11
$\begingroup$

I've learned in my probability courses that the cumulative distribution function $F$ of a random variable $X$ is right continuous. Is it possible to prove that?

$\endgroup$
17
$\begingroup$

To prove the right continuity of the distribution function you have to use the continuity from above of $P$, which you probably proved in one of your probability courses.

Lemma. If a sequence of events $\{A_n\}_{n\geq 1}$ is decreasing, in the sense that $A_n\supset A_{n+1}$ for every $n\geq 1$, then $P(A_n)\downarrow P(A)$, in which $A=\cap_{n=1}^\infty A_n$.

Let's use the Lemma. The distribution function $F$ is right continuous at some point $a$ if and only if for every decreasing sequence of real numbers $\{x_n\}_{n\geq 1}$ such that $x_n\downarrow a$ we have $F(x_n)\downarrow F(a)$.

Define the events $A_n=\{\omega : X(\omega)\leq x_n\}$, for $n\geq 1$. We will prove that $$\bigcap_{n=1}^\infty A_n=\{\omega:X(\omega)\leq a\}\, .$$

In one direction, if $X(\omega)\leq x_n$ for every $n\geq 1$, since $x_n\downarrow a$, we have $X(\omega)\leq a$.

In the other direction, if $X(\omega)\leq a$, since $a\leq x_n$ for each $n\geq 1$, we have $X(\omega)\leq x_n$, for every $n\geq 1$.

Using the Lemma, the result follows: $$ F(x_n) = P\{X\leq x_n\} = P(A_n) \downarrow P\left( \cap_{n=1}^\infty A_n \right) = P\{X\leq a\} = F(a) \, . $$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.