Writing a Distribution Function in terms of $F$ If you have a distribution function, $F(x)$, of a random variable $X$, how would you write this $\mathbb{P}(X =i )$ in terms of $F$?
Keep in mind that $F(x) = \mathbb{P}(X\leq x)$ and $X$ will take on one value of $1,2,3$.
I can't figure out where to begin. Is it simply just writing an equation or is it a lot more challenging?
 A: $$
\Pr(X=i) = F(i) - \lim_{x\uparrow i} F(x).
$$
This is sometimes denoted $F(i) - F(i-)$.
Proof:  In the first place, if $x<i$ then
\begin{align}
F(i) = \Pr( X\le i) & = \Pr\Big( X\le x \text{ or } x < X \le x \Big) \\[10pt]
& = \Pr(X\le x) + \Pr(x<X\le i) = F(x) + \Pr(x<X\le i);
\end{align}
and therefore
$$
\Pr(x<X\le i) = F(i) - F(x),
$$
and so we have
$$
\Pr(X=i) \le F(i) - F(x). \tag 1
$$
Next notice that if $x<y$ then $F(x) = \Pr(X\le x) \le \Pr(X\le y) = F(y)$, so $F(x)$ gets bigger or at least stays the same as $x$ gets bigger.  Therefore $F(i)-F(x)$ gets smaller or stays the same as $x$ gets bigger.  Hence $\lim_{x\uparrow i} (F(i) - F(x))$ exists, and since $(1)$ is true of all $x\le i$, we have
$$
\Pr(X=i) \le \lim_{x\uparrow i} (F(i) - F(x)).
$$
But could $\Pr(X=i)$ be less that the limit?
\begin{align}
\Pr(X\le i) & = \Pr(X=i) + \Pr(X<i) \\[10pt]
& = \Pr(X=i) \\[10pt]
& {}\qquad{} + \underbrace{\Pr(X\le i-1) + \Pr(i-1<X\le i-\tfrac 1 2) + \Pr(i-\tfrac 1 2 < X\le i - \tfrac 1 3) + \cdots}
\end{align}
Here the thing to notice is that the series over the $\underbrace{\text{underbrace}}$ adds up to $\lim\limits_{x\uparrow i} F(x)$.
A: If $X$ is continuous, for every $i \in \mathbb{R}$, $\mathbb{P}(X = i) = 0$.
Now if $X$ is discrete, say over the nonnegative integers $\{0, 1, \dots\}$, let's denote $\mathbb{P}\left(X = k\right) = p_k$, $k \in \{0, 1, \dots\}$.
Then $\displaystyle F(x) = \mathbb{P}\left(X \leq x\right) = \sum\limits_{i=0}^{x}p_k$, assuming $x$ is a nonnegative integer.
Hint. What is $F(x)-F(x-1), x \geq 1$?
