hw10_soln

hw10_soln - Probability and Stochastic Processes: A...

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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,5.5.3 5.7.5 5.8.2 5.8.3 5.9.1 5.9.2 5.7.6 and 5.9.3 Problem 5.5.3 The joint PDF of X and Y and the region of nonzero probability are © ¦ §   5x2 dy dx x 2 1 5x6 12 ¤  !¤  10 5x5 dx 12 1 0 ¤ ¤ ¥£ ¢ 1  ¤   0, the variance of X and the second moment are both x  ¤ EX 2 5x 2 10 2 dy dx 5x7 14 ¤ ¤ "  Var X x2 1 2 1 1 ¤ ¡ EX Since E X  ¨ ¨ (a) The first moment of X is 10 14 ¤ "  (b) The first and second moments of Y are x2 5 5x2 dy dx 2 14 0 2 5x 5 x2 y2 dy dx 2 26 0 1 EY    $¤ 1 ¤ ¤ $#  1 1 ¤ E Y2 y Therefore, Y has variance ¤  # EX ( '  ¤ " Y EY ¤ #  5x7 dx 14 1 0 5 14 Y is (¦ ¤ " ¢  1 57 0 0576 % 2 Cov X Y ¤ ( ( '  ¤ "  ¤ " ¢  ( ( Var Y 0 7719 %  Var X  ¤ # ¢  (e) By Theorem 5.10, the variance of X ( 0  ¤ ) Y 0 5x2 dy dx 2 Y is EX Var X 1 xy E XY . Thus, ¤  (d) The expected value of the sum X x2 1 0576 ¤ E XY § ' Cov X Y EXEY ¤ #    E XY  0, Cov X Y 2  ¤ #  (c) Since E X § Var Y 5 14 % &¤ 5 26  ¨¢ x2 x2 ¨ 1 y  5x2 2 1 x 10 0 otherwise fX Y x y ( Problem 5.7.5 Random variables X and Y have joint PDF 1 1 2 ¨ 3 ¦ 12 1xy 0 otherwise ¨ ¨ § ¡ fX Y x y ¤ ¥£ ¢ 5 64 1, the marginal PDF of Y is dx  1 ¤ ¤ ¡ £¢ ¤  ¥£ ∞ y 1 2 ¡ fX Y x y dx y 12 ( ∞ § ¡ fY y ¦£ y ¨ 1 ¨ (a) For The complete expression for the marginal PDF of Y is 1y1 otherwise § ¦£ 12 ¨ ( ¡ y 0 ¨ ¡ ¤ ¥£ fY y (b) The conditional PDF of X given Y is § 9 ¡ ¤ ¥£ 8 7 (c) Given Y y, the conditional PDF of X is uniform over value is E X Y y y 1 2. ¨ 0 1xy otherwise ¨ £ ¤ ¡ £¢ ¡ 1 1y fX Y x y fY y fX Y x y 1 y . Hence the conditional expected ¢ § ¦£ ¡ § ¤ ¤8 ¤ " X2 E X2 , we can use Theorem 5.8 to write Problem 5.8.2 ( '  E X2 0 Var X2 . Since X1 and X2 are independent, Var X1 Var X2 § ¤ B  ( '  ¤ "@£ £ X2 E X1  § ¡ ( ( ¡ §  ¤ " ¤ ) Var X1 X2 ¤ " § 1 2 Var X2 E 2 Var X  ¤ "@£  § X2 E X1 § X2 X2 ¤ # § §¤ &" §  (b) By Theorem 4.6(f), Var Theorem 5.15 says that ¤ " § Var X1 E X1 § A  X2 ¤ #   E X1 ¡ (a) Since E  Problem 5.8.3 Random variables X1 and X2 are independent and identically distributed with the following PDF: ¦ x2 0 x 2 0 otherwise ¨ ¨ ¤ ¥£ ¡ fX x ¡ (a) Since X1 and X2 are identically distributed they will share the same CDF FX x . ¨ F ¦ ¤ ED D 2 0 x 2 ¨ fX x dx ¨ 0 0 x 24 0 x 1 x 2 £ ¤  C£ ¡ FX x x (b) Since X1 and X2 are independent, we can say that ¡ 1 FX1 1 FX2 1 £ ¤ # 1 P X2 ¡ ¨  ¨ P X1 ¤ G¥£ 1 ¡  ¤ " ¨ ¢ ¨  FX 1 1 16 2 ¡ max X1 X2 ,  ¨ ¥£ 1 1 X2 ¨ 1 P X2 P X1 ¢ 1 1 ¨ ¤ # P max X1 X2  ¢  ¡ ¢ ¤ ¤ C£ FW 1 ¡ £ (c) For W 1 X2 ¤ @£  P X1 Since X1 and X2 are independent, ¡ FX 1 2 1 16 ¤ @£  ¤ G" ¨  P X1 ¦ ¨  ¤ ¥£ ¡ FW 1 (d) ¨ P X1 w X2 ¢ w ¨  ¤ " ¨ C£ ¡ P max X1 X2 w  ¢  ¤ C£ ¡ FW w Since X1 and X2 are independent, F 0 w w4 16 0 1 w 2 ¤ I£  ¡ FX w ¦ w ¨ ¤ H# w P X2 ¨ ¨  ¨ P X1 0 w 2 ¨  ¤ C£ ¡ FW w 2 Problem 5.9.1 ce S R TS R Q P P ¡ fX Y x y x2 8 y2 18 ¤ C£ ¢ The omission of any limits for the PDF indicates that it is defined over all x and y. We know that fX Y x y is in the form of the bivariate Gaussian distribution so we look to Definition 5.10 and attempt to find values for σY , σX , E X , E Y and ρ.  £¢  ¡ (a) First, we know that the constant is 1 ¤ 2πσX σY 1 ρ2 § c ¡ Because the exponent of fX Y x y doesn’t contain any cross terms we know that ρ must be zero, and we are left to solve the following for E X , E Y , σX , and σY : y § x2 8 EY σY  2 ¤  § From which we can conclude that EX EY ¤ #  U¤ ¤ Putting all the pieces together, we find that c U &¤ ¤ Q  σX σY ¤ (b) Since ρ 0 8 18 1 24π . 0, we also find that X and Y are independent. 3 2 ¤  EX σX  £¢ x y2 18 Problem 5.9.2 2x2 4xy 4y2 P T 9 ce S ¡ fX Y x y ¤ ¥£ ¢ Proceeding as in Problem 5.9.1 we attempt to find values for σY , σX , E X , E Y and ρ.  ¡ ¤ 81 ¡ ρ2 ¡ §  § EY § § The first two equations yield E X ρ2 y2 £ ¤ 81 £  ¤ EY 2 σY 2ρ σX σY ρ2 x2 41 § y 2 EX σX £ x  (a) First, we try to solve the following equations 0 ¤ #  ¤ Q  (b) To find the correlation coefficient ρ, we observe that σY ¦ ¤ 1 U¦ σX σY 2 ρ2 2. 12 ¤ ¡ 2, now we can solve for σX and σY . 81 ¦ ¤ § ¤ U¦ 1 U¦ £ ¤ (c) Since ρ 1 ¤ Using σX and σY in the third equation yields ρ 1 ¦ ρ2 ¡ 41 § 1 £ σX (d) From here we can solve for c. ¤ 2πσX σY 1 ρ2 § V¤ (e) X and Y are dependent because ρ 1 ¤ c 2 π 0. Problem 5.7.6 the joint PDF of X and Y is 0 x2 y2 otherwise r2 ¨ ( ¨ £ ¡¦ 1 πr2 0 ¡ fX Y x y ¤ C£ ¢ (a) The marginal PDF of X is 2 r2 x2 πr2 W ¤ 0 rxr otherwise § W X 0 1 dy πr2 ¨ 2 ¨ r2 x2 ¤ C£ ¡ fX x The conditional PDF of Y given X is y2 r2 x2 otherwise ¨ £ ¤ ¡ £¢ ¡ ¡¦ U § U § 0 ¤ ¤ ) ¤8 4 x U¢ EY X § § ¤ Y£ 8 ¡ 7 (b) Given X x, we observe that over the interval r2 x2 r2 Since the conditional PDF fY X y x is symmetric about y 0, x2 , Y has a uniform PDF.  £ x2 § 1 2 r2 0 fX Y x y fX x fY X y x £8 ¡ 7 ¤ Problem 5.9.3 that σ2 X 0 2 σY ¤ µY ¤ ¤ µX 1 ¤ From Theorem 5.18, the conditional expectation of Y given X is ¦ 2 x2 xy y2 3 P 5 e RS 9 ¤ C£ ¡ ¤ # 8  ¤ C 8  3π2 ¤ ( 1 ρX ¦ µX X 2. Hence ρ In the problem statement, we learn that E Y X joint PDF is ¡ U fX Y x y σY X σX ¡ ρ § µY ¤ ¥£ ˜ µY X EY X 1 2. From Definition 5.10, the ¤ ¥£ ¢ ...
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This note was uploaded on 02/11/2012 for the course EEE 352 taught by Professor Ferry during the Spring '08 term at ASU.

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