Proving a CDF is cadlag
By Sarah Richards
I'd like to know if this proof is correct. Thank you for your help.
Let $X$ be a random variable. A function $F_X: \mathbb{R} \to [0,1]$ defined by:$$F_X(x) = \mathbb{P}(X \leq x) $$is called the cumulative distribution function of X. By the definition of cadlag, it is a right continuous function with a limit on the left. That is,
$\bullet$ The limit on the left, $\, \lim_{s\uparrow t} F_X(s) = F_X(t^-) $ exists.
$\bullet$ The limit on the right, $\,\lim_{s\downarrow t} F_X(s) = F_X(t^+)$ exists and equals $F_X(t)$.
To begin, we show the continuous limit. For some decreasing sequence, $\{ x_n:\, \, x_n \downarrow t\}$, the sequence of events $\{X \leq x_n \}$, is a decreasing sequence of sets.$$\implies \lim_{n \to \infty} \boldsymbol{1}\{X \leq x_n \} = \boldsymbol{1} \left \{ \bigcap_{n=1}^\infty X \leq x \right\} = \boldsymbol{1}\{X \leq t\}$$$$\implies \lim_{s\downarrow t} F_X(s) = \lim_{n \to \infty} F_X(x_n) = \lim_{n \to \infty } \mathbb{P}(X \leq x_n) = \mathbb{P}(X\leq t) = F_X(t) $$By definition, $F_X$ is right-continuous. Now for the other direction. Naturally we notice that for a sequence of sets, $\{X \leq x_n \}$, which is decreasing the complement is correspondingly increasing.$$\boldsymbol{1}\left \{\bigcup_{n =1}^\infty X \leq x_n \right \} = \lim_{n \to \infty} \boldsymbol{1}\{X \leq x_n \} = \boldsymbol{1}\{X < x \} \implies \mathbb{P}(X < x) = \lim_{n \to \infty} \mathbb{P}(X \leq x_n)$$
If the sequence of numbers now approaches upwards, $\{ x_n:\, \, x_n \uparrow t\}$
$$ \lim_{s\uparrow t} F_X(s) = \lim_{n \to \infty } F_X(x_n) =\lim_{n \to \infty} \mathbb{P}(X \leq x_n) = \mathbb{P}(X < t)$$I'm most concerned about having to define some behavior of the function at this point for the second part. Please let me know how this may be improved.
$\endgroup$1 Answer
$\begingroup$You said. $$\lim_{s\uparrow t} F_X(s) = 1 - \lim_{s \downarrow t }F_X(s) $$ This is not correct. This would mean that if $F$ is a continuous CDF then $F(t)=1-F(t)$ or $F(t)=\frac 1 2$ for all $t$!.
If $x_n$ strictly increases to $x$ then $(X \leq x_n)$ increases to $X<x$ and $P(X \leq x_n) \to F(x-)$
Other parts of your answer seem OK.
$\endgroup$ 1![How can I play the content from another account on my new Xbox 360 console? [duplicate]](https://edu.smbbmcl.edu.pk/uploads/how-can-i-play-the-content-from-another-account-on-my-new-xbox-360-console-duplicate_720.jpg)
