lecture8

# lecture8 - Lecture-8 One Function of Two Random Variables...

This preview shows pages 1–8. Sign up to view the full content.

1 Lecture-8 One Function of Two Random Variables Given two random variables X and Y and a function g ( x , y ), we form a new random variable Z as Given the joint p.d.f how does one obtain the p.d.f of Z ? Problems of this type are of interest from a practical standpoint. For example, a receiver output signal usually consists of the desired signal buried in noise, and the above formulation in that case reduces to Z = X + Y . ). , ( Y X g Z = ), , ( y x f XY ), ( z f Z (1)

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

View Full Document
2 It is important to know the statistics of the incoming signal for proper receiver design. In this context, we shall analyze problems of the following type: Referring back to (1), to start with ) , ( Y X g Z = Y X + ) / ( tan 1 Y X - Y X - XY Y X / ) , max( Y X ) , min( Y X 2 2 Y X + (29 ( 29 [] ∫∫ = = = = z D y x XY z Z dxdy y x f D Y X P z Y X g P z Z P z F , , ) , ( ) , ( ) , ( ) ( ) ( ξ (2) (3)
3 where in the XY plane represents the region such that is satisfied. Note that need not be simply connected (Fig. 8.1). From (3), to determine it is enough to find the region for every z , and then evaluate the integral there. We shall illustrate this method through various examples. z D z y x g ) , ( ) ( z F Z z D z D X Y z D z D Fig. 8.1

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

View Full Document
4 Example 8.1: Z = X + Y. Find Solution: since the region of the xy plane where is the shaded area in Fig. 8.2 to the left of the line Integrating over the horizontal strip along the x -axis first (inner integral) followed by sliding that strip along the y -axis from to (outer integral) we cover the entire shaded area. (29 ∫∫ + -∞ = - -∞ = = + = , ) , ( ) ( y y z x XY Z dxdy y x f z Y X P z F (4) z D z y x + . z y x = + - + y z x - = x y Fig. 8.2 ). ( z f Z
5 We can find by differentiating directly. In this context, it is useful to recall the differentiation rule in (7.15) - (7.16) due to Leibnitz. Suppose Then Using (6) in (4) we get Alternatively, the integration in (4) can be carried out first along the y -axis followed by the x -axis as in Fig. 8.3. ) ( z F Z ) ( z f Z = ) ( ) ( . ) , ( ) ( z b z a dx z x h z H (5) (29 + - = ) ( ) ( . ) , ( ), ( ) ( ), ( ) ( ) ( z b z a dx z z x h z z a h dz z da z z b h dz z db dz z dH (6) ∫∫ + - + - + - - - - = + - - = = . ) , ( ) , ( 0 ) , ( 1 ) , ( ) ( dy y y z f dy z y x f y y z f dy dx y x f z z f XY XY XY y z XY Z (7)

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

View Full Document
6 In that case and differentiation of (8) gives ∫∫ + -∞ = - -∞ = = , ) , ( ) ( x x z y XY Z dxdy y x f z F (8) + -∞ = + -∞ = - -∞ = - = = = . ) , ( ) , ( ) ( ) ( x XY x x z y XY Z Z dx x z x f dx dy y x f z dz z dF z f (9) If X and Y are independent, then and inserting (10) into (8) and (9), we get ) ( ) ( ) , ( y f x f y x f Y X XY = . ) ( ) ( ) ( ) ( ) ( + -∞ = + -∞ = - = - = x Y X y Y X Z dx x z f x f dy y f y z f z f (10) (11) x z y - = x y Fig. 8.3
7 The above integral is the standard convolution of the functions and expressed two different ways. We thus reach the following conclusion: If two r.vs are independent, then the density of their sum equals the convolution of their density functions.

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 04/24/2011 for the course ECE 6303 taught by Professor Voltz during the Spring '11 term at NYU Poly.

### Page1 / 33

lecture8 - Lecture-8 One Function of Two Random Variables...

This preview shows document pages 1 - 8. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online