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.

Leave a Reply Cancel reply

Enter your comment here...

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

Email (required) (Address never made public)

Name (required)

Website

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

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

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

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

Cancel

Connecting to %s

	DSL for Generative M… on DSL for Generative Models…
	Measurable Spaces… on Measurable Spaces – Prob…
	Measurable Spaces… on Measurable Spaces – Prob…
	Measurable Spaces… on Measurable Spaces – Prob…
	Ballot Theorem… on Ballot Theorem – 2 …