hw6_soln

# hw6_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,3.5.3 3.6.3 3.6.7 3.7.2 and 3.7.3 n1 1 1n p∑ © ¥ ¦ k p 1 p n1 p ¤ n ¤ 1 ¤ ¨ ¥ ¤ ¡ § ¥ k1 nk ¨ ∑ PN K ¥ ¡ n ¤ PN n , the marginal PMF of N satisﬁes 1 © 12 ¢¢¢ £££¡ Problem 3.5.3 First we observe that for n ¦ Thus, the event B has probability  ¥ p ¤ ¨ ¤ 1 1 p ¨  p 9p 1 2 ¥  ¥ ¨ ¤ ¥ ¤   n 10 1  ££ n ¨ ∑ PN p ¥ ∞ PB 9 ¦ From Theorem 3.11, PN K n k PB "! § n n 10 11 otherwise ;k ¢¢¢ £££¡ ¡ \$ n 10 p 1 ©¥ p ¡ ¢¢¢ %£££¡ ¨ ¥ ¤ ¡ § ¤ 1 0 ¡ 0 nk B otherwise # PN K B n k n ¤ ¤ The conditional PMF PN B n b could be found directly from PN n using Theorem 2.19. However, we can also ﬁnd it just by summing the conditional joint PMF. n 10 p ©¥ p n 10 11 otherwise ¨ ¤ 1 0 ¡ ¥ ¡ ¤ §  ¥ k1 nk ¢¢¢ £££¡ ¥& ∑ PN K B ¥ n ¤ PN B n ¦ ¤  From the conditional PMF PN B n , we can calculate directly the conditional moments of N given B. Instead, however, we observe that given B, N N 9 has a geometric PMF with mean 1 p. That is, for n 1 2 , \$ ¥ ' ¨  ¢¢¢ £££¡ ¤ PN B n 9  9B ¥ &   n ¤ PN 1 p ¨  ¥ ¤ n1 p 9 and we can calculate the conditional expectations 1 ¨ Var N B 1p 9 p p2 \$ £¥ &' ' & 9B 9 ¤ &' & ' Var N ENB  9B &  Var N B EN \$  0' ENB   )( N n ¥ Hence, given B, N B © ¡ PN &  Note that further along in the problem we will need E N 2 B which we now calculate. & ¤  1 &    1 ENB 81  Var N B 2 17 2 p p ¥ 2 & E N2 B 2 & ¤ For the conditional moments of K , we work directly with the conditional PMF PN K B n k . § 1 10 p n pn n n 10 ∑k k1 ¦ ©¥ ¦ ¨ ¤  &  ¦ ¦ ¤ 1 2, \$ £¥ nn ∑ n 10 k 1 ∞ 10 p ¤ ¥ ¨ ¤  2  & n1 p n1 © 1 p 1 EN 2 1 1 2p 1B  n  ∑  & ¦ ∞ EKB 5   1k pn n ¨ Since ∑n k 1 ©¥ ∑ ∑k  ¦ We now can calculate the conditional expectation of the sum. 1 2p   ¤ 10 p 9 \$ 1p ¤ EKB 5  4¥  & \$ ENB 3 2p   3 & KB 14  &   EN  The conditional second moment of K is p 1 EN 6  n 10 ¤ © ¥ p 1 2N  ¨ ¤¥ 1 ¥ 1 ¤ ¤  ¥ ¤  ¤  1 2n 6 ¥  ¤ ¦ & n 10 1 6, we obtain \$ £¥ n k1  & ¦ ¦ ∑ n 10 ∑ k2 1B  5¥ & ∞ 1 2n 10 p n pn n ¦ ¨ nn 1 ¦ ©¥ 2 ∞ ∑ n 10 k 1 1k E K2 B pn n ¤ Using the identity ∑n k 1 ¨ n ∑ ∑ k2 ©¥ ∞ E K2 B ¦ Applying the values of E N B and E N 2 B found above, we ﬁnd that 1 6 2 3 p2  ENB 2 37 6p  &  E N2 B 3 2 3 31  & & E K2 B &   & Thus, we can calculate the conditional variance of K .  ¥ 2 & 10 p 5 12 p2 ¤ pn n 2 ¨ EKB 7 6p  ¨ E K2 B Var K B 6 2 3  & &  To ﬁnd the conditional correlation of N and K , n 10 ¨ ∑ n 1 p ¥ n1 p∑k k1 ¦ ¦ & n 10 k 1 ∞  ¦ ¦ 1 2, \$ £¥ 2 p 1 EN N 2  n 10 ¤ © ¥ p  ¨ 1 1B &  5¥ ¤¥  2 1 1 p2  n 10 nn 9 p  ∑ ¤  ¤ ∞ ¦ E NK B ¤ nn ¨ &  1k ©¥ Since ∑n k 1 ∑ nk ¤ ∑ n © ∞ E NK B 45 ¡ n ¥ ∞ EKB ¦ Problem 3.6.3 (a) First we observe that A takes on the values SA 1 1 while B takes on values from SB 0 1 . To construct a table describing PA B a b we build a table for all possible values of pairs A B . The general form of the entries is 8 ¨ 7 6 ¡ ¥ ¤ ¡ § ¥ ¥ b1 PB A 1 1 PA 1 PB A 1 1 PA 1 ¨ ¥ ¤ ¥ ¨ 2& ¤ ¤ ¥& ¤ ¥  ¤ ¨ ¥ ¤ ¥  ¤ b0 PB A 0 1 PA 1 PB A 0 1 PA 1 ¤ ¨ 2& ¥& ¤ ¤ ¥ ¤ ¡  8 ¡ 6 § ¨  ¡ ¤ PA B a b a 1 a1 Now we ﬁll in the entries using the conditional PMFs PB A b a and the marginal PMF PA a . This yields ¥&  b1 23 13 12 23 \$ ¥ \$ ¥ ¤ ¥ ¤ \$ ¤ ¥ \$ \$ ¥ ¤ ¥ ¤ \$ b0 13 13 12 23 ¥ \$ ¥ ¤ ¥ ¡ ¤ \$ ¥ ¤ PA B a b a 1 a1 ¨ ¤ § which simpliﬁes to \$ ¥ \$ \$ ¤ ¡ \$ § ¨ 1, then the conditional expectation of B is ¤ b1 PB A 1 1 ¥&    b0 ¤ ∑ bPB A 1 12 ¥& 1 EBA \$ (b) If A b 0b 1 19 29 13 13 PA B a b a 1 a1 &  ¦ ¤ (c) Before ﬁnding the conditional PMF PA B a 1 , we ﬁrst sum the columns of the joint PMF table to ﬁnd 0 1 49 b 59 b \$  ¥ \$ ¤ 1 is 25 a 35 a ¥ \$ 1 1 \$ ¤ PA B a 1 PB 1 ¡ § ¤ ¥ ¤ ¤ PA B a 1 ¨ The conditional PMF of A given B ¥& PB b ¥&  (d) Now that we have the conditional PMF PA B a 1 , calculating conditional expectations is easy. ¤ ¥& ¨  ¤ ¥& \$  2 \$ 1 35 35 15 1 ¤ \$ 25  ¥ © 9¦ © 9¦ § a2 PA B a 1 ¤ 11 ∑ 125 \$  §  & a aPA B a 1 11 ¥ 1 a \$   E A2 B ∑ 1 ¥& EAB & The conditional variance is then ¨ 1 ¤ ¥ 2 &  ¤ 3 EAB 15 \$ ¨ 1 ¥ E A2 B 2 & 1 24 25 \$  Var A B &  (e) To calculate the covariance, we need ¤ \$ ¥ ¨ ¤ 13 §  4¥ ¤ ¤ 11 1 3 \$ ¥  4¥ ¤ ¤ 10 13 19 ¥ \$ ¥  ¥ ¡ ¤ \$ ¤ ¥ ¥ § 11 29 \$ ¤ ab 59 ¥ ¤ \$ ¤ ¦ ¨ A4¥   ¦ ¤ § © @¦ ¤ 10 19 ¤ 1 1b 0 \$ © @¦ 1 159 ¥ 049 \$ b ∑ ∑ abPA B \$ ¥  a 123  4¥ ∑ bPB b0 E AB ¤ 1 EB 113 \$ aPA a 11 ¥ a \$ ∑ EA  ¨ The covariance is just 2 27 B ¥ ¤ \$ ¨  \$  ¨ 1   ¡  Problem 3.6.7 The key to solving this problem is to ﬁnd the joint PMF of M and N . Note that N the joint event M m N n has probability M. For n 8 © ¥ p nm1 p n22 p © ¥ ¨ p p1 ¨ ¨  ¡  ¤ ¤ 1 m1 ¥ p dv ©   ££ 1 d v dd ¤ P dd  ££ n © © mN © 6 PM m1 calls  n © ¡ m1 calls A complete expression for the joint PMF of M and N is 1; n ¨ n m m 12 otherwise ¡ ¢¢¢ D£££¡ ¡ ©¥ ¤ ¨ ¥ ¤ ¡ § , the marginal PMF of N satisﬁes n 11 ¨ ¤ p ¥ n22 ¤ © ¥ ¨ ¤ © ¥ m1 p ¨ 1 p ¥ ∑ n22 © n1 ¢¢¢ £££¡ ¡ ¤ PN n ¦ , the marginal PMF of M satisﬁes ¥ ¨ m1 © ¥ p m1  p 1 p ¨ 1 p ¥ ¤ ¤ ¦  ¨ 1 E p n22 m  ££A ¥ nm1 2 p © ¤ 1 ¤ ∑ ¨ ∞ ¥ ¡ PM m © 12 ¢¢¢ £££¡ Similarly, for m 1m  n 2 p2 ¡ 23 p p ¤ The complete expressions for the marginal PMF’s are ©¥ ¨ 4 n 2 p2 n 23 otherwise p ¡ © ¤ ¥ 11 m 12 otherwise ¢¢¢ £££¡ m 1p ¤ ¥ ¨ ¨ ¥ ¥ n 0 p ¡ ¤ ¤ ¤ PN n 1 0 ¢¢¢ £££¡ PM m p  For n 1 0 2 ¢¢¢ £££¡ PM N m n C ¤ 13 59 \$ 19 ¥ EAEB ¨ E AB \$ Cov A B m, Not surprisingly, if we view each voice call as a successful Bernoulli trial, M has a geometric PMF since it is the number of trials up to and including the ﬁrst success. Also, N has a Pascal PMF since it is the number of trials required to see 2 successes. The conditional PMF’s are now easy to ﬁnd. © ©¥ ¨ ¤ ¥ ¤ ¡ § ¤ ¥ ¥ ¤ &  The interpretation of the conditional PMF of N given M is that given M m, N has a geometric PMF with mean 1 p. The conditional PMF of M given N is m N where N  2 ' n m 1m otherwise ' n m 1p  p ¡ 1 0  PM N m n PM m ¢¢¢ £££¡ PN M n m \$ m1 n otherwise ¥ 1 ¡ ¢¢¢ D£££¡ ¨ ¤ 1n 0 ¨ ¥ \$ ¤ PM N m n PN n ¡ § ¤ ¥ ¤ PM N m n 1 ¥&  Given that call N n was the second voice call, the ﬁrst voice call is equally likely to occur in any of the previous n 1 calls. ¨ Problem 3.7.2 Using the following probability model \$ ¥ \$ ¤ PY k ¥ ¤ PX k 34 k 0 1 4 k 20 0 otherwise We can calculate the requested moments.  2 75 10  \$  2E X 5 ¨ 1 4 20 ¥ ¥   ¨    ¤ EX 2 5 5 ¤  EX   \$   3 \$    Y 34 0 \$  EX 1 4 20  Var X 340 EX  Since X and Y are independent, Theorem 3.14 yields ¤ 150 ¤ PX x PY y and ¤ ¤ 20 20 220 20 PX 20 PY 20  ¥ 1012 G ¥ ¤ 2 75 ¥ ¢ ¥ ¤ ¤ ¡  ¡ § § x 0 20 y 0 20 ¤ ¥ XY 2XY PX Y x y ¥ ¥  ∑∑ 2 Var X ¥   E XY 2XY  F Since X and Y are independent, PX Y x y Var Y  Var X  Y  Var X § ¦ § ¦ Problem 3.7.3 (a) Normally, checking independence requires the marginal PMFs. However, in this problem, the zeroes in the table of the joint PMF PX Y x y allows us to verify very quickly that X and Y are dependent. In particular, PX 1 1 4 and PY 1 14 48 but ¤ 1 PY 1 ¥ ¨ ¤ H § ¥ 5 PX ¥ ¤ ¥ \$ 0 ¥ ¡ 11 \$ ¥ ¤ ¡ ¤ ¨ ¨ ¤ PX Y § ¤ ¤ (b) To ﬁll in the tree diagram, we need the marginal PMF PX x and the conditional PMFs PY X y x . By summing the rows on the table for the joint PMF, we obtain ¥ x 4 2 4 ¤ \$ \$ \$ \$ ¤ ¤ ¤ ¥& 13 y 101 0 otherwise \$ %¥ ¡ ¤ ¡ ¨ \$ \$ ¥ \$ \$ \$ PX Y x y PX x to write ¥ ¨ \$ § §  ¨ \$ ¥& ¥& ¨  ¡ \$ \$ \$ ¨ I& ¤ ¤  PY X y 1 PY X y 0 ¡ ¥ 12 y 01 0 otherwise 1 ¡ 34 y 1 14 y 0 0 otherwise ¤ PX 1 1 1 ¥ 1y 0y 1 3 16 1 16 0 16 16 16 0 18 18 Now we use the conditional PMF deﬁnition PY X y x PY X y  y ¥& PX Y x y x 1 x0 x1  Now we can us these probabilities to label the tree. The generic solution and the speciﬁc solution with the exact values are 13 Y0 c aa Y1 Q Y0 Q Y1 Q ` bb ` a Q a a a c 12 c b 12 1 Q ` c b b b Q P c VV V V V P V R SQ Q Q c Q Q 6 c` Y1 X1 Y 13 13 14 Y0 X0 12 1 Y0 14 R SQ ` `` U Y ` RT X ` a ` Q UY TX aa a U Y T aX a UY TX bb b bb UY TX PY X 1 1 P PY X 0 1 1 `` R SQ PP U PRT b Q P P UT V X1 Q T VV U PX 1 Y1 cP PY X 1 0 Y0 R SQ PY X 0 0 1 X c X0 Y 14 Y Q U RY T X W WW WW U W R Y R T X PX 0 10 R SQ PY X Y0 PP 1 W PY X 0 W 1 34 1 W X Y cW 1 1 WW PX 1 R SQ PY X V V P V ...
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