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Random Variables and Signals

Topic 6: Random Variables: Distributions



How do we find , compute and model P(x ∈ A) for a random variable X for all A ∈ B(R)? We use three different functions:

  1. the cumulative distribution function (cdf)
  2. the probability density function (pdf)
  3. the probability mass function (pmf)

We will discuss these in this order, although we could come at this discussion in a different way and a different order and arrive at the same place.

Definition $ \quad $ The cumulative distribution function (cdf) of X is defined as

$ \begin{align} F_X(x) &= P_X((-\infty,x])\;\forall x\in \mathbb R \\ &= P(X^{-1}((-\infty, x])) \\ &=P(\{ \omega\in\mathcal S:\;X(\omega)\leq x \}) \end{align} $

Notation $ \quad $ Normally, we write this as

$ F_X(x) = P(X\leq x) $

So $ F_X(x) $ tells us P$ P_X(A) $ if A = (-∞,x] for some real x.
What about other A ∈ B(R)? It can be shown that any A ∈ B(R) can be written as a countable sequence of set operations (unions, intersections, complements) on intervals of the form (-∞,x$ _n $], so can use the probability axioms to find $ P_X(A) $ from $ F_X $ for any A ∈ B(R). This is not how we do things in practice normally. This will be discussed more later.

Can an arbitrary function $ F_X, $ be a valid cdf? No, it cannot.
Properties of a valid cdf:
1.

$ \lim_{x\rightarrow\infty}F_X(x) = 1\;\;\mbox{and}\;\;\lim_{x\rightarrow -\infty}F_X(x) = 0 $

This is because

$ \lim_{x\rightarrow\infty}F_X(x) = P(\{ \omega\in\mathcal S:\;X(\omega)\leq\infty \})=1 $

and

$ \lim_{x\rightarrow -\infty}F_X(x)= P(\varnothing)= 0 $

2. For any $ x_1,x_2 $R such that $ x_1<x_2 $,

$ F_X(x_1)\leqF_X(x_2) $

i.e. $ F_X(x) $ is a non decreasing function.

3. $ F_X $ is continuous from the right , i.e.

$ F_X(x^+) \equiv \lim_{\epsilon\rightarrow0,\epsilon>0}F_X(x+\epsilon)=F_X(x)\;\;\forall x\in\mathbb R $

Proof: First, we need some results from analysis and measure theory:
(i) For a sequence of sets, $ A_1, A_2,... $, if $ A_1 $$ A_2 $ ⊃ ..., then

$ \lim_{n\rightarrow\infty}A_n = \bigcap_{n=1}^{\infty}A_N $

(ii) If $ A_1 $$ A_2 $ ⊃ ..., then

$ P(\lim_{n\rightarrow\infty}A_n) = \lim_{n\rightarrow\infty}P(A_n) $

(iii) We can write $ F_X(x^+) $ as

$ F_X(x^+) = \lim_{n\rightarrow\infty}F_X(x+\frac{1}{n}) $

Now let

$ A = \{X\leq x+\frac{1}{n})\} $

Then

$ \begin{align} F_X(x^+) &= \lim_{n\rightarrow\infty}P(X\leq\frac{1}{n}) \\ &=\lim_{n\rightarrow\infty}P(A_n) \\ &=P(\lim_{n\rightarrow\infty}A_n) \\ &=P(\bigcap_{n=1}^{\infty}A_n) \\ &=P(\bigcap_{n=1}^{\infty}\{X\leq x+\frac{1}{n})\})\\ &=F_X(x) \end{align} $




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