Homework 7

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Homework 7

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

Problem 4.5
Let us denote by nS the event that n was produced by the source and sent over the channel, and by nC the event that n was observed at the output of the channel. Then
1)
$P (1C ) = P (1C |1S )P (1S ) + P(1C |0C )P(0C ) = .8 ⋅ .7 + .2 ⋅ .3 = .62$
where we have used the fact that P(1S) = .7, P(0C) = .3, P(1C|0C) = .2 and P(1C|1S) = 1 - .2 = .8
2)
$P (1C,1S ) P (1C |1S)P (1S) .8 ⋅ .7 P (1S|1C ) = ---------- = ---------------- = ------= .9032 P (1C ) P (1C ) .62$

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Problem 4.6
1) X can take four different values. 0, if no head shows up, 1, if only one head shows up in the four flips of the coin, 2, for two heads and 3 if the outcome of each flip is head.
2) X follows the binomial distribution with n = 3. Thus
$( ( ) |{ 3 k 3- k P(X = k ) = k p (1 - p ) for 0 ≤ k ≤ 3 |( 0 otherwise$

3)
$( ) ∑k 3 m 3- m FX (k) = m p (1 - p) m=0$
Hence
$( |||| 0 3 k < 0 |||| (1 - p) k = 0 { (1 - p)3 + 3p(1 - p)2 k = 1 FX (k ) = || (1 - p)3 + 3p(1 - p)2 + 3p2(1 - p) k = 2 |||| (1 - p)3 + 3p(1 - p)2 + 3p2(1 - p) + p3 = 1 k = 3 ||( 1 k > 3$

4)
$3 ( ) ∑ 3 k 3-k 2 3 P(X > 1) = k p (1 - p) = 3p (1 - p) + (1 - p) k=2$
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Problem 4.7
1) The random variables X and Y follow the binomial distribution with n = 4 and p = 1∕4 and 1∕2 respectively. Thus
$( ) ( )0 ( )4 4 ( ) ( )4 p(X = 0) = 4 1- 3- = 3-- p (Y = 0) = 4 1- = -1- 0 4 4 28 0 2 24 ( ) ( 1)1 (3 )3 334 ( ) (1 )4 4 p(X = 1) = 4 -- -- = ---- p (Y = 1) = 4 -- = --- ( 1 ) 4 4 28 ( 1 ) 2 24 4 ( 1)2 (3 )2 332 4 (1 )4 6 p(X = 2) = -- -- = --8- p (Y = 2) = -- = -4- ( 2) 4 4 2 ( 2 ) 2 2 4 ( 1)3 ( 3)1 3 ⋅ 4 4 (1 )4 4 p(X = 3) = 3 4- 4- = -28- p (Y = 3) = 3 2- = 24- ( ) ( ) ( ) ( ) ( ) 4 1- 4 3- 0 -1- 4 1- 4 -1- p(X = 4) = 4 4 4 = 28 p (Y = 4) = 4 2 = 24$
Since X and Y are independent we have $3 3-2-6- -81-- p(X = Y = 2) = p(X = 2)p(Y = 2 ) = 28 24 = 1024$

2)
$p (X = Y) = p(X = 0)p(Y = 0) + p(X = 1)p(Y = 1) + p(X = 2)p(Y = 2) +p (X = 3)p(Y = 3) + p(X = 4)p(Y = 4) 4 3 2 4 2 2 = -3--+ 3--⋅ 4 + 3--⋅ 2 + 3-⋅ 4 + -1--= 886-- 212 212 212 212 212 4096$

3)
$p(X > Y ) = p (Y = 0) [p(X = 1) + p(X = 2) + p(X = 3) + p(X = 4)] + p (Y = 1) [p(X = 2) + p(X = 3) + p(X = 4)] + p (Y = 2) [p(X = 3) + p(X = 4)] + p (Y = 3) [p(X = 4)] 535-- = 4096$

4) In general p(X + Y ≤ 5) = ∑ _l=0⁵ ∑_m=0^lp(X = l -m)p(Y = m). However it is easier to find p(X + Y ≤ 5) through p(X + Y ≤ 5) = 1 -p(X + Y > 5) because fewer terms are involved in the calculation of the probability p(X + Y > 5). Note also that p(X + Y > 5|X = 0) = p(X + Y > 5|X = 1) = 0.
$p(X + Y > 5) = p(X = 2)p(Y = 4) + p(X = 3)[p (Y = 3) + p(Y = 4)] + p(X = 4)[p(Y = 2) + p (Y = 3) + p(Y = 4)] -125- = 4096$
Hence, p(X + Y ≤ 5) = 1 - p(X + Y > 5) = 1 - 125 _ 4096
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Problem 4.8
1) Since lim _x→∞F_X(x) = 1 and F_X(x) = 1 for all x ≥ 1 we obtain K = 1.
2) The random variable is of the mixed-type since there is a discontinuity at x = 1. lim _ϵ→0F_X(1 - ϵ) = 1∕2 whereas lim _ϵ→0F_X(1 + ϵ) = 1
3)
$1 1 1 3 P (--< X ≤ 1) = FX (1) - FX (-) = 1 - --= -- 2 2 4 4$

4)
$1 - 1 1 1 1 P (--< X < 1) = FX (1 ) - FX (-) = --- --= -- 2 2 2 4 4$

5)
$P (X > 2) = 1 - P(X ≤ 2) = 1 - FX (2) = 1 - 1 = 0$
Problem 4.10
1) The random variable X is Gaussian with zero mean and variance σ² = 10^-8. Thus p(X > x) = Q(x σ) and
$( ) -4 10- 4 p(X > 10 ) = Q ----4 = Q (1) = .159 ( 10 ) -4 4-×-10--4 -5 p (X > 4 × 10 ) = Q 10- 4 = Q (4) = 3.17 × 10 -4 -4 p(- 2 × 10 < X ≤ 10 ) = 1 - Q (1) - Q(2) = .8182$

2)
$-4 - 4 p(X > 10- 4|X > 0) = p(X-->--10--,-X->--0)= p(X-->-10---)= .159- = .318 p(X > 0) p(X > 0) .5$

3) y = g(x) = xu(x). Clearly f_Y(y) = 0 and F_Y(y) = 0 for y < 0. If y > 0, then the equation y = xu(x) has a unique solution x₁ = y. Hence, F_Y(y) = F_X(y) and f_Y(y) = f_X(y) for y > 0. F_Y(y) is discontinuous at y = 0 and the jump of the discontinuity equals F_X(0).
$FY (0+ ) - FY (0- ) = FX (0) = 1 2$
In summary the PDF f_Y(y) equals
$1- fY (y) = fX(y)u (y ) + 2δ(y)$
The general expression for finding f_Y(y) can not be used because g(x) is constant for some interval so that there is an uncountable number of solutions for x in this interval.
4)
$∫ ∞ E [Y ] = yfY(y )dy ∫- ∞ [ ] ∞ 1- = - ∞ y fX (y)u(y) + 2δ (y ) dy 1 ∫ ∞ y2- σ = √------ ye- 2σ2dy = √---- 2πσ2 0 2π$

5) y = g(x) = |x|. For a given y > 0 there are two solutions to the equation y = g(x) = |x|, that is x_1,2 = ±y. Hence for y > 0
$fX(x1) fX (x2) fY(y) = ---------+ --------- = fX (y) + fX (- y) |sgn (x1 )| |sgn(x2)| --2---- -2y2σ2 = √ 2πσ2 e$
For y < 0 there are no solutions to the equation y = |x| and f_Y(y) = 0. $∫ ∞ y2 E[Y ] = √--2--- ye- 2σ2dy = √2σ-- 2πσ2 0 2π$

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