EE126: Problem Set # 7: Solutions

Transcription

1 EE26: Problem Set # 7: Solutions Instructor: Jean Walrand TA: S. Mubaraq Mishra Assigned: October 7, 23 Due: October 24, 23 Problem. Let X, Y, Z, V be independent N(, ) random variables. Find the p.d.f. of X 2 + Y 2 + Z 2 + V 2. This problem is definitely tricky and hard to do without checking reference texts unless you know the first step. First, we claim that W : X 2 +Y 2 d Ex(/2). To see this we calculate the characteristic function of W. We find E(e iuw ) 2π e iu(x2 +y 2 ) 2π e (x2 +y2 )/2 dxdy e iur2 2π e r2 /2 rdrdθ e iur2 e r2 /2 rdr 2iu d[eiur2 r2 /2 ] 2iu. On the other hand, if W d Ex(λ), then E(e iuw ) e iux λe λx dx λ λ iu λ iu.. Comparing these expressions shows that X 2 + Y 2 d Ex(/2) as claimed. Second, we note that U : X 2 + Y 2 + Z 2 + V 2 is the sum of two independent Ex(/2) random variables. This sum is not exponentially distributed. We can find its density as the convolution of the densities of the two Ex(/2) random variables. That is, with λ /2, f U (u) u λe λx λe λ(u x) dx λ 2 ue λu.

2 Problem 2. Let {X n, n } be Gaussian N(, ) random variables. Assume that Y n+ ay n + X n for n where Y is a Gaussian random variable with mean zero and variance σ 2 independent of the X n s and a <. a. Calculate var(y n ) for n. Show that var(y n ) γ 2 as n for some value γ 2. b. Find the values of σ 2 so that the variance of Y n does not dependent on n. a. We see that var(y n+ ) var(ay n + X n ) a 2 var(y n ) + var(x n ) a 2 var(y n ) +. Thus, we α n : var(y n ), one has Solving these equations we find Since a <, it follows that b. The obvious answer is σ 2 γ 2. α n+ a 2 α n + and α σ 2. var(y n ) α n a 2n σ 2 + a2n, for n. a2 var(y n ) γ 2 : as n. a2 2

3 Problem 3. Let the X n s be as in Problem 2. a.calculate E[X + X 2 + X 3 X + X 2, X 2 + X 3, X 3 + X 4 ]. b. Calculate E[X + X 2 + X 3 X + X 2 + X 3 + X 4 + X 5 ]. a. We know that the solution is of the form Y a(x + X 2 ) + b(x 2 + X 3 ) + c(x 3 + X 4 ) where the coefficients a, b, c must be such that the estimation error is orthogonal to the conditioning variables. That is, E((X + X 2 + X 3 ) Y )(X + X 2 )) E((X + X 2 + X 3 ) Y )(X 2 + X 3 )) E((X + X 2 + X 3 ) Y )(X 3 + X 4 )). These equalities read 2 a (a + b) 2 (a + b) (b + c) (b + c) c, and solving these equalities gives a 3/4, b /2, and c /4. b. Here we use symmetry. For k,..., 5, let Y k E[X k X + X 2 + X 3 + X 4 + X 5 ]. Note that Y Y 2 Y 5, by symmetry. Moreover, Y +Y 2 +Y 3 +Y 4 +Y 5 E[X +X 2 +X 3 +X 4 +X 5 X +X 2 +X 3 +X 4 +X 5 ] X +X 2 +X 3 +X 4 +X 5. It follows that Y k (X + X 2 + X 3 + X 4 + X 5 )/5 for k,..., 5. Hence, E[X + X 2 + X 3 X + X 2 + X 3 + X 4 + X 5 ] Y + Y 2 + Y (X + X 2 + X 3 + X 4 + X 5 ). 3

4 Problem 4. Let the X n s be as in Problem 2. Find the j.p.d.f. of (X +2X 2 +3X 3, 2X + 3X 2 + X 3, 3X + X 2 + 2X 3 ). These random variables are jointly Gaussian, zero mean, and with covariance matrix Σ given by 4 Σ 4. 4 Indeed, Σ is the matrix of covariances. For instance, its entry (2, 3) is given by E((2X + 3X 2 + X 3 )(3X + X 2 + 2X 3 )) We conclude that the j.p.d.f. is f X (x) We could calculate Σ and Σ... (2π) 3/2 Σ /2 exp{ 2 xt Σ x}. Problem 5. Let X, X 2, X 3 be independent N(, ) random variables. Calculate E[X + 3X 2 Y] where Y [ ] X X 2 X 3 By now, this is familiar stuff... The solution is Y : a(x +2X 2 +3X 3 )+b(3x +2X 2 +X 3 ) where a and b are such that E((X +3X 2 Y )(X +2X 2 +3X 3 )) 7 (a+3b) (4a+4b) (9a+3b) 7 4a b and E((X +3X 2 Y )(3X +2X 2 +X 3 )) 9 (3a+9b) (4a+4b) (3a+b) 9 a 4b. Solving these equations gives a /2 and b 7/2. 4

5 Problem 6. Find the j.p.d.f. of (2X +X 2, X +3X 2 ) where X and X 2 are independent N(, ) random variables. Hence, These random variables are jointly Gaussian, zero-mean, with covariance Σ given by Σ [ ]. where f X (x) 2π Σ exp{ /2 2 xt Σ x} π exp{ 2 xt Σ x} Σ 25 [ ]. Problem 7. The random variable X is N(µ, ). Find an approximate value of µ so that P (.5 X.) P ( X 2). We write X µ + Y where Y is N(, ). We must find µ so that g(µ) : P (.5 µ Y. µ) P ( µ Y 2 µ). We do a little search using a table of the N(, ) distribution or using a calculator. I find that µ.65. 5

6 Problem 8. Let X be a N(, ) random variable. Calculate the mean and the variance of cos(x) and sin(x). We know that E(e iux ) e u2 /2 and e iθ cos(θ) + i sin(θ). Therefore, E(cos(uX) + i sin(ux)) e u2 /2, so that E(cos(uX)) e u2 /2 and E(sin(uX)). In particular, E(cos(X)) e /2 and E(sin(X)). Problem 9. Let X be a N(, ) random variable. Define Find the p.d.f. of Y. Y { X, if X X, if X >. By symmetry, X is N(, ). 6