Distribution of a Random Variable « Mathematics Prelims

November 1, 2008

Distribution of a Random Variable

Let $(\Omega, \mathcal{F}, P)$ be a probability space and $X : \Omega \to \mathbb{R}$ a random variable. Define a function $F(x)$ as the probability $X$ takes on a value less than or equal to $x$ .

$\displaystyle F(x) := P(X^{-1}((-\infty, x])$

That is, $F(x)$ is the measure of the set of $\omega \in \Omega$ such that $X(\omega) \leq x$ . Notice that $F(x)$ is an increasing function as $a \leq b$ implies $(-\infty, a] \subseteq (-\infty, b]$ and so

$\displaystyle X^{-1}((-\infty, a]) = \{ \omega \in \Omega : X(\omega) \leq a \}$

$\displaystyle \, \, \subseteq \{ \omega \in \Omega : X(\omega) \leq b \}$

$\displaystyle \, \, = X^{-1}((-\infty, b])$

And since a measure is monotonic (i.e., $A \subseteq B \implies P(A) \leq P(B)$ ), we have that $F(a) \leq F(b)$ .

Recall that if $(E_n)_{n \in \mathbb{N}}$ is a decreasing sequence of measurable sets (i.e., $E_n \supseteq E_{n+1}$ ), then $P\left(\bigcap_{n \in \mathbb{N}} E_n\right) = \lim_{n \to \infty} P(E_n)$ , provided there exsists an $E_n$ with $P(E_n)$ finite (in our case $P$ is a probability measure, so this is given to us anyway). We can use this fact to show that our $F$ function is right-continuous.

Let $(x_n)_{n \in \mathbb{N}}$ be a real-valued sequence that decreases to $x$ . Then we have that

$\displaystyle \lim_{n \to \infty} F(x_n) = \lim_{n \to \infty} P(X^{-1}((-\infty, x_n]))$

$\displaystyle = P \left( \bigcap_{n = 1}^\infty X^{-1}((-\infty, x_n])\right)$

$\displaystyle = P \left( X^{-1} \left( \bigcap_{n=1}^\infty (-\infty, x_n] \right) \right)$

$\displaystyle = P(X^{-1}((-\infty, x]))$

$\displaystyle = F(x)$

This shows that $\displaystyle \lim_{x \to x_0^+} F(x) = F(x_0)$ and so $F$ is right-continuous.

Additionally, $\lim_{x \to \infty} F(x) = 1$ and $\lim_{x \to -\infty} F(x) = 0$ .

The function $F$ that we’ve described is known as the cumulative distribution function (or just distribution function) of our random variable $X$ . If $F(x)$ only takes on countably many values, then we say that $X$ is a discrete random variable. If $F(x)$ is a continuous function, we say $X$ is a continuous random variable.

Now that we’ve seen how to construct this distribution function given a random variable $X$ , the natural question to ask is if we have a function that satisfies the properties of a distribution that we’ve listed, is there a corresponding random variable?

Suppose that $F(x)$ is an increasing, right-continuous function with $\lim_{x \to -\infty} F(x) = 0$ and $\lim_{x \to \infty} F(x) = 1$ . Using the ideas from the Lebesgue-Stieltjes measure article, we have that $F(x)$ gives us a measure and sigma-algebra on $\mathbb{R}$ . Let $P_F$ and $\mathcal{F}_F$ be the measure and sigma-algebra, respectively, that we get from $F(x)$ . It follows from the fact that $\lim_{x \to \infty} F(x) = 1$ and $\lim_{x \to -\infty} F(x) = 0$ that $P_F(\mathbb{R}) = 1$ , and so $(\mathbb{R}, \mathcal{F}_F, P_F)$ is a probability space.

Define $X : \mathbb{R} \to \mathbb{R}$ as $x \mapsto x$ . Note that this is a random variable as

$\displaystyle X^{-1}((-\infty, x]) = (-\infty, x]$

To see that $X$ is a measurable function (random variable), let $A \subseteq \mathbb{R}$ and $\epsilon > 0$ . There exists a sequence $(I_n)_{n \in \mathbb{N}}$ of intervals that covers $A$ and

$\displaystyle P_F^*(A) \leq \sum_{n=1}^\infty \ell_F(I_N) \leq P_F^*(A) + \epsilon$

Define $J_n := I_n \cap (-\infty, t]$ and $K_n := I_n \cap (t, \infty)$ ; note that $\bigcup_n J_n$ and $\bigcup_n K_n$ cover $A \cap (-\infty, t)$ and $A \cap (t, \infty)$ , respectively. Also notice that $\ell_F(I_n) = \ell_F(J_n) + \ell_F(K_n)$ . This gives us

$\displaystyle P_F^*(A) \leq \sum_{n=1}^\infty \ell_F(J_n) + \sum_{n=1}^\infty \ell_F(K_n) \leq P_F^*(A) + \epsilon$

$\displaystyle \implies P_F^*(A) \leq P_F^*(A \cap (-\infty, t]) + P_F^*(A \cap (t, \infty)) \leq P_F^*(A) + \epsilon$

$\displaystyle \implies P_F^*(A) = P_F^*(A \cap (-\infty, t]) + P_F^*(A \cap (-\infty, t]^\complement)$

so $X$ is a measurable function (random variable). This means that any function $F$ which is increasing, right-continuous, and $\lim_{x \to -\infty} F(x) = 0$ and $\lim_{x \to \infty} F(X) = 1$ , is the distribution of some random variable.

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Mathematics Prelims

November 1, 2008

Distribution of a Random Variable

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