Revision as of 12:48, 12 November 2013 by Mhossain (Talk | contribs)

Back to all ECE 600 notes


Random Variables and Signals

Topic 11: Two Random Variables: Joint Distribution




Two Random Variables

We have been considering a single random variable X and introduces the pdf f$ _X $, and pmf p$ _X $, conditional pdf f$ _X $(x|M), the conditional pmf p$ _X $(x|M), pdf f$ _Y $ or pmf p$ _Y $ when Y = g(X), expectation E[g(X)], conditional expectation E[g(X)|M], and characteristic function $ \Phi_X $. We will now define similar tools for the case of two random variables X and Y.
How do we define two random variables X,Y on a probability space (S,F,P)?


Fig 1: Mapping from S to X($ \omega $) and Y($ \omega $).


So two random variables can be viewed aw a mapping from S to R$ ^2 $, and (X,Y) is an ordered pair in R$ ^2 $. Note that we could draw the picture this way:

Fig 2: Mapping from S to X($ \omega $) and Y($ \omega $). Note that this model does not capture the joint behavior of X and Y and is hence incomplete.


but this would not capture the joint behavior of X and Y. Note also that if X and Y are defined on two different probability spaces, those two spaces can be combined to create (S,F,P).

In order for X and Y to be a valid random variable pair, we will need to consider regions D ⊂ R$ ^2 $.

$ B(\mathbb{R}^2) = \sigma (\{\mbox{all open rectangles in }\mathbb{R}^2\}) $

We need {(X,Y) ∈ O} ∈ F for any open rectangle O ⊂ R$ ^2 $, then {(X,Y) ∈ D} ∈ F ∀D ∈ B(R$ ^2 $).
But (X($ \omega $),Y($ \omega $)) ∈ O if X($ \omega $) ∈ A and Y($ \omega $) ∈ B for some A, B ∈ B(R), so {(X,Y) ∈ 0} = X$ ^{-1} $(A) ∩ Y$ ^{-1} $(B)
If X and Y are valid random variables then

$ \begin{align} &X^{-1}(A) \in \mathcal F \\ &Y^{-1}(B) \in \mathcal F \\ &\forall A,B\in B(\mathbb R) \end{align} $

So,

$ \begin{align} &X^{-1}(A)\cap Y^{-1}(B) \in \mathcal F \\ \Rightarrow &\{(X,Y)\in O\}\in\mathcal F \end{align} $

So how do we find P((X,Y) ∈ D) for D ∈ B(R$ ^2 $)?

We will use joint cdfs, pdfs, and pmfs.



Joint Cumulative Distribution Function

Knowledge of F$ _X $(x) and F$ _Y $(y) alone will not be sufficient to compute P((X,Y) ∈ D) ∀D ∈ B(R$ ^2 $), in general.

Definition $ \qquad $ The joint cumulative distribution function of random variables X,Y defined on (S,F,P) is F$ _{XY} $(x,y) ≡ P({X ≤ x} ∩ {Y ≤ y}) for x,y ∈ R.
Note that in this case, D ≡ D$ _{XY} $ = {(x',y') ∈ R$ ^2 $: x' ≤ x, y' ≤ y}

Fig 3: The shaded region represents D


Properties of F$ _{XY} $:


$ \bullet\lim_{x\rightarrow -\infty}F_{XY}(x,y) = \lim_{y\rightarrow -\infty}F_{XY}(x,y) = 0 $
$ \begin{align} \bullet &\lim_{x\rightarrow \infty}F_{XY}(x,y) = F_Y(y)\qquad \forall y\in\mathbb R \\ &\lim_{y\rightarrow \infty}F_{XY}(x,y) = F_X(x)\qquad \forall x\in\mathbb R \end{align} $
F$ _X $ and F$ _Y $ are called the marginal cdfs of X and Y.
$ \bullet P(\{x_1 < X\leq x_2\}\cap\{y_1<Y\leq y_2\}) = F_{XY}(x_2,y_2)-F_{XY}(x_1,y_2)-F_{XY}(x_2,y_1)+F_{XY}(x_1,y_1) $



The Joint Probability Density Function

Definition $ \qquad $ The joint probability density function of random variables X and Y is

$ f_{XY}(x,y) \equiv \frac{\partial^2}{\partial x\partial y}F_{XY}(x,y) $

∀(x,y) ∈ R$ ^2 $ where the derivative exists.

It can be shown that if D ∈ B(R$ ^2 $), then,

$ P((X,Y)\in D)=\int\int_Df_{XY}(x,y)dxdy $

where D ≡ D$ _{XY} $ = {(x',y') ∈ R$ ^2 $: x' ≤ x, y' ≤ y}


Properties of f$ _{XY} $:

$ \bullet f_{XY}(x,y)\geq 0\qquad\forall x,y\in\mathbb R $
$ \bullet \int\int_{\mathbb R}f_{XY}(x,y)dxdy = 1 $
$ \bullet F_{XY}(x,y) = \int_{-\infty}^{y}\int_{-\infty}^xf_{XY}(x',y')dx'dy'\qquad\forall(x,y)\in\mathbb R^2 $
$ \begin{align} \bullet &f_X(x) = \int_{-\infty}^{\infty}f_{XY}(x,y)dy \\ &f_Y(y) = \int_{-\infty}^{\infty}f_{XY}(x,y)dx \end{align} $ are the marginal pdfs of X and Y.



The Joint Probability Mass Function

If X and Y are discrete random variables, we will use the joint pdf given by

$ p_{XY}(x,y) = P(X=x,Y=y)\qquad \forall(x,y)\in\mathcal R_X \times\mathcal R_Y $

Note that if X is continuous and Y discrete (or vice versa), we will be interested in

$ P(\{X\in A\}\cap\{Y=y\}),\;\;A\in B(\mathbb R),\;y\in\mathcal R_y $

We often use a form of Bayes' Theorem, which we will discuss later, to get this probability.



Joint Gaussian Random Variables

An important case of two random variables is: X and Y are jointly Gaussian if their joint pdf is given by

$ f_{XY}(x,y)=\frac{1}{2\pi\sigma_X\sigma_Y\sqrt{1-r^2}}exp\left\{-\frac{1}{2(1-r^2)}\left[\frac{(x-\mu_X)^2}{\sigma_X^2}-\frac{2r(x-\mu_X)(y-\mu_Y)}{\sigma_X\sigma_Y}+\frac{(y-\mu_y)^2}{\sigma_Y^2}\right]\right\} $

where μ$ _X $, μ$ _Y $, σ$ _X $, σ$ _Y $, r ∈ R; σ$ _X $$ _Y $ > 0; -1 <r <1.

It can be shown that is X and Y are jointly Gaussian then X is N(μ$ _X $, σ$ _X $$ ^2 $) and Y is N(μ$ _Y $, σ$ _Y $$ ^2 $) (proof)


Special Case

We often model X and Y as jointly Gaussian with μ$ _X $ = μ$ _Y $ = 0, σ$ _X $ = σ$ _Y $ = σ, r = 0, so that

$ f_{XY}(x,y) = \frac{1}{2\pi\sigma^2}e^{-\frac{x^2+y^2}{2\sigma^2}} $


Example $ /qquad $ Let X and Y be jointly Gaussian with μ$ _X $ = μ$ _Y $ = 0, σ$ _X $ = σ$ _Y $ = σ, r= = 0. Find the probability that (X,Y) lies within a distance d from the origin.

Let

$ D_d = \{(x,y)\in\mathbb R^2:\;x^2+y^2\leq d^2\} $


Fig 4:The shaded region shows D$ _d $ = {(x,y)∈R$ ^2 $: x$ ^2 $+y$ ^2 $ ≤ d}


Then

$ P((X,Y)\in D_d) = \int\int_{D_d}\frac{1}{2\pi\sigma^2}e^{-\frac{x^2+y^2}{2\sigma^2}}dxdy $

Use polar coordinates to make integration easier: let

$ \begin{align} r&=x^2+y^2 \\ \theta &= \tan^{-1}(\frac{x}{y}) \end{align} $

Then

$ \begin{align} P((X,Y)\in D_d) &= \int_{\pi}^{\pi}\int_{0}^{d}f_{XY}(r\cos\theta,r\sin\theta)rdrd\theta \\ &= \int_{\pi}^{\pi}\int_{0}^{d} \frac{r}{2\pi\sigma^2}e^{-\frac{r^2}{2\sigma^2}}drd\theta \\ &= 1-e^{-\frac{d^2}{2\sigma^2}} \end{align} $


So the probability that (X,Y) lies within distance d from the origin looks like the graph in figure 5 (as a function of d).

Fig 5: P({X$ ^2 $,Y$ ^2 $ ≤ d}) plotted as a function of d



References



Questions and comments

If you have any questions, comments, etc. please post them on this page



Back to all ECE 600 notes

Alumni Liaison

Followed her dream after having raised her family.

Ruth Enoch, PhD Mathematics