Homework 8

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Homework 8

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ECE 460: Communication and Information Theory
Prof. B.-P. Paris
Homework 8
Solution

Problem 4.17
1) f_X,Y(x,y) is a PDF so that its integral over the support region of x, y should be one.
$∫ 1 ∫ 1 ∫ 1∫ 1 fX,Y(x,y)dxdy = K (x + y)dxdy 0 0 [0∫ 10∫ 1 ∫ 1∫ 1 ] = K xdxdy + ydxdy [ 0 0 0 0] 1 2||1 1 1 2||1 1 = K --x ||y |0 + -y ||x|0 2 0 2 0 = K$
Thus K = 1.
2)
$p(X + Y > 1) = 1 - P (X + Y ≤ 1) ∫ 1∫ 1-x = 1 - (x + y)dxdy 0 0 ∫ 1 ∫ 1- x ∫ 1 ∫ 1- x = 1 - x dydx - dx ydy ∫01 0 ∫ 10 0 = 1 - x(1 - x)dx - 1(1 - x)2dx 0 0 2 2- = 3$

3) By exploiting the symmetry of f_X,Y and the fact that it has to integrate to 1, one immediately sees that the answer to this question is 1/2. The “mechanical” solution is:
$∫ ∫ p (X > Y ) = 1 1(x + y)dxdy 0 y ∫ 1∫ 1 ∫ 1∫ 1 = xdxdy + ydxdy ∫0 y | ∫ 0 y| 11- 2||1 1 ||1 = 0 2x |ydy + 0 yx |ydy ∫ 1 ∫ 1 = 1(1 - y2)dy + y (1 - y )dy 0 2 0 1- = 2$

4)
$p(X > Y |X + 2Y > 1) = p(X > Y, X + 2Y > 1 )∕p (X + 2Y > 1)$
The region over which we integrate in order to find p(X > Y,X + 2Y > 1) is marked with an A in the following figure.

Thus
$∫ 1 ∫ x p(X > Y,X + 2Y > 1) = (x + y)dxdy 13 1-2x ∫ 1 [ 1 - x 1 2 1 - x 2 ] = 1 x(x - -----) + --(x - (-----) ) dx ∫ 3 ( 2 2) 2 = 1 15-x2 - 1x - 1- dx 13 8 4 8 49 = ---- ∫108∫ p(X + 2Y > 1) = 1 1 (x + y)dxdy 0 1-2x ∫ 1 [ 1 - x 1 1 - x ] = x(1 - -----) + --(1 - (------)2) dx ∫0 ( 2 )2 2 1 3- 2 3- 3- = 0 8 x + 4x + 8 dx 3 1 ||1 3 1 ||1 3 ||1 = --× -x3 || + --× --x2|| + -x|| 8 3 0 4 2 0 8 0 7- = 8$
Hence, p(X > Y |X + 2Y > 1) = (49∕108)∕(7∕8) = 14∕27
5) When X = Y the volume under integration has measure zero and thus
$P (X = Y ) = 0$

6) Conditioned on the fact that X = Y , the new p.d.f of X is
$--fX,Y-(x,x)--- fX|X=Y (x ) = ∫1fX,Y (x,x)dx = 2x. 0$
In words, we re-normalize f_X,Y(x,y) so that it integrates to 1 on the region characterized by X = Y . The result depends only on x. Then p(X > 1 2|X = Y ) = ∫ _1∕2¹f_X|X=Y(x)dx = 3∕4.
7)
$∫ 1 ∫ 1 1- fX(x ) = 0 (x + y)dy = x + 0 ydy = x + 2 ∫ 1 ∫ 1 1 fY(y ) = (x + y)dx = y + xdx = y + -- 0 0 2$

8) F_X(x|X + 2Y > 1) = p(X ≤ x,X + 2Y > 1)∕p(X + 2Y > 1)
$∫ ∫ p(X ≤ x,X + 2Y > 1) = x 1 (v + y)dvdy 0 1-2v ∫ x[3 3 3] = --v2 + -v + -- dv 0 8 4 8 1- 3 3- 2 3- = 8x + 8 x + 8x$
Hence, $3x2 + 6x + 3 3 6 3 fX (x|X + 2Y > 1) = --8-----8----8- = -x2 + --x + -- p (X + 2Y > 1) 7 7 7$

$∫ 1 E [X |X + 2Y > 1] = xfX (x|X + 2Y > 1)dx ∫01( ) = 3x3 + 6-x2 + 3x 0 7 7 7 3 1 4||1 6 1 3||1 3 1 2||1 17 = --× -x || + --× -x || + --× --x || = --- 7 4 0 7 3 0 7 2 0 28$
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Problem 4.21

1) f_X,Y(x,y) is a PDF, hence its integral over the supporting region of x, and y is 1.
$∫ ∫ ∫ ∫ ∞ ∞ f (x,y)dxdy = ∞ ∞ Ke -x-ydxdy 0 y X,Y 0 y ∫ ∞ -y ∫ ∞ -x = K e e dxdy ∫0∞ y |∞ = K e-2ydy = K (- 1-)e -2y|| = K 1- 0 2 |0 2$
Thus K should be equal to 2.
2)
$∫ x - x-y - x - y||x -x - x fX(x ) = 2e dy = 2e (- e )|| = 2e (1 - e ) ∫0∞ 0|∞ fY (y ) = 2e-x- ydy = 2e-y(- e-x)|| = 2e -2y y |y$

3)
$- x - x -2y -x-y - y - x fX(x)fY (y) = 2e (1 - e )2e = 2e 2e (1 - e ) ⁄= 2e- x-y = fX,Y(x,y)$
Thus X and Y are not independent.
4) If x < y then f_X|Y(x|y) = 0. If x ≥ y, then with u = x - y ≥ 0 we obtain
$-x-y f (u) = f (x|y) = fX,Y(x,y)-= 2e-----= e-x+y = e-u U X |Y fY(y) 2e -2y$

5)
$∫ ∞ ∫ ∞ E [X |Y = y] = xe -x+ydx = ey xe- xdx y[ y ] ||∞ ∫ ∞ = ey - xe-x|| + e- xdx y y = ey(ye -y + e-y) = y + 1$

6) In this part of the problem we will use extensively the following definite integral
$∫ ∞ 1 x ν-1e-μxdx = -ν(ν - 1)! 0 μ$

$∫ ∞ ∫ ∞ -x-y ∫ ∞ - y ∫ ∞ - x E [XY ] = 0 y xy2e dxdy = 0 2ye y xe dxdy ∫ ∞ ∫ ∞ ∫ ∞ = 2ye -y(ye-y + e-y)dy = 2 y2e-2ydy + 2 ye-2ydy 0 0 0 = 2 1-2! + 2 1-1! = 1 23 22$
$∫ ∞ - x - x ∫ ∞ -x ∫ ∞ -2x E [X ] = 2 xe (1 - e )dx = 2 xe dx - 2 xe dx 0 0 0 = 2 - 2 1--= 3- ∫ 22 2 ∞ -2y 1-- 1- E [Y ] = 2 0 ye dy = 222 = 2 2 ∫ ∞ 2 -x - x ∫ ∞ 2 -x ∫ ∞ 2 -2x E [X ] = 2 x e (1 - e )dx = 2 x e dx - 2 x e dx 0 0 0 = 2 ⋅ 2! - 2-12! = 7 ∫ 23 2 2 ∞ 2 -2y 1-- 1- E [Y ] = 2 0 y e dy = 2 232! = 2$
Hence, $3 1 1 COV (X, Y ) = E [XY ] - E [X ]E [Y ] = 1 - --⋅--= -- 2 2 4$
and
$COV (X, Y ) 1 ρX,Y = -----2----------2-1∕2-----2----------2-1∕2 = √--- (E [X ] - (E [X ]) ) (E [Y ] - (E[Y ]) ) 5$

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Problem 4.24
The following is a program written in Fortran to compute the Q function

REAL*8

x,t,a,q,pi,p,b1,b2,b3,b4,b5

PARAMETER

(p=.2316419d+00, b1=.31981530d+00,

b2=-.356563782d+00, b3=1.781477937d+00,

b4=-1.821255978d+00, b5=1.330274429d+00)

C-

pi=4.*atan(1.)

C-INPUT

PRINT*,

’Enter -x-’

READ*,

C-

t=1./(1.+p*x)

a=b1*t + b2*t**2. + b3*t**3. + b4*t**4. + b5*t**5.

q=(exp(-x**2./2.)/sqrt(2.*pi))*a

C-OUTPUT

PRINT*,

C-

STOP

END

The results of this approximation along with the actual values of Q(x) (taken from text Table 4.1) are tabulated in the following table. As it is observed a very good approximation is achieved.


x	Q(x)	Approximation


1.	1.59 × 10^-1	1.587 × 10^-1
1.5	6.68 × 10^-2	6.685 × 10^-2
2.	2.28 × 10^-2	2.276 × 10^-2
2.5	6.21 × 10^-3	6.214 × 10^-3
3.	1.35 × 10^-3	1.351 × 10^-3
3.5	2.33 × 10^-4	2.328 × 10^-4
4.	3.17 × 10^-5	3.171 × 10^-5
4.5	3.40 × 10^-6	3.404 × 10^-6
5.	2.87 × 10^-7	2.874 × 10^-7

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Problem 4.30
1) f_X,Y(x,y) is a PDF and its integral over the supporting region of x and y should be one.
$∫ ∞ ∫ ∞ -∞ -∞ fX,Y(x,y)dxdy ∫ 0 ∫ 0 x2+y2 ∫ ∞ ∫ ∞ x2+y2 = K-e---2--dxdy + K-e- -2--dxdy -∞∫ -∞ π ∫ 0 0∫ π ∫ K-- 0 - x2 0 - y2 K-- ∞ - x2- ∞ - y2 = π -∞ e 2 dx -∞ e 2 dx + π 0 e 2 dx 0 e 2 dx K [ 1√ --- ] = --- 2(-- 2π)2 = K π 2$
Thus K = 1
2) If x < 0 then
$∫ 0 1-- x2+y2- 1-- x2∫ 0 - y2 fX(x ) = -∞ πe 2 dy = πe 2 - ∞ e 2 dy 1 x21 √ --- 1 x2 = --e--2-- 2π = √----e- 2- π 2 2π$
If x > 0 then
$∫ ∞ 1 - x2+y2- 1 - x2∫ ∞ - y2 fX (x) = -e 2 dy = -e 2 e 2 dy 0 π2 √ --- π 2 0 = 1e- x2-1- 2π = √1--e- x2- π 2 2π$
Thus for every x, f_X(x) = 1 __ e^{-x²
2} which implies that f_X(x) is a zero-mean Gaussian random variable with variance 1. Since f_X,Y(x,y) is symmetric to its arguments and the same is true for the region of integration we conclude that f_Y(y) is a zero-mean Gaussian random variable of variance 1.
3) f_X,Y(x,y) has not the same form as a binormal distribution. For xy < 0, f_X,Y(x,y) = 0 but a binormal distribution is strictly positive for every x, y.
4) The random variables X and Y are not independent for if xy < 0 then f_X(x)f_Y(y)0 whereas f_X,Y(x,y) = 0.
5)
$1 ∫ 0 ∫ 0 x2+y2 1 ∫ ∞ ∫ ∞ x2+y2 E[XY ] = -- XY e---2--dxdy + -- e---2--dxdy π ∫- ∞ -∞ ∫ π 0∫ 0 ∫ 1- 0 - x22 0 - y22- 1- ∞ - x22 ∞ - y22 = π - ∞ Xe dx -∞ Y e dy + π 0 Xe dx 0 Y e dy 1 1 2 = --(- 1 )(- 1) +--= -- π π π$
Thus the random variables X and Y are correlated since E[XY ]0 and E[X] = E[Y ] = 0, so that E[XY ] - E[X]E[Y ]0.
6) In general f_X|Y(x,y) = f_X,Y(x,y) f_Y(y) . If y > 0, then
$( { 0∘ -- 2 x < 0 fX|Y(x,y ) = ( 2e- x2- x ≥ 0 π$
If y ≤ 0, then
$( { 0 x > 0 fX|Y(x,y ) = ( ∘ 2-- x22 πe x < 0$
Thus
$∘ -- 2 fX|Y(x,y) = 2-e- x2 u (xy ) π$
which is not a Gaussian distribution.

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