hw11_soln

# hw11_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,7.1.2 7.2.3 7.3.1 7.4.1 7.4.3 7.7.2 and 7.8.1 Problem 7.1.2 X2 , Theorem 7.1 says that the expected value of the difference is ¨ § ¤ ¨ § ¨ ¤§ ¡ ¨ § ¨ § X1 ¦ ¥£¡ ¤¢ (a) Since Y EY E X1 E X2 EX EX 0 (b) By Theorem 7.2, the variance of the difference is ¡¨ § ¡ w µw e λ µx 0 λµ λµ e µw e 0 When µ we have dx dx λw w0 otherwise λ, the previous derivation is invalid because of the denominator term λ ¤ λµe µw x µe µ. For µ ¤         ¦¢ ¢ ¦¢ λx λe x dx  © 0 fX x fY w 2 Var X Y by convolving the two exponential dis-    ¦¤ ¨§ ∞ w X2 X ∞ fW w ¨ ¤§ Problem 7.2.3 By using Theorem 7.5, we can ﬁnd the PDF of W tributions. For µ λ, Var X1 ¨§ Var Var Y λ, ∞       ¦ ¤ ¢ ¦ ¢  ¦ ¢ fW w ∞ w fX x fY w λe 0 λ2 e λx λw λ2 we 0 Note that when µ Erlang PDF. λe x dx λw x dx w dx 0 λw w0 otherwise λ, W is the sum of two iid exponential random variables and has a second order 1 0, a zero mean Laplace random variable X has PDF ∞ ax ¦ ¢ The moment generating function of X is ∞ ¡     ¦¢ ae s a x 2s a a 2 a2 0 1 a2 ∞ ∞ a esx e 0 ∞ ae s a x 2s a ax dx 0 1 ¤¤¡ s a 2 esx eax dx  0 a 2 E esX ∞ ¤   ¡ ¡ φX s x  ¤   a e 2 fX x   Problem 7.3.1 For a constant a s a ¤ s2 Problem 7.4.1 Ki has PMF ¤ pk 0 k1 otherwise 1 p 0 ¦ ¢ PK k (a) The MGF of K is ¡ ¤ § ¨¦ ¢ § ¦ ¢ ¡ ¥¥£¡  ¦¢ K1 K2 E esK 1 p pes p pes Kn has MGF φM s φK s n 1 n ¨ ¡ (b) By Theorem 7.10, M ¡¤ φK s (c) Although we could just use the fact that the expectation of the sum equals the sum of the expectations, the problem asks us to ﬁnd the moments using φM s . In this case, ¦ ¡ ¤¢ ! s0 pes p n1 pes ! n1 ¦¢ ¦¢ d φM s ds ¨ § EM s0 np (d) The second moment of M can be found via ¦ ¨ § "¤ ¢ EM 2 2 p ¦ ¤ ¢ ¨ § E M2 Var M 1 1 np 1 p pes n1s e n Var K ¨§ 1p ¦ np n pes pe2s p ¡ ¤ £¡ ¢ 11 ¦ np n ¨ ¡ ¦ ¤ ¢§ ¡ ¤ ¢¦! ¤ ¢ The variance of M is s0 ! ¦¢ d φM s ds E M2 s0 Problem 7.4.3 ¦ ¢ 2 k! k 0 1 2 otherwise Ki    ¦¢ ¡ K2 ¡ %¥¥£¡ K1 # And let Ri  ¥¥ \$ \$ \$ 2k e 0 PKi k (a) From Example 7.11, we ﬁnd that the Poisson random variable K has MGF φK s (b) The MGF of Ri is the product of the MGFs of the Ki ’s. ¦¢ ! ¦¢ n1 s 1 .  i ∏ φK φRi s s e2 e s e2i e 1 (c) Since the MGF of Ri is of the same form as that of the Poisson with parameter, α 2i. Therefore we can conclude that Ri is in fact a Poisson random variable with parameter α 2i. That is,  ¥¥ \$ \$ \$ r! r 0 1 2 otherwise ¦¢ (d) Because Ri is a Poisson random variable with parameter α are then both 2i. #¦¢ 2i 2i r e 0 PRi r 2i, the mean and variance of Ri Problem 7.7.2 Knowing that the probability that voice call occurs is 0.8 and the probability that a data call occurs is 0.2 we can deﬁne the random variable Di as the number of data calls in a single telephone call. It is obvious that for any i there are only two possible values for Di , namely 0 and 1. Furthermore for all i the Di ’s are independent and identically distributed withe the following PMF.  08 d 0 02 d 1 0 otherwise  ¦ ¢ PD d From the above we can determine that   ¤  ¨ §  ¨ § ED 02 Var D 02 0 04 0 16 With the previous descriptions, we can answer the following questions. ¨ § 20 § (§ ¦ ¥¢ ¤ ¦ ¢  ¨ ( ¤ ¤  ¦ # ¢ ¦ # ¤¥¢ ¤  ¤ ¨  § & ' ¨ § ¨ § (b) Var K100 (c) P K100 (d) P 16 100Var D 18 1 K100 24 Φ Φ 18 20 4 24 20 4 16 1 Φ 4 Φ 12 16 20 4 3 Φ12 Φ1 Φ 0 6915 1  ¤¦ ¢ 100E D ¨ (a) E K100 2Φ 1 1 0 6826 Problem 7.8.1 In Example 7.12, we learned that a sum of iid Poisson random variables is a Poisson random variable. Hence Wn is a Poisson random variable with mean E Wn nE K n. Thus Wn has variance Var Wn n and PMF ¨ §  ¥¥ \$ \$ \$ # n ¨ § nw e 0 ¨ § ¦ ¢ PWn w w! w 0 1 2 otherwise All of this implies that we can exactly calculate ¦ ¢ ¨ § n # P Wn n nn e PWn n n! Since we can perform the exact calculation, using a central limit theorem may seem silly; however for large n, calculating nn or n! is difﬁcult for large n. Moreover, it’s interesting to see how good the approximation is. In this case, the approximation is n 05 n n n Φ 05 n n 2Φ 1 & Φ & ¤ ¡ n ¤ ) 0¨ ( Wn & ¤ ¤ ( § ¨ § Pn n 2n ¤ P Wn 1 The comparison of the exact calculation and the approximation are given in the following table. ¨ § P Wn n n 1 n 4 n 16 n 64 exact 0 3679 0 1954 0 0992 0 0498 approximate 0 3829 0 1974 0 0995 0 0498       4   ...
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