handout_week6_a - Example P X 2 Y 2< 20 = Pairs of...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Stochastic Signals and Systems Multiple Random Variables Virginia Tech Fall 2008 Conditional Expected Values Given that the expected value of a function of a rv is given by E [ Y ] = Z -∞ yf Y ( y ) dy , we obtain the conditional expectation of Y given X = x as E [ Y | x ] = Z -∞ yf Y ( y | x ) dy . More generally, the conditional mean of g ( Y ) is defined as E [ g ( Y ) |A ] = Z -∞ g ( y ) f Y ( y |A ) dy .
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Example The joint pdf of ( X , Y ) is given by f X , Y ( x , y ) = 6 ( 1 - x - y ) for values of x and y for which ( x , y ) lies within the triangle as shown in the figure below. Find the conditional expected values of X and X 2 given Y = y . Pairs of Discrete Random Variables The joint probability mass function of ( X , Y ) specify the probabilities of the event { X = x k } ∩ { Y = y k } : p X , Y ( x j , y k ) = P ±² X = x j ³ ∩ { Y = y k } ´ = P ± X = x j , Y = y k ´ j = 1 , 2 , . . . k = 1 , 2 , . . . Thus the joint pmf gives the probability of the occurrence of the pairs ( x j , y k ) . The probability of any event A is the sum of the pmf over the outcomes in A : P [( X , Y ) A ] = XX ( x j , y k ) A p X , Y ( x j , y k ) .
Background image of page 2
Pairs of Discrete Random Variables
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 4
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Example: P [ X 2 + Y 2 < 20 ] = Pairs of Discrete Random Variables • The fact that the probability of the sample space S is 1 gives ∞ X j = 1 ∞ X k = 1 p X , Y ( x j , y k ) = 1 . • The marginal probability mass function of X is obtained by: p X ( x j ) = P [ X = x j ] = P [ X = x j , Y = anything ] = ∞ X k = 1 p X , Y ( x j , y k ) Similarly, p Y ( y k ) = ∞ X j = 1 p X , Y ( x j , y k ) • Conditional expected values: E [ Y | x ] = X y j y j p Y ( y j | x ) . Pairs of Discrete Random Variables For all nonnegative integers k , m , let P [ X = k , Y = m ] = ( 1-v 1 ) ( 1-v 2 ) v k 1 v m 2 ; X and Y might be the numbers of photoelectrons counted at two photodetectors. Find the marginal probability mass function of X ....
View Full Document

{[ snackBarMessage ]}

Page1 / 4

handout_week6_a - Example P X 2 Y 2< 20 = Pairs of...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online