hw7_soln

hw7_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,4.1.1 4.2.1 4.2.2 4.3.2 4.3.5 4.3.7 4.4.4 4.4.6 4.4.8 and 4.4.11 Problem 4.1.1 x 1 x  ¥ ¥ ¦¤ 1 x 1 1 12 ¨ ©¡ § 0 x 1 ¤ ¢ £¡ FX x Each question can be answered by expressing the requested probability in terms of FX x . ¡ (a) FX 1 2 ¥ 1 1 ¢ £¡ ¨ 12 ¥ PX 34 ¨ ¢  ¨  ¥ 1 14 ¢ 12 ¨ PX ¢  ¨  (b) This is a little trickier than it should be. Being careful, we can write X 34  ¤ 12 PX §  ¨ ¨ P ¥ ¢  ¥ 34 12 PX ¥  ¨ ¢  ¨ ¤ X 34 ¢  12 ¨ ¨ P ¥ Since the CDF of X is a continuous function, the probability that X takes on any specific value is zero. This implies P X 3 4 0 and P X 12 0. (If this is not clear at this point, it will become clear in Section 4.6.) Thus, 34 FX 3 4 ¥ !¡ ¨ ¢  ¨ X FX  ¤ 12 12 ¥ P ¢ "¡ ¨ ¨ ¥ ¢ 34 58 ¨ ¢ ¨ ¢ ¨ ¤ X ¥ ¢   12 ¢ ¨ ¨ ¥ P (c) 12 PX ¥ ¢  ¨ # %# $ ¢ ¨ (¤  ¥ ¢ '¡ ¨ ¢ & ¨ ¥ ¤  § a 1 2 ¢ ¢ "¡  ¢    ¢  ¨  # )# $ ¢ "¡  0 1 14 12 1, we need to satisfy 08 0 ¥  ¨ 0 6. ¢  ¨ FX a 34 ¥¨ 1. For a a 12 ¨ PX ¢ ¨ ¢ Thus a ¥  PX ¢ £¡ ¨ 1, we must have a 12 ¨ ¨ PX ¨ ¨ (d) Since FX 1 12 ¥ ¢   PX 12 ¢ & ¨  Note that P X 1 2 FX 1 2 3 4. Since the probability that P X P X 1 2 . Hence P X 12 FX 1 2 1 4. This implies 12 0, P X ¥ ¦¤  PX 12 ¢ & ¨ 12 ¢  ¨ X  12 ¥  ¨ P ¥ ¤  12 ¨ PX Problem 4.2.1 cx 0 x 2 0 otherwise   ¢ £¡ fX x (a) From the above PDF we can determine the value of c by integrating the PDF and setting it equal to 1. 1 2 ¢ 12x 0 2 ¢ 3 2  ¨ 12 dx 1 16 ¢   X 14 ¨ 2¢ ¦   12 1x 0 2 dx ¢ ¨ ¥ (c) P 1 ¨ X ¢ (b) P 0 1 1 2. ¨ Therefore c 2c ¢ cx dx 0 (d) The CDF of X is found by integrating the PDF from 0 to x. 0 x 2 2   fX x dx ¨  1 ¢ !4 4 ¢ £¡ 0 ¤ FX x 0 x 24 0 x 1 x x Problem 4.2.2 From the CDF, we can find the PDF by direct differentiation. The CDF and correponding PDF are   ¥ ¥ ¦¤ 1 fX x 12 1x1 0 otherwise ¢ £¡ ¨ ©¡ x 1 ¨ 1 x 1 ¥ x 12  § 0 x 1  ¢ "¡ FX x Problem 4.3.2 (a) Since the PDF is uniform over [1,9] Var X 9 1 12 2 ¢ ¢   ¢ 2 § ¢   16 3 X then x 12 3@ 1 ¢¡ 76  ¢ 98¡ EhX 5 9 8 1 dx ¢ 1 12 ¨ ¢ "¡ hEX 5¨ 5 ¨ 1 5 ¥ (b) Define h X 9 ¡ 1 EX  (c) 12 @ ¢ B  ¡ 2 8 1 dx ¥ 1 x EX 2 ¢ 9 ln9 8 ¥ 2 14 ¨ EY 1 ¥  E Y2 ¢ A¡ ¢   Var Y EhX ¨ EY ¢   Problem 4.3.5 The CDF of Y is y 1 y  ¥ ¥ ¤ 12 1 ¨ ©¡ § 0 y 1 y 1 1 ¤ ¢ £¡ FY y (a) We can find the expected value of Y by first find the PDF by differentiating the above CDF. ¢ "¡ 12 1y1 0 otherwise EY 1   ¥ ¨ fY y And 1 ¢   @ 1 y2 dy 1 3 12 E Y2 EY 2 1 3 1 13 ¢ 0 ¨ ¥¨ ¢ ¢  ¢ @ ¥ Var Y ¨ E Y2 ¨ (b) 0 ¢ y 2 dy 1 ¢ C  Problem 4.3.7 find the PDF of U by taking the derivative of FU u . The CDF and corresponding PDF are ¨ fU u 5 ¢ "¡ 0 ¤ ¤¥ ¥ ¥ ¤ ¥ ¤ 3 u 5 0 u 5 18 5u 3 0 3u3 38 3 u 5 0 u5 ¨ ¨ ©¡ 383 u 3 ¨ ©¡ ¥ § §¨ ¨ ¢ D¡ (a) The expected value of U is 3 5 3 @ 5 @ 3 16 ¢ ¢ ¡ @ § 1  @ ¢ 16 5 3 25 3u § u2 1 u du 8 § 1 ∞ u fU u du @ ∞ 1 EU 3u du 8 3 2 ¥ ¢ (b) The second moment of U is 3 5 ¢ 3 @ u3 @ ¢ 1 @ ¢ 49 3 2 ¢ E  ¡ 3 ¨ EU @ ¥ E U2 24 5 37 3. 8 5 3 5 u3 ¨ ¡ The variance of U is Var U u2 du 8 § 1 @ ∞ u2 fU u du § ∞ 1 E U2 3 0 ¤ ¤¥ ¥ ¥ ¤ 5 u u 5 3 ¥ ¤ u ¡ FU u 0 u58 14 14 3u 1 3u2 du 8 ¢   e ln 2 U . This implies that 2u ln 2 1 ln 2 u e ln2 F e ln 2 u du ¢G G ¢ F 1 ¢ 2u du 1 G F ¢ (c) Note that 2U The expected value of 2U is then 2u du 8 3 3 3 2u du 8 3 2u 8 ln 2 @ § 1 @ 5 2u 5 H 3 § ¢ 5 @ @ ¢ 1 ¢ @ ¢ 8 ln 2 5 3 2307 13 001 256ln 2 0 ¡ ∞ 2u fU u du 1 ∞ H E 2U ¢ Problem 4.4.4 x 5, the CDF is 5  ¥ (¤ 5 ¢ fX τ d τ x § ¥ ¡ 5 5  ¢ £¡ ¢ £¡ 1 @ ¢ £¡  ¢ £¡ ¥ ¦¤ FX x 1. For x  ¨ 5, FX x ¥ 0. For x  5, FX x 1 10 5x5 0 otherwise  (b) For x ¥ fX x 5 5 is ¡I (a) The PDF of a continuous uniform random variable distributed from 10 The complete expression for the CDF of X is x 5 10 5 x x 5  ¨ ©¡ § ¢ £¡ FX x 0 x 1 5 (c) the expected value of X is 1 x dx 5 10 5 0 5 ¢ x2 20 5 ¢ @ @ Another way to obtain this answer is to use Theorem 4.7 which says the expected value of X is 5 5 EX 0 2 ¥ P§ ¢ ¢   (d) The fifth moment of X is 0 5 ¢ e 10 14 84 0 ¢ @ 4 e5 ¥ 5 5 5 @ ex 10 5 @ 1 @ ex dx 5 10 1 ¢ 5 ¢ The expected value of eX is x6 60 ¢ x5 dx 5 10 5 @ Problem 4.4.6 Given that 3@¡¨ x2 12e 0 x0 otherwise  ¢ £¡ fX x (a) ¢  1 dx e ¢ x2 12e 12 e 1 0 2387 0 1 2 3@¡¨ 2 ¥3@ X ¢@  P1  (b) The CDF of X may be be expressed as x2 x x 0 0 1 2. By Theorem 4.9, the expected ¨ ¢  ¢ "¡ ¢ 1 a2 e ¥ 3@¡¨ 2 (c) X is an exponential random variable with parameter a 1 a 2. value of X is E X 4. ¢ ¨ ¢ Q  (d) By Theorem 4.9, the variance of X is Var X 0 1  0 0 ¤ x x x 2 dτ 12e ¢ x 0 3@ 0 ¤ FX x ¨ ¢ Q  Problem 4.4.8 The integral I1 is dx ¥ R¢ λx e λx @ @ ¢ 0 λe ∞ 0 1 ¢ 1, we have dt dv 1 ¢ We define u and dv as shown above in order to use the integration by parts formula u dv v du. Since 2 2 λn 1 xn 1 dx n 2! ¥ ¢ @ v e λx @ ¡ ¥ @ ¢ du we can write 5 ¥ @ @ ¡ @ ¥ 1 0 λn 1 xn 1 e n 2! ¡ § @ ¥ S¢ § ¢ 1. 0 ∞ 1 In ∞ 0 1! λx @ n e @ ¢  1 for all n v du 0 λn 1 xn 1 @ ¥# ¢ Hence, In ∞ 1 uv ∞ 0 In λx dx ¢ ¥ u ¡ 0 λx uv ¥ λn 1 xn 1 λe n 1! @ ∞ @ In @  For n ∞ 1 I1 Problem 4.4.11 r over the entire integral, we can write 1 r fX x dx  r ¡ ¡ 1  £¡ r ∞ rP X ¢ x fX x dx r  ∞  0 and x  (a) Since fX x (b) We can write the expected value of X in the form § ¡ 1 ¢   0 ∞ x fX x dx r ¡ x fX x dx 1 r EX Hence, ¢ ¡ 1   r EX ¥ T  x fX x dx r x fX x dx 0 ¡ ∞ 1 r rP X  Allowing r to approach infinity yields 2¥ ∞ x fX x dx ¡ 1 ¢ § %8¡ # 0   2  U    ¥ ¢  A¡ ¥ ¢ 1 By applying part (a), we now observe that x fX x dx ¢ ¡ 1 0 ¥ ¢ 6 ∞ FX x ∞ 0 EX  0 and this implies x 1 FX x dx lim rP X r∞ ¢ %A¡ #  A¡ ¢  ¥ ¥ 1  0 U 1 0 U ¢ %8¡ # ∞ ¥ r X FX r ¥ 8¡ ∞ rP lim r 1 r∞ ¢ U By part (b), limr ∞ 0  FX x r  x1 0. Thus, ¢ v du by defining u 1 ¥ ∞ 0 0 FX x ¡ U FX x 0.  x1 r X ∞ rP uv EX ¢  ¥ V  U FX x dx u dv EX ¡ 0, we must have limr ∞ 0 x fX x dx ¢ r∞0 (c) We can use the integration by parts formula and dv dx. This yields 1 r ¥ T  0 for all r lim ¢ Q  r EX 1 r∞ Since rP X r lim rP X ...
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