11
$\begingroup$

Definition: The c.d.f. $F$ of a random variable $X$ is a function defined for each real number $x$ as follows:$$F(x)=\Pr(X\leq x) \text{ for } -\infty<x<\infty$$


Let $$F(x^-)=\lim_{y\rightarrow x,\,y<x}F(y)$$ and $$F(x^+)=\lim_{y\rightarrow x,\,y>x}F(y)$$


Property of cumulative distribution function: A c.d.f. is always continuous from the right; that is , $F(x)=F(x^+)$ at every point $x$.


Proof: Let $y_1>y_2>\dots$ be a sequence of numbers that are decreasing such that $$\lim_{n\rightarrow \infty}y_n=x.$$Then the event $\{X\leq x\}$ is the intersection of all the events $\{X\leq y_n\}$ for $n=1,2,\dots$ .Hence, $$F(x)=\Pr(X\leq x)=\lim_{n\rightarrow \infty} \Pr(X\leq y_n)=F(x^+).$$


Now I think the left inequality can also be proved in the similar way as:

Let $y_1<y_2<\dots$ be a sequence of numbers that are increasing such that $$\lim_{n\rightarrow \infty}y_n=x.$$Then the event $\{X\leq x\}$ is the union of all the events $\{X\leq y_n\}$ for $n=1,2,\dots$ .Hence, $$F(x)=\Pr(X\leq x)=\lim_{n\rightarrow \infty}\Pr(X\leq y_n)=F(x^-).$$

Where am I wrong?

$\endgroup$
  • 2
    $\begingroup$ Look at $F=1_{[\alpha,\infty)}$ (point mass at $\alpha$). $F$ is a cdf, but is not continuous from the left. You can see that if $y_n \uparrow \alpha$, but $y_n < \alpha$, then $\lim_n F(y_n) = 0$, but $F(\alpha) = 1$. $\endgroup$ – copper.hat Oct 11 '13 at 6:51
22
$\begingroup$

You write:

Then the event $\{X\leq x\}$ is the union of all the events $\{X\leq y_n\}$ for $n=1,2,\dots$.

This is the faulty step. To wit:

If $y_n\lt x$ for every $n$ and $y_n\to x$, then $\bigcup\limits_n\{X\leqslant y_n\}$ is equal to $\{X\lt x\}$, not to $\{X\leqslant x\}$.

You might want to check that $x$ is in $(-\infty,y_n]$ for no $n$ whatsoever hence $x$ is not in $\bigcup\limits_n(-\infty,y_n]$, and in fact $\bigcup\limits_n(-\infty,y_n]$ is equal to $(-\infty,x)$, not to $(-\infty,x]$.

$\endgroup$

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.