10. Joint Moments and Joint Characteristic Functions

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1 10. Joint Moments and Joint Characteristic Functions Following section 6, in this section we shall introduce various parameters to compactly represent the inormation contained in the joint p.d. o two r.vs. Given two r.vs and and a unction g( x, y, deine the r.v g (, (10-1 Using (6-, we can deine the mean o to be E ( z ( z dz. (10-1

2 However, the situation here is similar to that in (6-13, and it is possible to express the mean o g(, in terms o ( x, y without computing (z. To see this, recall rom (5-6 and (7-10 that P z z z ( z z Pz g(, z z ( x, y D where D z is the region in xy plane satisying the above inequality. From (10-3, we get As z covers the entire z axis, the corresponding regions are nonoverlapping, and they cover the entire xy plane. z ( x, y xy (10-3 z ( z z g( x, y ( x, y xy. (10-4 ( x, y D z D z

3 By integrating (10-4, we obtain the useul ormula or (10-5 E( z ( z dz g( x, y ( x, y dxdy. E[ g (, ] gxy (, ( xydxdy,. I and are discrete-type r.vs, then Since expectation is a linear operator, we also get (10-6 Eg [ (, ] g( xi, y j P ( x i, y j. (10-7 i j E ag k k(, aeg k [ k(, ]. k k (10-8 3

4 I and are independent r.vs, it is easy to see that and W h( are always independent o each other. In that case using (10-7, we get the interesting result E[ g( h( ] g( x h( y g( x ( x dx ( x h( y ( y dxdy ( y dy However (10-9 is in general not true (i and are not independent. g( In the case o one random variable (see (10-6, we deined the parameters mean and variance to represent its average behavior. How does one parametrically represent similar cross-behavior between two random variables? Towards this, we can generalize the variance deinition given in (6-16 as shown below: 4 E[ g( ] E[ h( ]. (10-9

5 Covariance: Given any two r.vs and, deine Cov (, E( (. (10-10 By expanding and simpliying the right side o (10-10, we also get Cov (, E( It is easy to see that E( E( E(. (10-11 Cov (, Var ( Var (. (10-1 To see (10-1, let so that U a, Var ( U E a a ( Var ( ( a Cov (, Var ( 0. (

6 The right side o (10-13 represents a quadratic in the variable a that has no distinct real roots (Fig Thus the roots are imaginary (or double and hence the discriminant must be non-positive, and that gives (10-1. Using (10-1, we may deine the normalized parameter or Cov (, Var ( Var ( Cov ( Var (, Var ( Cov (, Cov (, (10-15, 1 Var(U 1, (10-14 and it represents the correlation coeicient between and. Fig a

7 Uncorrelated r.vs: I 0, then and are said to be uncorrelated r.vs. From (11, i and are uncorrelated, then E ( E ( E ( Orthogonality: and are said to be orthogonal i E ( From ( (10-17, i either or has zero mean, then orthogonality implies uncorrelatedness also and vice-versa. Suppose and are independent r.vs. Then rom (10-9 with g(, h(, we get and together with (10-16, we conclude that the random variables are uncorrelated, thus justiying the original deinition in ( Thus independence implies uncorrelatedness. 0. E ( E ( E (., (10-16 (

8 Naturally, i two random variables are statistically independent, then there cannot be any correlation between them ( 0. However, the converse is in general not true. As the next example shows, random variables can be uncorrelated without being independent. Example 10.1: Let U(0,1, U(0,1. Suppose and are independent. Deine = +, W = -. Show that and W are dependent, but uncorrelated r.vs. Solution: z x y, w x y gives the only solution set to be Moreover and J ( z, x z w, y z w. 0 z, 1 w 1, z w, z w, z w w 1/. 8

9 Thus (see the shaded region in Fig. 10. W ( z, w 1 /, 0, 0 z w, 1 w 1, z w otherwise,, z w, w z, ( z and hence 1 Fig. 10. ( z W ( z, w dw or by direct computation ( = + z z -z z- 1 1 dw z, dw z, 0 z 1, 1 z, 9

10 and W ( w z, 0 z 1, ( z ( z ( z z, 1 z, 0, otherwise, W ( z, w dz 1 dz 1 w, 0, 1 w w w 1, otherwise. Clearly W ( z, w ( z W ( w. Thus and W are not independent. However (10-19 (10-0 and and hence E( W E ( ( E( E( 0, E( W E( 0, Cov (, W E( W E( E( W 0 implying that and W are uncorrelated random variables. (10-1 (10-10

11 Example 10.: Let a b. Determine the variance o in terms o, and. Solution: and using (10-15 E( E( a b a b Var( E ( E a( b( a E( abe ( ( b E( a ab b. (10-3 In particular i and are uncorrelated, then 0, and (10-3 reduces to Thus the variance o the sum o uncorrelated r.vs is the sum 11 o their variances ( a b 1. a b. (10-4

12 Moments: represents the joint moment o order (k,m or and. Following the one random variable case, we can deine the joint characteristic unction between two random variables which will turn out to be useul or moment calculations. Joint characteristic unctions: The joint characteristic unction between and is deined as Note that k m E [ ] x k y m ( x, y dx dy, (10-5 j ( 1 j ( 1 (, E e e ( x, y dxdy. ( (, (0,0 1. 1

13 I and are independent r.vs, then rom (10-6, we obtain Also j1 j (, Ee ( Ee ( ( (. 1 1 (10-8 ( (,0, ( (0,. 1 1 (

14 Joint characteristic unctions are useul in determining the p.d. o linear combinations o r.vs. For example, with and as independent Poisson r.vs with parameters 1and respectively, let. (10-56 Then But rom (6-33 ( ( (. (10-57 j 1 e ( e, ( e (10-58 ( 1 ( e 1 so that 1 ( ( ( j e 1 e P( 1 (10-59 i.e., sum o independent Poisson r.vs is also a Poisson random variable. 14 j

15 More on Gaussian r.vs : From Lecture 7, and are said to be jointly Gaussian as N ( i their joint p.d. has the orm in (7-,,,,, 3. In that case, by direct substitution and simpliication, we obtain the joint characteristic unction o two jointly Gaussian r.vs to be j( 1 (, Ee ( e 1 1 j ( 1 ( 1 1. ( j ( (,0 e 15

16 Example 10.3: Let and be jointly Gaussian r.vs with parameters N (,,,,. Deine a b. Determine (z. Solution: In this case we can make use o characteristic unction to solve this problem. j j( a b ja jb ( Ee ( Ee ( Ee ( ( a, b. (

17 From (10-30 with w1 and w replaced by aw and bw respectively we get where 1 1 j( a b ( a ab b j ( e e, a a b, ab b. (10-33 (10-34 (10-35 Notice that (10-33 has the same orm as (10-31, and hence we conclude that a b is also Gaussian with mean and variance as in ( (10-35, which also agrees with (10-3. From the previous example, we conclude that any linear combination o jointly Gaussian r.vs generate a Gaussian r.v. 17

18 In other words, linearity preserves Gaussianity. We can use the characteristic unction relation to conclude an even more general result. Example 10.4: Suppose and are jointly Gaussian r.vs as in the previous example. Deine two linear combinations a b, W c d. (10-36 what can we say about their joint distribution? Solution: The characteristic unction o and W is given by W j( 1 W j1( a b j( c d (, Ee ( Ee ( 1 j ( a1c j ( b1d E( e ( a c, b d. ( As beore substituting (10-30 into (10-37 with w1 and w replaced by aw1 + cw and bw1 + dw respectively, we 18 get

19 where and 1 j ( 1 W ( 1 W 1 W W 1 e (,, W W W a c a c ac b d ( ad ab cd From (10-38, we conclude that and W are also jointly distributed Gaussian r.vs with means, variances and correlation coeicient as in ( (10-43.,, bc W b d,, bd. (10-38 (10-39 (10-40 (10-41 (10-4 (

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