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# NPT19_jointRV2doc - Conditional Joint Probability Density Function

By Dean Elliott,2014-11-26 12:17
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NPT19_jointRV2doc - Conditional Joint Probability Density Function

Conditional Distributions

We discussed the conditional CDF and conditional PDF of a random variable conditioned

on some events defined in terms of the same random variable. We observed that

PXxB?，，?：；?? /0FxBPB，?；？；？XPB；？

and

d //fxBFxB；？；？XXdx

We can define these quantities for two random variables. We start with the conditional

probability mass functions of two random variables.

Conditional probability mass functions

Suppose and are two discrete jointly random variable with the joint PMF YX

pxy(,). The conditional PMF of given is defined as YXxXY,

pyxPYyXx(/)({}/{})，，，YX/

PXxYy({}{})，，， PXx{}

pxy(,)XY, provided ()0，?pxXpx()X

Similarly we can define the conditional probability mass function

pxy(/)XY/

Remarks

) The conditional PMF satisfies the properties of the probability mass

functions.

) From the definition of conditional probability mass functions, we can

define two independent random variables. Two discrete random variables

X and Y are said to be independent if and only if

;??(x,y)RR, XY

pyxpy(/)() YXY/

so that

pxypxpy(,)()()XYXY,

Bayes’ rule for discrete random variables

px() Suppose and are two discrete jointly random variables. Given and YXX

pyx(/)pxy(/) we can determine the a posteriori probability mass function by YX/XY/

using the Bayes’ rule as follows:

pxyPXxYy(/)({}/{})，，，XY/

PXxYy({}{})，，， PYy{}

pxy(,)XY, py()Y

pxy(,)XY, ( Using the total probability law)~pxy(,)XY,~?xRX

Example Consider the random variables and with the joint probability mass YX

function as presented in the following table

0 1 2 py()XY Y 0 0.25 0.1 0.15 0.5

1 0.14 0.35 0.01 0.5

0.39 0.45 px()X

The marginal probabilities are as shown in the last column and the last row

p(0,1)XY,p(0/1)YX/p(1)X 0.14 =0.39

Conditional Probability Distribution function

Consider two continuous jointly random variables and with the joint YX

Fxy(,).probability distribution function We are interested to find the conditional XY,

distribution of one of the random variables on the condition of a particular value of the

other random variable.

We cannot define the conditional distribution of the random variable on the Y

{}Xxcondition of the event by the relation

FyxPYyXx(/)(/)，?，YX/

PYyXx(,)?， =PXx()

as PXx()0，， in the above expression. The conditional distribution function is defined

in the limiting sense as follows:

FyxlimPYyxXxx(/)(/)，?！?？((?x0YX/

PYyxXxx(,)?！?？( =lim(?x0PxXxx()！?？(

y

fxuxdu(,)(?XY, ? =l im(?x0fxx()(X

y

fxudu(,)?XY,? =fx()X

Conditional Probability Density Function

f(y/Xx)f(y/x) is called conditional density of given YX.Y/XY/X

Let us define the conditional distribution function .

The conditional density is defined in the limiting sense as follows

f(y/Xx)lim(F(y(y/Xx)F(y/Xx))/(yY/X(y?0Y/XY/X

lim(F(y(y/xX?x(x)F(y/xX?x(x))/(y (1)(y?0,(x?0Y/XY/X

(Xx)lim(xX?x(x)Because (x?0

The right hand side in equation (1) is lim((/)(/))/FyyxXxxFyxXxxy？(！！？(，！！？(((?(?yxYXYX0,0//

lim((/))/，！?？(！?？((PyYyyxXxxy(?(?yx0,0

lim((,))/()，！?？(！?？(！?？((PyYyyxXxxPxXxxy(?(?yx0,0

lim(,)/()，((((fxyxyfxxy (?(?yxXYX0,0,

(,)/()fxyfxXYX,

；，fyxfxyfx(/) (,)/()YXXYX/,

(2)

Similarly we have

；，fxyfxyfy(/) (,)/()XYXYY/,

(3)

2(,),xy?) Two random variables are statistically independent if for all

fyxfy(/) () YXY/

or equivalently

) (4) (,)()()fxyfxfyXYXY,

Bayes rule for continuous random variables:

From (2) and (3) we get the Bayes rule for two continuous joint random varibles

(,)fxyXY,fxy(/) XY/fy()Y

fxfyx()(/)XYX/；，fxy(/)XY/fy()Y

fxy(,)XY, =?fxydx(,)?XY,，?

(4) fyxfx(/)()YXX/ =?fufyxdu()(/)?XYX/，?

Given the joint density function we can find out the conditional density function.

Example: For random variables X and Y, the joint probability density function is given

by

1xyfxyxy(,) 1, 1，??XY, 4

0 otherwise

fxfyfyx(), () and (/). Find the marginal density Areindependent? XY and XYYX/

11xyfxdy() X?41

1 2

Similarly

1fyy() -11 ，??Y2

and

fxy(,)XY,fyx(/)YX/ fx()X

1xy ?fy；？Y2

Hence, and are not independent. YX

Bayes’ Rule for mixed random variables

Let be a discrete random variable with probability mass function and be a px()XYXcontinuous random variable defined on the same sample space with the conditional

probability density function In practical problems we may have to estimate fyx(/).YX/

the conditional PMF of given the observed value We can define this Yy.X

conditional PMF also in the limiting sense

pxyPXxyYyy(/)lim(/)，，！?？(XYy/0(?

PXxyYyy(,)，！?？(

=lim(?y0PyYyy()！?？(

pxfyxy()(/)(XYX/ =lim(?y0fyy()(Y

pxfyx()(/)XYX/pxfyx()(/) = XYX/xfy()Y

Example pxfyx()(/)XYX/ =

XY +

V y [n]

is a binary random variable with X 1 with probability p?X ? 1 with probability 1-p?

is the Gaussian noise with mean V2 0 and variance . x [n]

Then

pxfyx()(/)XYX/pxy(1/) ，，XY/pxfyx()(/)XYX/x 22，，(1)/2ype2222，，，？(1)/2(1)/2yy，，？，pepe(1)

Independent Random Variables

Let and be two random variables characterized by the joint distribution function XY

F(x,y)P{X?x,Y?y} X,Y

2(and the corresponding joint density function f(x,y)F(x,y)X,YX,Y(x(y

2;??(,), {}xyXxThen and are independent if and are {}Yy?XY

independent events. Thus,

FxyPXxYy(,){,}，??XY,

={}{}PXxPYy??

()()FxFy XY

2(Fxy(,)XY,；，fxy(,)XY,((xy

()()fxfyXY

and equivalently

f(y)f(y) Y/XY

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