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3. Several Random Variables 3.1 Two Random Variables 3.2 Conditional Probability--Revisited 3.3 Statistical Independence 3.4 Correlation between Random Variables Standardized (or zero mean normalized) random variables 3.5 Density Function of the Sum of Two Random Variables 3.6 Probability Density Function of a Function of Two Random Variables 3.7 The Characteristic Function Concepts Two Dimensional Random Variables Probability in Two Dimensions, Conditional Probability--Revisited Statistical Independence Two Dimensional Statistics, Correlation between Random Variables Density Function of the Linear Combination of Two Random Variables Multi-input Electrical Circuits Simulating Convolution Integrals Notes and figures are based on or taken from materials in the course textbook: Probabilistic Methods of Signal and System Analysis (3rd ed.) by George R. Cooper and Clare D. McGillem; Oxford Press, 1999. ISBN: 0-19-512354-9. B.J. Bazuin, Spring 2015 1 of 25 ECE 3800 Examples of Joint Density Functions Exercise 3-2.1 Two R.V., X and Y, have a joint pdf of the form f X ,Y x, y k x 2 y , for 0 x 1 and 0 y 1 Find everything …. k Coefficient 1 1 1 f x, y dx dy k x 2 y dx dy 0 0 1 1 x 1 1 k 2 y x dy k 2 y dy 2 2 0 0 0 1 2 1 1 y2 1 1 3 1 k y 2 k 2 k 2 0 2 2 2 2 k 2 3 Marginal density functions of X and Y 1 2 2 x2 f Y y x 2 y dx 2 y x 3 3 2 0 0 2 1 1 4 fY y 2 y y 3 2 3 3 1 1 2 2 y2 f X x x 2 y dy x y 2 3 3 2 0 0 2 f X x x 1 3 1 Joint Distribution of X and Y y x 2 FX ,Y x, y u 2 v du dv 3 0 0 2 x2 FX ,Y x, y 2 v x dv 3 2 0 y FX ,Y x, y 2 x2 y x y 2 , 3 2 for 0 x 1 and 0 y 1 Notes and figures are based on or taken from materials in the course textbook: Probabilistic Methods of Signal and System Analysis (3rd ed.) by George R. Cooper and Clare D. McGillem; Oxford Press, 1999. ISBN: 0-19-512354-9. B.J. Bazuin, Spring 2015 2 of 25 ECE 3800 The conditional Probability that X is greater than ½ given that Y=½. f X ,Y x, y f X |Y x | Y y fY y 2 x 2 y 2 x 2 y f X |Y x | Y y 3 1 4 1 4 y y 3 3 2 x 2 y 1 1 4 y dx 1 1 FX ,Y X | Y 2 2 1 2 1 1 FX ,Y X | Y 2 2 1 3 x 1 dx 1 2 2 1 2 2 x2 1 3 x 1 dx 3 2 x 1 2 1 2 1 2 1 1 1 2 3 5 2 7 7 1 FX ,Y X | Y 1 2 3 2 8 2 3 2 8 3 8 12 2 The conditional Probability that Y is less than or equal to ½ given that X=½. f X ,Y x, y fY|X y | X x f X x 2 x 2 y x 2 y fY|X y | X x 3 2 x 1 x 1 3 1 1 FY | X Y | X 2 2 x 2 y 1 x 1 dx 1 2 1 1 FY | X Y | X 2 2 1 1 2 0 1 2 y 2 dx 1 1 2 1 2 3 1 4 y dx 1 0 1 2 y2 1 1 0 3 1 4 y dx 3 y 4 2 0 2 1 1 1 1 1 1 1 FY | X Y | X 4 1 3 8 3 2 3 2 2 Notes and figures are based on or taken from materials in the course textbook: Probabilistic Methods of Signal and System Analysis (3rd ed.) by George R. Cooper and Clare D. McGillem; Oxford Press, 1999. ISBN: 0-19-512354-9. B.J. Bazuin, Spring 2015 3 of 25 ECE 3800 Exercise 3-3.1 f X ,Y x, y k exp x y 1, for 0 x and 1 y There is no a, unless the problem was supposed to be stated as … f X ,Y x, y k exp x y a , for 0 x and 1 y but then k and a are not necessarily computed separately as 1 and 1.! Overall, the correct pdf is f X ,Y x, y exp x y 1, for 0 x and 1 y f X ,Y x, y exp x exp y 1, for 0 x and 1 y f X x exp x , for 0 x f Y y exp y 1, for 1 y Correlation EX Y x y f x, y dx dy E X Y x y exp x y 1 dx dy 1 0 expax ax 1 a2 x expax dx exp0 0 1 dx E X Y exp 1 y exp y 2 1 1 E X Y exp 1 y exp y 1 dx 1 exp 1 1 1 exp 1 exp 1 2 2 E X Y exp 1 2 1 E X Y E X 1 E Y 1 1 1 1 1 2 Notes and figures are based on or taken from materials in the course textbook: Probabilistic Methods of Signal and System Analysis (3rd ed.) by George R. Cooper and Clare D. McGillem; Oxford Press, 1999. ISBN: 0-19-512354-9. B.J. Bazuin, Spring 2015 4 of 25 ECE 3800 Exercise 3-3.2 Assume X and Y independent f X x 0.5 exp x 1 , for x f Y y 0.5 exp y 1 , for y Find PrX Y 0 ? That is, the product of the random variables is positive. PrX Y 0 PrX 0 PrY 0 PrX 0 PrY 0 PrX Y 0 1 Fx 0 1 FY 0 Fx 0 FY 0 Find the distribution for X and Y based on the ranges defined for the absolute value FX x x 0.5 exp x 1 dx For for x 1 and for 1 x for x 1 FX x for 1 x x x 0.5 expx 1 dx FX x 0.5 0.5 exp1 x dx 1 exp1 x 1 1 FX x 0.5 0.5 1 exp1 x exp x 1 1 FX x 0.5 exp x 1 F X x 0 .5 x F X x 0 .5 0 .5 x FX 0 FY 0 0.5 exp 1 0.1839 PrX 0 PrY 0 1 Fx 0 1 FY 0 0.6660 PrX 0 PrY 0 Fx 0 FY 0 0.0338 PrX Y 0 1 Fx 0 1 FY 0 Fx 0 FY 0 0.6660 0.0338 0.6998 Notes and figures are based on or taken from materials in the course textbook: Probabilistic Methods of Signal and System Analysis (3rd ed.) by George R. Cooper and Clare D. McGillem; Oxford Press, 1999. ISBN: 0-19-512354-9. B.J. Bazuin, Spring 2015 5 of 25 ECE 3800 Exercise 3-4.1 – Again the text is wrong Two R.V. have means of 1 and variances of 1 and 4, respectively. Their correlation coefficient is 0.5. Find everything ….. EX 1 EY 1 X2 1 Y2 4 E X 2 X2 E X 1 1 2 E Y 2 Y2 E Y 4 1 5 2 2 EX Y X Y 1 X Y 2 EX Y X Y X Y E X Y 1 1 2 1 1 2 2 E X Y E X 2 2 X Y Y 2 E X 2 2 E X Y E Y 2 2 E X Y E X Y E X Y 2 2 2 5 11 2 X2 Y 2 2 2 X Y for EX Y EX EY 2 11 2 7 2 Alternately, X2 Y X2 Y2 2 X Y 1 2 X2 Y 1 4 2 1 2 7 Differences E X Y E X 2 2 E X Y E Y 2 2 4 5 3 2 X2 Y 3 0 2 2 for E X Y E X EY 0 or X2 Y X2 Y2 2 X Y 1 4 2 3 Notes and figures are based on or taken from materials in the course textbook: Probabilistic Methods of Signal and System Analysis (3rd ed.) by George R. Cooper and Clare D. McGillem; Oxford Press, 1999. ISBN: 0-19-512354-9. B.J. Bazuin, Spring 2015 6 of 25 ECE 3800 Density Function of the Sum of Two Random Variables The summing (differencing) of random variables creates a new random variable. Expect that the resulting probability distribution and density function are different (except when Gaussians are involved). Z X Y We want to find Pr Z z FZ z Pr X Y z For the experiment, let Z X Y Figure in two dimensions with the line z=x+y …. We know the joint density function, then any probability can be computed as y F x, y x f u, v du dv We know that x z y , which says that x z y and we can allow y as the limitation in x is incorporated in the first inequality. Therefore, FZ z FXY z y, z y f u, y du dy If we assume that X and Y are statistically independent, f x, y f X x f Y y : z y FZ z f Y y f X x dx dy Taking the derivative to find the density function results in … a convolution! d FZ z d FZ z dx f Z z f Y y f X z y 1 dy dz dx dz f Z z f y f z y dy Y X Notes and figures are based on or taken from materials in the course textbook: Probabilistic Methods of Signal and System Analysis (3rd ed.) by George R. Cooper and Clare D. McGillem; Oxford Press, 1999. ISBN: 0-19-512354-9. B.J. Bazuin, Spring 2015 7 of 25 ECE 3800 Interpreting the math: this is the convolution of the two density functions! The sum of random variables density function may be formed/derived as either of the two equations (equivalent convolution results) f Z z f X x fY z x dx f Z z f Y y f X z y dy Thus, there are two equivalent forms for the density function. f Z z f X x fY z x dx fY y f X z y dy Notes and figures are based on or taken from materials in the course textbook: Probabilistic Methods of Signal and System Analysis (3rd ed.) by George R. Cooper and Clare D. McGillem; Oxford Press, 1999. ISBN: 0-19-512354-9. B.J. Bazuin, Spring 2015 8 of 25 ECE 3800 Example #1: Z X Y Uniform densities in X and Y f X x 1, for 1 1 1 1 and f Y y 1, for y x 2 2 2 2 Forming the density in z f Z z f X x fY z x dx fY y f X z y dy f Z z 0.5 f X z y dy 0.5 Lines of x=z-y y=0.5 -0.5<x<0.5 y=-0.5 For 1 z 0 , f Z z z 0.5 z 0.5 0.5 0.5 0.5 0.5 f X z y dy 1 dy y 0.5 z 0.5 z 0.5 0.5) 1 z For 0 z 1 , f Z z f X z y dy 1 dy y z 0.5 0.5 z 0.5) 1 z 0.5 z 0 .5 z 0 .5 The convolution of two rectangles makes a triangle. 1 z, f Z z 1 z, 0, 1 z 0 0 z 1 else Notes and figures are based on or taken from materials in the course textbook: Probabilistic Methods of Signal and System Analysis (3rd ed.) by George R. Cooper and Clare D. McGillem; Oxford Press, 1999. ISBN: 0-19-512354-9. B.J. Bazuin, Spring 2015 9 of 25 ECE 3800 Application Example: This describes the error in the least significant bit of numbers that are rounded to integers! Initially a uniform distribution between +/- 0.5. After summing numbers that are rounded off, the density function of the error changes … the maximum possible error increases and the shape “distributes”. Notes and figures are based on or taken from materials in the course textbook: Probabilistic Methods of Signal and System Analysis (3rd ed.) by George R. Cooper and Clare D. McGillem; Oxford Press, 1999. ISBN: 0-19-512354-9. B.J. Bazuin, Spring 2015 10 of 25 ECE 3800 Example #2: Z X Y Uniform densities in X and exponential density in Y f X x 1, for 0 x 1 and f Y y exp y for 0 y Forming the density in z f Z z f X x fY z x dx fY y f X z y dy f Z z 1 fY z x dx 0 Lines of y=z-x x=1 0<y<∞ 0<x<1 For 0 z 1 , f Z z z z f Y z x dx exp z x dx exp z x 0 exp z z exp z 0 z 0 f Z z 1 exp z For 1 z , f Z z 1 1 fY z x dx exp z x dx exp z x 0 exp z 1 exp z 1 0 0 f Z z exp z e1 1 Therefore, 1 exp z f Z z exp z e 1 for 0 z 1 for 1 z Notes and figures are based on or taken from materials in the course textbook: Probabilistic Methods of Signal and System Analysis (3rd ed.) by George R. Cooper and Clare D. McGillem; Oxford Press, 1999. ISBN: 0-19-512354-9. B.J. Bazuin, Spring 2015 11 of 25 ECE 3800 Example #3: Z X Y Independent Gaussian densities in X and Y f X x x 2 X exp and f Y y 2 2 X 2 X 1 y 2 Y exp 2 2 Y 2 Y 1 Forming the density in z f Z z f X x fY z x dx fY y f X z y dy x 2 z x 2 1 X Y exp exp f Z z dx 2 2 2 X 2 Y 2 X 2 Y f Z z 1 x X 2 z x Y 2 exp dx 2 2 Y2 2 2 X Y 2 X 1 1 To solve, complete the square in x and remember that the Gaussian pdf integrates to 1.0. The remaining elements of the equation (terms not in x) transfer outside the integral. Once completed, the result should be: f Z z z X Y 2 exp 2 2 2 X2 Y2 2 X Y 1 This is a Gaussian that has a variance that is the square root of the sum of the variances of the individual Gaussians (larger – spreading out, but still Gaussian) and the mean is the sum of the individual means. This result (and the extension to multiple random variables) is a major reason why Gaussian Noise models are so widely used! This is also part of an empirical proof of the central limit theorem (Section 2-5). The MATALB example demonstrates convolved exponential pdfs becoming Gaussian-like. Notes and figures are based on or taken from materials in the course textbook: Probabilistic Methods of Signal and System Analysis (3rd ed.) by George R. Cooper and Clare D. McGillem; Oxford Press, 1999. ISBN: 0-19-512354-9. B.J. Bazuin, Spring 2015 12 of 25 ECE 3800 %% % Gauseeian Convolution % p. 140 Cooper and McGillam % clear close all; x=0:0.1:5; f=exp(-x); g=f; figure; plot(x,f); hold on; for ii = 1:10 g=0.1*conv(f,g); y=0.1*(0:length(g)-1); plot(y,g); end xlabel('y'); ylabel('g(y)') axis([0 20 0 1]); hold off; figure plot(y,g); xlabel('y'); ylabel('g(y)') 1 0.25 0.9 0.8 0.2 0.7 0.15 g(y) g(y) 0.6 0.5 0.4 0.1 0.3 0.2 0.05 0.1 0 0 2 4 6 8 10 y 12 14 16 18 20 0 0 10 20 30 y 40 50 Estimation theory … reminder xˆ EX xˆ EX | Y Notes and figures are based on or taken from materials in the course textbook: Probabilistic Methods of Signal and System Analysis (3rd ed.) by George R. Cooper and Clare D. McGillem; Oxford Press, 1999. ISBN: 0-19-512354-9. B.J. Bazuin, Spring 2015 13 of 25 ECE 3800 60 HW 3-2.1 A signal X has a Rayleigh density function and a mean value of 10 and is added to a noise, N, that is uniformly distributed with a mean value of zero and a variance of 12. X and N are statistically independent and can be observed only as Y=X+N. a) Find, sketch, and label the conditional probability density function f(x|y), as a function of x for y=0, y=6, and y=12. r2 r , for 0 r f R r 2 exp Rayleigh: 2 2 R E R E R 10 2 and R 2 E R 2 2 2 therefore 10 2 2 and 2 2 400 Zero mean uniform implies f N n Uniform: N2 2 A2 12 1 , 2 A for A n A 12 , then, 2 A 12 2 , and A 6 2 The density function in y based on convolution becomes f Y y f N y r f R r dr f N n f R y n dn But … since both X and N are bounded, the integration bounds and range of allowable values must be carefully considered. N: -6<n<6 Notes and figures are based on or taken from materials in the course textbook: Probabilistic Methods of Signal and System Analysis (3rd ed.) by George R. Cooper and Clare D. McGillem; Oxford Press, 1999. ISBN: 0-19-512354-9. B.J. Bazuin, Spring 2015 14 of 25 ECE 3800 f Y y f N y r f R r dr Selecting fY y 1 12 y 6 0 x2 exp 2 2 2 x y 6 x2 1 x 2 exp 2 12 y 6 2 x2 1 f Y y 1 exp 2 12 2 x2 1 1 exp 2 12 2 fY y y 6 2 1 1 exp 2 12 2 2 1 y 6 exp 12 2 2 dx, for 6 y 6 dx, for 6 y y 6 0 for 6 y 6 y 6 , y 6 for 6 y for 6 y 6 y 6 2 exp 2 2 , for 6 y Knowing the density of Y, we need to determine the conditional density, 1 , f Y , X y | x f N y x 12 0, for 6 y x 6 else Then f Y , X x | y f Y , X x | y f x fY y f N y x f X x fY y For y=0 … the allowed range of x is 0<=x<6, x2 exp 2 2 2 1 f Y , X x | y 0 12 1 y 6 2 1 exp 2 2 12 x , 0x6 Notes and figures are based on or taken from materials in the course textbook: Probabilistic Methods of Signal and System Analysis (3rd ed.) by George R. Cooper and Clare D. McGillem; Oxford Press, 1999. ISBN: 0-19-512354-9. B.J. Bazuin, Spring 2015 15 of 25 ECE 3800 x2 exp 2 2 2 , f Y , X x | y 0 6 2 1 exp 2 2 x 0 x6 For y=6 … the allowed range of x is 0<=x<12, x2 exp 2 2 2 1 f Y , X x | y 6 12 1 6 6 2 6 6 2 exp exp 2 2 2 12 2 x x2 exp 2 2 2 f Y , X x | y 6 , 12 2 1 exp 2 2 , 0 x 12 x 0 x 12 For y=12 … the allowed range of x is 6<=x<18, x2 exp 2 2 2 1 f Y , X x | y 6 12 1 12 6 2 12 6 2 exp exp 2 2 2 12 2 x x2 exp 2 2 2 f Y , X x | y 6 6 2 182 exp exp 2 2 2 2 , 6 x 18 x 6 x 18 , x2 exp 2 2 2 f Y , X x | y 6 6 2 182 6 2 exp 1 exp 2 2 2 2 x , 6 x 18 Notes and figures are based on or taken from materials in the course textbook: Probabilistic Methods of Signal and System Analysis (3rd ed.) by George R. Cooper and Clare D. McGillem; Oxford Press, 1999. ISBN: 0-19-512354-9. B.J. Bazuin, Spring 2015 16 of 25 ECE 3800 b) If an observation yields a value of y=12, what is the best estimate of the true value of X? xˆ EX | y 12 xˆ E X | y 12 x f x | y dx x2 exp 18 2 2 2 xˆ E X | y 12 x 6 2 182 6 2 6 exp 1 exp 2 2 2 2 x xˆ E X | y 12 x2 exp 6 2 2 2 18 x2 x2 exp 6 2 2 2 18 x2 62 dx 6 exp 2 2 62 6 exp 2 2 2 2 xˆ dx x2 2 exp 2 182 62 6 62 2 exp 1 exp 2 2 2 2 18 1 x2 dx x exp 2 2 xˆ E X | y 12 400 18 6 1 6 dx x2 exp 2 2 2 18 2 18 2 18 exp 2 2 x2 dx 18 6 2 Q Q 18 2 6 18 2 Q Q 18 exp 2 2 6 2 182 6 2 1 exp exp 2 2 2 2 127.324 and 200 10 2 7.9788 18 2 6 Q Q 2 10 10 2 10 2 288 36 exp 1 exp 400 400 36 324 6 exp 400 18 exp 400 xˆ 3.1093 20 0.2140 7.3892 10.9434 0.7537 0.8959 0.6752 Notes and figures are based on or taken from materials in the course textbook: Probabilistic Methods of Signal and System Analysis (3rd ed.) by George R. Cooper and Clare D. McGillem; Oxford Press, 1999. ISBN: 0-19-512354-9. B.J. Bazuin, Spring 2015 17 of 25 ECE 3800 Probability Density Function of a Function of Two Random Variables Functions of random variables create new random variable. As before, expect that the resulting probability distribution and density function are different. Assume that there is a function that combines two random variables and that the functions inverse exists. Z 1 X , Y and W 2 X , Y and the inverse X 1 Z ,W and Y 2 Z ,W The original pdf is f x, y with the derived pdf in the transform space of g z, w . Then it can be proven that: Pr z1 Z z 2 , w1 W w2 Pr x1 X x2 , y1 Y y 2 or equivalently w2 z 2 y2 x2 w1 z1 y1 x1 g z, w dz dw f x, y dx dy Empirically, since the density function must integrate to one for infinite bounds, the “transformed” portion of one density must have the same “volume” as the original density function. Using an advanced calculus theorem to perform a transformation of coordinates. x g z , w f 1 z, w, 2 z, w z y z w2 z 2 w2 z 2 w1 z1 w1 z1 x w f z , w, z, w J 1 2 y w g z, w dz dw f 1 z, w, 2 z, w J dz dw Notes and figures are based on or taken from materials in the course textbook: Probabilistic Methods of Signal and System Analysis (3rd ed.) by George R. Cooper and Clare D. McGillem; Oxford Press, 1999. ISBN: 0-19-512354-9. B.J. Bazuin, Spring 2015 18 of 25 ECE 3800 Example #4: Z X Y and let W X , an arbitrary selection Then, z x y and w x describes the forward transformation, and y z and x w describes the inverse transformation. w The Jocobian is x J z y z x 0 w 1 y w w 1 z 1 1 1 w w w2 Therefore, z 1 g z , w f w, w w and integrating for all w to find z, g z g z , w dw 1 z f w, dw w w As may be expected, the integral may need to be solved using numerical methods. Notes and figures are based on or taken from materials in the course textbook: Probabilistic Methods of Signal and System Analysis (3rd ed.) by George R. Cooper and Clare D. McGillem; Oxford Press, 1999. ISBN: 0-19-512354-9. B.J. Bazuin, Spring 2015 19 of 25 ECE 3800 Example #5: Compute the area where the nominal 10 cm square IC die size varies by independent uniform random variables allowing a +/- 0.5% tolerance. Then: f XY x, y 1 1 0 .1 0 .1 for 9.95 x, y 10.05 Derive the area density function ( Z X Y ). Then, z x y and w x describes the forward transformation, and y z and x w describes the inverse transformation, and w z 1 g z , w f w, w w Therefore, g z 1 z f w, dw w w z 10 1 w 10 w dw 100 rect rect 0 . 1 0 . 1 w z 10 1 w 10 w dw g z 100 rect rect w 0.1 0 .1 for 9.95 w 10.05 and 9.95 z 10.05 w or 9.95 w z 10.05 w z 9.95 z 1 100 dw 100 ln 2 w 9 . 95 9.95 g z 10.05 z 1 100 dw 100 ln w 10.05 2 z 10.05 for 9.95 2 z 9.95 10.05 for 9.95 10.05 z 10.05 2 Integrating to find the distribution Notes and figures are based on or taken from materials in the course textbook: Probabilistic Methods of Signal and System Analysis (3rd ed.) by George R. Cooper and Clare D. McGillem; Oxford Press, 1999. ISBN: 0-19-512354-9. B.J. Bazuin, Spring 2015 20 of 25 ECE 3800 z z 100 dz ln 2 9 . 95 9.95 2 G z z z 0.5 100 dz ln 10.05 2 9.9510.05 for 9.95 2 z 9.95 10.05 for 9.95 10.05 z 10.05 2 z z 9.95 2 for 9.95 2 z 9.95 10.05 100 z ln 2 9.95 z z G z z ln 10.05 2 for 9.95 10.05 z 10.05 2 0.5 100 9.95 10.05 1 ln 9.95 10.05 10.05 2 To determine the area within 0.5% bounds, Pr 100 0.995 z 100 1.005 Pr 100 0.995 z 100 1.005 G100.5 G 99.5 100.5 100.5 z ln 10.05 2 Pr 100 0.995 z 100 1.005 100 9.95 10.05 1 ln 9.95 10.05 10.05 2 99.5 99.5 9.95 2 100 z ln 9.95 2 Pr 100 0.995 z 100 1.005 0.8751 0.1251 0.7500 Notes and figures are based on or taken from materials in the course textbook: Probabilistic Methods of Signal and System Analysis (3rd ed.) by George R. Cooper and Clare D. McGillem; Oxford Press, 1999. ISBN: 0-19-512354-9. B.J. Bazuin, Spring 2015 21 of 25 ECE 3800 The Characteristic Function Whenever you see convolution and transformation, expect to see equations related to either the Fourier Transform or Laplace Transform! The characteristic function of a random variable X is defined as: u Eexp j u X or u f x exp j u x dx The inverse of the characteristic function is then defined as: 1 f x 2 u exp j u x du Application: The density function of two summed random variables, where X u f X x exp j u x dx and Y u fY y exp j u y dy Since we already know f Z z f X x fY z x dx fY y f X z y dy We can compute Z u X u Y u and solve for the density function as 1 f Z z 2 X u Y u exp j u z du Notes and figures are based on or taken from materials in the course textbook: Probabilistic Methods of Signal and System Analysis (3rd ed.) by George R. Cooper and Clare D. McGillem; Oxford Press, 1999. ISBN: 0-19-512354-9. B.J. Bazuin, Spring 2015 22 of 25 ECE 3800 Example #6 Repeat example #2 using the characteristic functions Uniform densities in X and exponential density in Y f X x 1, for 0 x 1 and f Y y exp y for 0 y 1 exp j u x exp j u 1 X u 1 exp j u x dx j u j u 0 1 0 exp j u y y 1 1 Y u exp y exp j u y dy j u 1 j u 1 1 j u 0 0 Then Z u X u Y u exp j u 1 1 exp j u 1 j u 1 j u j u 1 j u Z u Solving for the density function 1 f Z z 2 f Z z 1 2 1 2 X u Y u exp j u z du for 0 z exp j u 1 exp j u z du j u 1 j u exp j u z 1 1 du 2 j u 1 j u exp j u z du j u 1 j u Find it for extra credit … but it should result in 1 exp z f Z z exp z e 1 for 0 z 1 for 1 z Notes and figures are based on or taken from materials in the course textbook: Probabilistic Methods of Signal and System Analysis (3rd ed.) by George R. Cooper and Clare D. McGillem; Oxford Press, 1999. ISBN: 0-19-512354-9. B.J. Bazuin, Spring 2015 23 of 25 ECE 3800 Computing the first moment using the characteristic function: Differentiating the characteristic function by u: u d u du f x exp j u x dx f x j x exp j u x dx Evaluating the function at u=0 d u du u 0 f x j x exp0 dx j x f x dx j EX Computing other moments is performed similarly, where: d n u du du u 0 f x j x n exp j u x dx d n u n n j x n f x dx j n x n f x dx j n E X n Notes and figures are based on or taken from materials in the course textbook: Probabilistic Methods of Signal and System Analysis (3rd ed.) by George R. Cooper and Clare D. McGillem; Oxford Press, 1999. ISBN: 0-19-512354-9. B.J. Bazuin, Spring 2015 24 of 25 ECE 3800 The Joint Characteristic Function Whenever you see convolution and transformation, expect to see equations related to either the Fourier Transform or Laplace Transform! The characteristic function of a random variable X and Y is defined as: u, v Eexp j u X v Y or u, f x, y exp j u x v y dx dy The inverse of the characteristic function is then defined as: f x, y 2 2 1 XY u, v exp j u x v y du dv Note that 2 u, v E X Y j 2 uv u v 0 2 XY u, v uv u v 0 Notes and figures are based on or taken from materials in the course textbook: Probabilistic Methods of Signal and System Analysis (3rd ed.) by George R. Cooper and Clare D. McGillem; Oxford Press, 1999. ISBN: 0-19-512354-9. B.J. Bazuin, Spring 2015 25 of 25 ECE 3800