Try that here. –  coffeemath Oct 17 '12 at 5:21

1 Answer 1

up vote 3 down vote accepted

For any $a$, $P(X=a)=P(X\le a)-P(X<a)$ is the ‘jump’ in $F_X$ and $a$, which is $$F_X(a)-\lim_{x\to a^-}F_X(a)\;.$$ This jump is of course $0$ if $F_X$ is continuous at $a$, but in this problem we’re interested in values of $a$ at which $F_X$ is not continuous. In particular,

$$\begin{align*} P\left(X=\frac12\right)&=P\left(X\le\frac12\right)-P(\left(X<\frac12\right)\\\\ &=F_X\left(\frac12\right)-\lim_{x\to\frac12^-}F_X(x)\\\\ &=1-\lim_{x\to\frac12^-}\frac32x\\\\ &=1-\frac32\cdot\frac12\\\\ &=\frac14\;. \end{align*}$$

Similarly, $$P\left(X<\frac14\right)=\lim_{x\to\frac14^-}F_X(x)=\lim_{x\to\frac14^-}\frac{x}2=\frac18\;.$$

But

$$\begin{align*} P\left(X=\frac38\right)&=P\left(X\le\frac38\right)-P\left(X<\frac38\right)\\\\ &=F_X\left(\frac38\right)-\lim_{x\to\frac38^-}F_X(x)\\\\ &=\frac32\cdot\frac38-\lim_{x\to\frac38^-}\frac32x\\\\ &=\frac32\cdot\frac38-\frac32\cdot\frac38\\\\ &=0\;, \end{align*}$$

because there is no ‘jump’ at $\frac38$: $F_X$ is continuous at $x=\frac38$.

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