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ECE 650 1
ECE 650 Homework Set 3 – Solutions
1. Note that the aimpoint error (say R) is a Rayleigh RV.
a. From the table in the back of Papoulis, the average is:
in51.222/
b. P{R > 4} = 1 – F(4) = exp(-2) = 0.1353.
MATLAB check: >> 1-raylcdf(4,2)
ans =
0.1353
2a. Step 1: Let S = X + Y. Find fS(s) = fX(s) * fY(s)
Step 2: Let W = S + Z. Find fW(w) = fS(w) * fZ(w); fix fS(); flip &
shift fZ() by w on axis:
fW(w) = fS(w) * fZ(w) = d)w(f)(f zs
Case 1: w < 0 or w > 6 fw(w) = 0
Case 2: 0 < w < 2
fw(w) =
d
4
1
2
1d)w(f)(f
w
0zs
w
0
= 16
w2
1/2
0 2 4 s
f(s)
1/2
0 w 4
fs() = ¼
fw(w-)
ECE 650 2
Case 3: 2 < w < 4
fw(w) =
d)w(f)(f zs
w
2w
d14
1
2
1d
4
1
2
1 w
2
2
2w
= 4
3w
4
3
8
w2
Over-all Answer (obtaining the rest by symmetry):
0 w < 0, w > 6
w2/16 0 w 2
4
3w
4
3
8
w2
2 < w 4
(6-w)2/16 4 < w 6
0 2 4 60
0.1
0.2
0.3
0.4
0.5
1/2
w-2 2 w 4
fs() = ¼ fw(w-) fs() = -¼ + 1
f(w) =
w
Histogram (blue) vs. pdf (red) for W
ECE 650 3
MATLAB Code: % ece_650_hw4_prob2
% First generate the 3 vectors and the sum;
hold off
x = unifrnd(0, 2, 10000, 1);
y = unifrnd(0, 2, 10000, 1);
z = unifrnd(0, 2, 10000, 1);
w = x + y + z;
%
% Then create the density histogram;
centers = 0 : .1 : 6;
freq = hist(w, centers);
relfreq = freq/sum(freq);
dens = relfreq/.1; % bin widths are .1
bar(centers, dens, 1)
hold on
%
%Then generate the theoretical pdf for w
piece1 = (centers .^2)/16;
piece1 = piece1 .* ((centers >= 0)& (centers <=2));
piece2 = -(centers .^2)/8 + .75*centers - .75;
piece2 = piece2 .* ((centers >2)& (centers <= 4));
piece3 = ((6-centers).^2)/16;
piece3 = piece3 .* ((centers > 4)&(centers <= 6));
theo = piece1 + piece2 + piece3;
plot(centers,theo,'ro')
3a. Joint pdf contains 2 delta’s: 1 at (x = 1, y = 1), with area (1/2); and 1 at
(x = -1, y = 0), with area (1/2). (Note: I’m using (x,y) equiv. to (x1,x2)).
x
y
f(x,y)
(1,1)
(-1,0)
(1/2)
(1/2)
ECE 650 4
b. Marginal pdf’s
c.
H T
X 1 -1
Y 1 0
Z 2 -1
f(z) contains 2 deltas, each of area (1/2); one is at z = 2; the
other is at z = -1. Note that x and y are not independent; hence,
we cannot just convolve their pdf’s to get the pdf for z.
4. (Papoulis 6-1, a & f) Note that the solution for (a) is equivalent to the
convolution of fX(z) with fY(z).
x
f(x)
y
f(y)
-1 0 1
(1/2) (1/2) (1/2) (1/2)
0 1
z
f(z)
-1 0 2
(1/2) (1/2)
ECE 650 5
5. fXY(x, y) = (1/8) e-(x/2 + y/4)
U(x) U(y),
Note: f(x, y) factors: fXY(x, y) = (1/2) e-x/2
U(x) (1/4) e
-y/4 U(y) = fX(x) fY(y)
Also note that each of the marginal densities is exponential.
a. f(x) = (1/2) e-x/2
U(x) b. f(y) = (1/4) e-y/4
U(y)
c. Since the pdf factors into the marginal densities, X and Y are independent,
so: f(x|y) = f(x)
d. Similarly, f(y|x) = f(y)
e. As shown above, X and Y are independent.
6. Based on the equation of the line: x + 2y = 2, define a new rv
Z = X + 2Y, with E(Z) = 0 and Z2 = X
2 + 4
Y
2 = 1 + 4 = 5.
Then Pr(X + 2Y > 2) = Pr(Z > 2) = 1- normcdf(2, 0, sqrt(5)) = .1855
7a. var(10X) = 100(2) = 200 7b. var(10X + 3) = 200
7c. var(-X) = 2 7d. E(X2) = 2+100 = 102
8. (Carol Ash) var(X – Y) = E{(X - Y)2} – {E(X – Y)}
2 = E{( X
2 – 2 X Y +
Y 2)} –{E(X) – E(Y)}
2 = E{ X
2} – 2E(X Y) + E{ Y
2} – {[E(X)]
2 –
2E(X)E(Y) +[E(Y)]2} = var(X) + var(Y) – 2{E(X Y) – E(X)E(Y)} =
var(X) + var(Y) – 2 cov(X, Y)
ECE 650 6
9. (Yates & Goodman, Ch. 9)
P(x,y) y = -3 y = -1 y = 1 y = 3 Marginals for
X
x = -1 1/6 1/8 1/24 0 Pr{X=-1} =1/3
x = 0 1/12 1/12 1/12 1/12 Pr{X =0} =1/3
x = 1 0 1/24 1/8 1/6 Pr{X =1} =1/3
Marginals
For Y
Pr{Y=-
3} =1/4
Pr{Y =-
1} =1/4
Pr{Y=1}
=1/4
Pr{Y=3}
=1/4
Check: Total
Prob = 1
fX(x) fY(y)
b. No, X and Y are not independent because (for example) f(x|y = -3)
f(x).
10. (Yates & Goodman, Ch. 5)
432
1
02
x
0
1
04321
x
0
1
0
x
0
1
0
dxdxdxkxdxdxdxdxk424
43
x
0
1
043
x
0
1
0x
22
1
0
dxdx12
kdxdxx
2
k 44
2
4k14
kx
4
kdxx
2
k 1
0x
2444
1
0 4
f(x2, x3) = )x1(x4dx1x4dxdx14dxdx4 324
1
x241
x
0
1
x41
3
2
3
= 4x2(1-x3), 0 < x2 < 1, 0 < x3 < 1
0 else
x -1 0 1
(1/3) (1/3) (1/3)
y -3 -1 1 3
1
(1/4) (1/4) (1/4) (1/4)
ECE 650 7
f(x3) =
1x0),x1(2dx)x1(x4dx)x,x(f 33
1
0x232
1
0x232
22
0 else
Solution starts on next page.
11. (Papoulis 4th
ed., 6.58)
ECE 650 8
ECE 650 9
12. (Papoulis 4th
ed., 6-71)
13. C =
2/31
11
32
22
)Cdet(
1C,
22
23 1 since det(C) = =
2. Using x for x1 and y for x2, we get:
f(x, y)
y
x
2/31
11]yx[
2
1exp
22
1
f(x, y) = (1/(2 sqrt(2)) exp(-1/2 x2 + xy - (3/4) y
2)
14. Note: both X1 and X2 have mean 2, var 16/12 = 4/3.
X1, X2 independent X1, X2 uncorrelated E{ X1 X2 } = E[X1] E[X2] = 2
2 = 4;
var(X1) = E(X12) – X1
2 4/3 = E(X12) – 4 E(X1
2) = E(X2
2) = 16/3;
R =
3/164
43/16
15. T: N(0, 9), observation: R = T + X, where X: U(-3,3), independent of T
a. var(R) = var(T) + var(X) = 9 + 62/12 = 12
dx
y)2/3(x
yx]yx[
2
1
(arg. of exp.)
= -1/2 [ x(x-y) + y(-x + (3/2)y) ]
= -1/2 x2 + xy - (3/4) y
2
ECE 650 10
b. cov(R, T) = E{( R - R)( T – T) T = 0, R = E(T) + E(X) = 0
cov(R, T) = E{ R T } = E{( T + X) T } = E{ T 2} +
E{ X T }
= 9 + E(X) E(T) = 9
c. T = AR + B.
A = cov(R, T) / var(R) = 9/12 = ¾
As a check: A = rrt t/r = [sqrt(3)/2] (3/sqrt(12)) = 3/4
B = T – A R = 0
T = ¾ R
16. input Y: N(0, 1), output (observable) X = Y + N, where noise N is N(0,
4). (Y, N independent); note Y = 0,X = Y + N = 0
a. For linear MMSE estimate of Y in terms of X;
A = Y/X = cov/X2
cov = E(XY) - XY= E((Y+N) Y) - 0 = E(Y2 + NY) = E(Y
2) + E(N)E(Y)
= 1;
Y = 1; var(X) = var(Y+N) = var(Y) + var(N) = 5
X = sqrt(5)
A = 1/sqrt(5)
B = Y – A X, where X = E(Y+N) = 0; so B = 0 – (1/sqrt(5)) 0 = 0;
Linear MMSE estimate: Y 5
1 X
b. Find the best (arbitrary) MMSE estimate of Y in terms of X; i.e. find the
function c(X) that minimizes the mean squared error.
Since X and Y are jointly Gaussian, the best estimate is the Linear MMSE
estimate given in part (a).
17. Z = XY, X & Y independent; find fz(z).
Let Z = XY; let W = Y; Or: z = g(x, y) = xy; w = h(x, y) = y;
Step 1: Find the Jacobian of the transformation: |J(x, y)| =
= = |y|
0
^
y
w
x
wy
z
x
z
10
xy
^
^
ECE 650 11
Step 2: Solve “backwards” for roots x and y (in terms of w, z): y = w,
x = z/w
Step 3: fzw(z, w) =
Step 4: Integrate to find fz(z):
fz(z) = dww
)w(f)w/z(fdw)w,z(f
yxzw
18.
0 elsewhere
c. f(y|x) = )x(f
)y,x(f, where f(x) =
1y
xy
21
x
xy2
y2dy)xy(2dy)y,x(f
= 2{ [1/2 +x] – [x2/2 + x
2] } = -3 x
2 + 2x + 1
So: f(y|x) = )x(f
)y,x(f= 1yx0,
1x2x3
)xy(22
d. mmse estimate of Y, given X = x.
E[Y|x] =
dy
1x2x3
)xyy(2dy)x|y(fy
2
2
1y
xy
23
2
1
x
22 2
xy
3
y
1x2x3
2dy)xyy(
1x2x3
2
w
)w(f)w/z(f
w
)w,w/z(f
)y,x(J
)y,x(f yxxyxy
1yx0),xy(2)y,x(f
ECE 650 12
= 3x6x9
2x3x5
2
x
3
x
2
x
3
1
1x2x3
22
333
2
19. X =
3
2
1
X
X
X
, where Xk = k Yk, and where the Yk are independent
Gaussian RV’s with mean k and unit variance.
a. mean vector
b. Correlation matrix, RX
Ind. implies uncorr.
Similarly, E( X 1 X 3) = E( X 3 X 1) = 9; E( X 2 X 3) = E( X 3 X 2) = 36;
E( X 22) = 20; E( X 3
2) = 90
Thus,
c. Finding f(x1,x2): Note cov( X 1, X 2) = R12 – X1X2 = 0, so X 1 and X 2 are
uncorrelated and therefore independent since Gaussian (Yk Gaussian
implies Xk = kYk Gaussian).
f(x1,x2) = f(x1)f(x2)
9
4
1
)Y(E3
)Y(E2
)Y(E
)Y3(E
)Y2(E
)Y(E
)X(E
)X(E
)X(E
3
2
1
3
2
1
3
2
1
)X(E)XX(E)XX(E
)XX(E)X(E)XX(E
)XX(E)XX(E)X(E
R2
32313
322
212
31212
1
X
4)2)(1(2)Y(E)Y(E2)Y2Y(E)XX(E
211)Y(E)X(E
212121
2Y
2Y
21
21 11
90369
36204
942
RX
ECE 650 13
d. correlation matrix, R, for random variables X1 and X2; picking off the
relevant entries from the given R matrix:
20. Let X be a R vector with mean X =
3
1 and covariance matrix CX =
22
25.
a. RX = CX + XXT
b. R12 = 5; Cov(X1, X2) = C12 = 2; 12 = cov/(X1X2)
= 2/sqrt(10) = sqrt(10)/5; X12X2
2 = 2
c. W: elt. white; find H, C, such that X = HW + C, to yield X described
above
C = X; H = E 1/2
, where E: columns are eigenvectors of CX
To find eigenvalues of CX: det (CX – I) = 0 = 6, 1
To find eigenvector x corresponding to = 6:
CX x = x becomes
204
42RX
8
)4x(
2
)1x(exp
4
1
2
)x(exp
2
1
2
)x(exp
2
1
22
21
2X
2X2
X2
X
2X1
X2
2
21
1
1
115
5631
3
1
22
25
ECE 650 14
Solving for x yields
Similarly, the eigenvector corresponding to = 1 is:
Thus, H = E 1/2
So X = HW + C
21. (old test question) Two Gaussian random variables X1 and X 2 have 0
means and variances 4 and 9 respectively. Their covariance CX1X2 is equal to
3.
a.
CX =
93
34
b.
CY = H CX Ht =
42
31
93
34
43
21
2
1
2
1
x6
x6
x
x
22
25
51
52
x
52
51
x
52
56
51
562
10
06
52
51
515
2
3
1
52
56
51
562
W
ECE 650 15
25266
6628
42
31
4524
152CY
c. Joint pdf, f(y1, y2), for the random variables Y1 and Y2.
YY
Y
1TC2
1
21 e2
1)y,y(f
where C =
25266
6628
= det(C) = 2700, C-1
=
0104.0244.
0244.0933.
YY
Y
0104.0244.
0244.0933.
2
1
21
T
e0031.)y,y(f
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