Random Variables – Problem (50/365)

Moving on to the next chapter “Random Variables – I”, take a look at the following problem. Show that the random variable \theta is continuous if and only if P(\theta = x) = 0 for all x \in \mathbb{R}.

(Forward direction) Suppose \theta is a continuous random variable, then its distribution function F is also continuous by definition. Hence, P(\theta = x) = F(x) - F(x-) = F(x) - \lim_{y \rightarrow x} F(y) = 0 by definition of continuity.

(Reverse direction) Suppose P(\theta = x) = 0 for all x \in \mathbb{R} and let F be the corresponding distribution function. Let \{A_n\} be a sequence of sets such that A_n \supseteq A_{n+1} such that \cap_{n=1}^\infty A_n = \{x\}. Then, P(\cap_{n=1}^\infty A_n) = \lim_n P(A_n) because P is countably additive. Thus, \lim_n P(A_n) = P(x) and since P(a,b] = F(b) - F(a) we see that \lim_{x \rightarrow c} F(x) = F(c) for any c which is the definition of continuity.

This entry was posted in Uncategorized and tagged , . Bookmark the permalink.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s