hw8_soln

hw8_soln - Probability and Stochastic Processes: A Friendly...

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Unformatted text preview: Probability and Stochastic Processes: A Friendly Introduction for Electrical and Computer Engineers Roy D. Yates and David J. Goodman Problem Solutions : Yates and Goodman,4.5.3 4.5.6 4.5.7 4.7.3 4.7.5 4.7.8 4.7.11 4.7.13 and 4.7.16 Problem 4.5.3 X is a Gaussian random variable with zero mean but unknown variance. We do know, however, that 01 ¡ ¢ ¥  FX 10 ©  FX 10  £ 10 10 σX 2Φ ¦  © ¦ ¥ £ X  1 0 55. Using Table 4.1 for the Gaussian CDF, we find that 10 σX   ¨ ¦  This implies Φ 10 σX or σX 66 6. 10  ¢© P function. 0 15 ¦ ¦ §¥ We can find the variance Var X by expanding the above probability in terms of the Φ ¨ £ ¤¡ 10 ¨ PX  © ¨ ¦ Problem 4.5.6 We are given that there are 100 000 000 men in the United States and 23 000 of them are at least 7 feet tall, and the heights of U.S men are independent Gaussian random variables with mean 5 10 .      (a) Let H denote the height in inches of a U.S male. To find σX , we look at the fact that the probability that P H 84 is the number of men who are at least 7 feet tall divided by the total number of men (the frequency interpretation of probability). Since we measure H in inches, we have ¥    Qx,  ¨ ¦ !  ©  ¦ !  © © 3 5 or σX ¦ ¦ ! ¨ ¦ From Table 4.2, this implies 14 σX 2 3 10  © Q 14 σX 4 " ¦ §¥ Φx ¦ 1 0 00023 ¦ x 70 84 σX Φ ¨ Since Φ 23 000 100 000 000 84  PH 4.  (b) The probability that a randomly chosen man is at least 8 feet tall is Q 65 © ¦ 70 4 ¨ 96 Q  96  ¦ §¥ PH Unfortunately, Table 4.2 doesn’t include Q 6 5 , although it should be apparent that the probability is very small. In fact, Q 6 5 4 0 10 11 . "  ¨ © # 1 ¨ ¦  ¨ © (c) First we need to find the probability that a man is at least 7’6”. β 2 87 10 7 . ¨ ¦  © ¦ ¦ ¥ Although Table 4.2 stops at Q 4 99 , if you’re curious, the exact value is Q 5  7 " 3 10 $ ! Q5  70 4 ¦ 90 Q  90 " PH ©  ¨ © Now we can begin to find the probability that no man is at least 7’6”. This can be modeled as 100,000,000 repetitions of a Bernoulli trial with parameter 1 β. The probability that no man is at least 7’6” is  100 000 000 94 10 # β 14 " % ¨ ¦ %   1 © (d) The expected value of N is just the number of trials multiplied by the probability that a man is at least 7’6”. 100 000 000 β 30 ¦   EN  ¦ &¥ Problem 4.5.7 N µ σ2 distribution, the integral we wish to evaluate is ¥ ' dw dw σ and ¦  © e ' ∞ x2 2 " dx " ' (¦ "    ∞ 3 ¦ 1 2π 2σ2  ) µ σ, we have dx ) © 4¦ I 2 " w 0 1" (a) Using the substitution x ∞ wµ " 2πσ2 e 2 ∞ ∞ 1 fW w dw 3  ∞ I ¦ (b) When we write I2 as the product of integrals, we use y to denote the other variable of integration so that x2 y2 2 32 5 0 1" e dx 1 2π ∞ ∞ e y2 2 dy " " ' " ' ∞ 3 dx dy " 0 2 e dθ ¦ 2π  2π 0 r2 2 r2 2 " ' ' 0 e ' 0 ∞ 3 2π " ' 1 2π 1 2π 1 2π r2 and dx dy 3 6 ¦ " ¦ ¦ ¦ I2 y2 ¦ ¦ (c) By changing to polar coordinates, x2 1 r dr d θ so that r dr d θ ∞ 0 ¦ ) ' ∞ x2 2 ) ∞ ∞ ∞ e ' 1 2π ∞ " 1 2π 3 I2 dθ Problem 4.7.3 X2 0. We can conclude that FY y © PX ¦ dx 1 ¦ 9x 0,  " 9e 8 9' © ¦ ! 0 y 9y e " ¦ ! ¦ y FY y 0, 8 £ From the exponential PDF of X , we see that for y 0. For y ¥ y £ P X2 ) FY y 0 for y ¦ ! 0 we see that Y 7 (a) Since X © Hence, the CDF of Y is 0 0 8  e y y 9y " 0 1 7 ¦ ! FY y © By taking the derivative of the CDF, we obtain the PDF 9y  " 8 92e 0 )   © ¦ ! fY y yy 0 otherwise © (b) Finding the moments of Y is easiest using the moments of X . Specifically, we know from Probn! 9n . Hence, E Y E X2 2 81. lem 4.4.10 that E X n ¦ §¥   24 94 4 94  2 ¦  ¦  @¥ EY 20 6561  ¦  ¦ ©  U.  ¦ §¥  ln 1 4! 94 . This implies  ¦ §¥ © ¦ Problem 4.7.5 have a sketch of the function X E Y2 ¦ Var Y E X4  (c) The second moment is E Y 2 4 X 2 0 0 0.5 U 1 (a) From the sketch, we observe that X will be nonnegative. Hence FX x 0 for x 0. Since U has a uniform distribution on 0 1 , for 0 u 1, P U u u. We use this fact to find the CDF of X . For x 0, £ £ e x "  U PU ¦ P1 £ ¦ ¥   © 1 and so x 1 ¦ "  e  FU 1 e ¦  £ ¦  "  © £ © The complete CDF can be written as  ¦ B 3 e x " 0 1 x x 7 FX x 0 0 x "  FX x £ x 1   U e x " ¥ x £ e ¦ §¥ 1 © 0, 0 ln 1 ¦ A For x P 7 FX x © (b) By taking the derivative, the PDF is x e 0 x0 otherwise " ¦ B fX x © Thus, X has an exponential PDF. In fact, since most computer languages provide uniform 0 1 random numbers, the procedure outlined in this problem provides a way to generate exponential random variables from uniform random variables.  ¦ 1¥ 1, E X ¥ ¦ (c) Since X is an exponential random variable with parameter a 1. Problem 4.7.8 Let X denote the position of the pointer and Y denote the area within the arc defined by the stopping position of the pointer. (a) If the disc has radius r, then the area of the disc is πr2 . Since the circumference of the disc is 1 and X is measured around the circumference, Y πr2 X . For example, when X 1, the shaded area is the whole disc and Y πr2 . Similarly, if X 1 2, then Y πr2 2 is half the area of the disc. Since the disc has circumference 1, r 1 2π and ¦ ¦ ¦  ¦  ¦ © C ¦ X 4π ¦ πr2 X  ¦ Y £ y PX ¦ X 4π P 4πy £ £ y ¦ §¥ ¦ §¥ PY ¦ ! © Therefore the CDF is 0 y £ 7 0 y 4πy 0 1 y £ ¦ ! FY y 1 4π 1 4π © (c) By taking the derivative of the CDF, the PDF of Y is 4π 0 y 41 π 0 otherwise £ 1 8π . ¦ 4πy dy © C 2 1 4π 0 0 © 3 D (§¥ ¦ Problem 4.7.11 The PDF of U is £ £  12 1u1 0 otherwise  ¦ ! fU u 4  ¦ B (d) The expected value of Y is E Y £ fY y FX 4πy © FY y  (b) The CDF of Y can be expressed as © 0 for w 0 fU u du © 0 ¦ §¥ ' 1  PU 12 ¦ 7 © 0  ¦ E ¦ ¥ PW 0. Next, we observe that the rectifier output W is a 7 Since W 0, we see that FW w mixed random variable since ¦ " The above facts imply that 0 12 £ ¦ ¥ ¦ ! © 1, w fU u du  © ' 1 w © 4¦ £ w 12 6 PU   ¦ §¥ ¦  7 7 FW w © " w 120 w £ 0 w 1 0 w 1 7 6   G £ © © ¦  FW w 1 for w 1. Hence, the complete expression £ £ 1, which implies FW w ¦ F Finally, U 1 implies W for the CDF is PW ¦ w 0 ¦ ¥ Next, we note that for 0 PW  FW 0 1 © By taking the derivative of the CDF, we find the PDF of W ; however, we must keep in mind that the discontinuity in the CDF at w 0 yields a corresponding impulse in the PDF. ¦ £ 120 w 1 otherwise 6    δw 0 £ © © ¦  fW w © From the PDF, we can calculate the expected value © 6 ' 0 w 2 dw  0  ¦ 6    © © 0 1 1 2 dw 14 ¦ wδw  1 ¦ §¥ ' EW Perhaps an easier way to find the expected value is to use Theorem 2.10. In this case, ∞ 1 u 1 2 du   © ' 0 14 ¦ ¦  ©  © ' ∞ g u fW w du  ¦ §¥ EW " As we expect, both approaches give the same answer. Problem 4.7.13 shown in the following figure: 3 Y 2 1 0 0 1 2 3 X 1  © ' ¦ §¥ 5 0 1 fX x dx ¦ 1 ' £ X 0 © £ P0 x 2 dx  12   ¦ F¥  ¦ PY 1. Thus, ¦ X 14  £ 1 2 if and only if 0 £ ¦ (a) Note that Y 12 ¦   ©  7 © ¦ B £  £ 7 12  PY ¦ ¥ ¦ 1 14  ¦ ¥ £ P0 ¦    © £ 2, © ' £ ¦ ¥ y2 4 2. The complete expression of the CDF is y 12 12 y 1 1y2 y2  0 14 y2 4 1 £ 7 £  7  7 ¦ ! © ¦ !  ¦ ! £ © FY y fX x dx 0  1 for y y y ¦ 2, FY y PX  7 FY y Lastly, since Y PY ¦ y 1 2. Also, FY 1 2  FY y Next, for 1 X 0 for y ¦ §¥ (b) Since Y 1 2, we can conclude that FY y 1 4. Similarly, for 1 2 y 1, © Problem 4.7.16 First, we must verify that F 1 u is a nondecreasing function. To show this, suppose that for u u , x F 1 u and x F 1 u . In this case, u F x and u F x . Since F x is nondecreasing, F x implies that x x . Hence, we can write Fx   ©  © ¦   © © ¦ "  © £ ¦ 6 PU Fx © x ¦¥ I £  U Fx © © 1   " PF  ¦  " ¦    H FX x © © © ¦ "  © ...
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