In the Wikipedia article on cumulative distribution functions (CDF), the following statement is given:
The probability that $X$ lies in the semi-closed interval $(a,b]$, where $a<b$, is therefore$$\mathbb{P}\left(a < X \mathbin{\color{darkblue}{\le}} b \right) = F_X(b) – F_X(a)
$$
It then proceeds to state the following:
[…] the "less than or equal to" sign, "$\mathbin{\color{darkblue}{\le}}$", is a convention, not a universally used one (e.g. Hungarian literature uses "$<$"), but the distinction is important for discrete distributions.
This got me thinking, if we have a discrete distribution, then what is $\mathbb{P}\left(a < X \mathbin{\color{red}{<}} b\right)$?
I believe the problem arises as a consequence of a CDF only having guaranteed right-continuity, but not overall continuity. This means that for a discreet case we could have $\lim\limits_{x \to b^{-}} F_X(x)$ not necessarily equal to $\lim\limits_{x \to b^{+}} F_X(x) = F_X(b)$.
I tried to come up with a formula for changing the "less than or equal" sign to "strictly less" but couldn't make it work. My guess is that $\mathbb{P}\left(a < X < b\right) = \lim\limits_{x \to b^-}F_X(x) – F(a)$ using a similar reasoning to the above.
I also tried to use the complimentary CDF, which hs the property that $\overline{F_X(x)} = 1 -F_X(x)$, but I couldn't see how to combine the two inequalities $a <X <b$ at the same time with this property of the complementary CDF.
Does anyone know if there's a formula for $\mathbb{P}\left(a < X < b\right)$ using CDF's in the general case? Or alternatively, if a closed form isn't possible, could someone explain why? Thank you!
Best Answer
Your guess is correct and quite easy to prove. So first of all, as $F_X$ is monotone and bounded, the limit
$$\lim_{x\to b^-}F_X(x)$$
exists and we can compute it using some convenient sequence. Indeed we have that
$$\lim_{x\to b^-}F_X(x)=\lim_{n\to\infty}F_X\left(b-\frac{1}{n}\right)=\lim_{n\to\infty}\mathbb{P}_X\left(\left(-\infty,b-\frac{1}{n}\right]\right)=\mathbb{P}_X((-\infty,b))=\mathbb{P}(X<b)$$
where we've used the continuity of measures on the increasing sequence of sets $\left\{\left(-\infty,b-\frac{1}{n}\right]\right\}_{n\in\mathbb{Z}^+}$ with limit $(-\infty,b)$. It then follows that
$$\mathbb{P}(a<X<b)=\mathbb{P}(X<b)-\mathbb{P}(X\leq a)=\lim_{x\to b^-}F_X(x)-F_X(a).$$
Notice that this works for any random variable $X$, regardless of if it's continuous, discrete or neither.