Joint, Marginal, and Conditional Distributions, Summaries of Pre-Calculus

Remarks: 1. 2. One way to remember these is by saying the words: the conditional distribution is the joint distribution divided by the marginal distribution.

Typology: Summaries

2022/2023

Uploaded on 03/01/2023

esha
esha 🇺🇸

3

(1)

224 documents

1 / 4

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
Joint, Marginal, and Conditional Distributions Page 1 of 4
Joint, Marginal, and Conditional Distributions
Problems involving the joint distribution of random variables X and Y use
the pdf of the joint distribution, denoted ).,(
,yxf YX This pdf is usually given,
although some problems only give it up to a constant. The methods for
solving problems involving joint distributions are similar to the methods for
single random variables, except that we work with double integrals and
2-dimensional probability spaces instead of single integrals and
1-dimensional probability spaces. We illustrate these methods by example.
Discrete Case: Analogous to the discrete single random variable case, we
have
1))()Pr((),(0 ,
=
=
=
yYxXyxf YX
The Continuous Case is illustrated with examples.
The Mixed Case (one of the random variables is discrete, the other is
continuous) is also illustrated with examples.
cdf of Joint Distribution – denoted ),(
,yxF YX
))()Pr((),(
,yYxXyxF YX
=
Notice that we can get the cdf’s for X and Y from the joint cdf as follows:
),()(
),()(
,
,
yFyF
xFxF
YXY
YXX
=
=
pf3
pf4

Partial preview of the text

Download Joint, Marginal, and Conditional Distributions and more Summaries Pre-Calculus in PDF only on Docsity!

Joint, Marginal, and Conditional Distributions

Problems involving the joint distribution of random variables X and Y use

the pdf of the joint distribution, denoted f X , Y ( x , y ). This pdf is usually given,

although some problems only give it up to a constant. The methods for

solving problems involving joint distributions are similar to the methods for

single random variables, except that we work with double integrals and

2-dimensional probability spaces instead of single integrals and

1-dimensional probability spaces. We illustrate these methods by example.

Discrete Case: Analogous to the discrete single random variable case, we

have

0 ≤ f (^) X , Y ( x , y )=Pr(( X = x )∧( Y = y ))≤ 1

The Continuous Case is illustrated with examples.

The Mixed Case (one of the random variables is discrete, the other is

continuous) is also illustrated with examples.

cdf of Joint Distribution – denoted F X , Y ( x , y )

F (^) X , Y ( x , y )=Pr(( Xx )∩( Yy ))

Notice that we can get the cdf’s for X and Y from the joint cdf as follows:

,

,

F y F y

F x F x

Y XY

X XY

= ∞

cdf of Joint Distribution (continued)

Discrete Case: We have ∑ ∑

= −∞=−∞

x

s

y

t

FX (^) , Y ( x , y ) fX , Y ( s , t )

Continuous Case: We have

,

2

,

, ,

F x y x y

f x y

F x y f stdtds

XY XY

x y XY XY

General Expectation using Joint Distribution

Discrete Case: We have = ∑∑ ⋅

x y

E [ h ( X , Y )] h ( x , y ) fX , Y ( x , y )

Continuous Case: We have ∫ ∫

−∞

−∞

E [ h ( X , Y )]= h ( x , y )⋅ fX , Y ( x , y ) dydx

The covariance of random variables X and Y is a generalization of variance

of a single random variable

Cov ( X , Y )= Cov ( Y , X )= E [( X −μ X )⋅( Y −μ Y )]= E [ X ⋅ Y ]− E [ X ]⋅ E [ Y ]

Remarks:

1. Var ( X )= Cov ( X , X )

2. Cov ( aX + bY , cZ + d )= a ⋅ c ⋅ Cov ( X , Z )+ b ⋅ c ⋅ Cov ( Y , Z )where a, b,

c, and d are constants and X, Y, and Z are random variables

3. Var ( X + Y )= Cov ( X + Y , X + Y )= Var ( X )+ Var ( Y )+ 2 ⋅ Cov ( X , Y )

Conditional Distributions for X , given Y and for Y , given X

, |

, |

f x

f x y f y x

f y

f x y f x y

X

XY YX x

Y

XY XY y

=

=

One way to remember these is by saying the words: the conditional

distribution is the joint distribution divided by the marginal distribution.

Also notice the probability interpretation when X and Y are discrete.

Independence of the jointly distributed random variables X and Y

If X and Y are independent, then each of the following will be true:

1. f X | Y = y ( x | y )= fX ( x ) and f Y | X = x ( y | x )= fY ( y )

and therefore we have f X , Y ( x , y )= fX ( x )⋅ fY ( y )

(Think of the probability interpretation in the discrete case.)

2. E [ X ⋅ Y ]= E [ X ]⋅ E [ Y ]and more generally,

E [ h ( X )⋅ g ( Y )]= E [ h ( X )]⋅ E [ g ( Y )]

3. Cov ( X , Y )= 0

4. Var ( X + Y )= Var ( X )+ Var ( Y )

Double Expectation Theorem (Very Important and Useful)

( ) [ ( | )] ( [ | ])

[ ] [ [ | ]]

Var X EVarX Y VarE X Y

E X EE X Y