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Continuous Random Variables: Joint PDFs, Conditioning, Expectation and Independence Berlin Chen Department of Computer Science & Information Engineering National Taiwan Normal University Reference: - D. P. Bertsekas, J. N. Tsitsiklis, Introduction to Probability , Sections 3.4-3.6 Multiple Continuous Random Variables (1/2) • Two continuous random variables X and Y associated with a common experiment are jointly continuous and can be described in terms of a joint PDF f X ,Y satisfying P X , Y B – f X ,Y f X ,Y x , y dxdy x , y B is a nonnegative function – Normalization Probability f X ,Y x , y dxdy 1 • Similarly, f X ,Y a , c can be viewed as the “probability per unit area” in the vicinity of a, c P a X a , c Y c aa cc f X ,Y x , y dxdy f X ,Y a , c 2 – Where is a small positive number Probability-Berlin Chen 2 Multiple Continuous Random Variables (2/2) • Marginal Probability P X A P X A and Y , X A f X ,Y x , y dydx – We have already defined that P X A • X A f X x dx We thus have the marginal PDF f X x f X ,Y x , y dy Similarly f Y y f X ,Y x , y dx Probability-Berlin Chen 3 An Illustrative Example • Example 3.10. Two-Dimensional Uniform PDF. We are told that the joint PDF of the random variables X and Y is a constant c on an area S and is zero outside. Find the value of c and the marginal PDFs of X and Y . The correspond ing uniform joint PDF on an area S is defined to be (cf. Example 3.9) 1 , f X ,Y x,y Size of area S 0, f X ,Y x,y if x,y S otherwise 1 for x,y S 4 for 1 x 2 for 1 y 2 f X x 14 f X,Y x,y dy 1 3 14 dy 4 4 for 2 x 3 f Y y 12 f X,Y x,y dx 1 1 12 dx 4 4 for 2 y 3 f X x 23 f X,Y x,y dy 1 1 23 dy 4 4 f Y y 13 f X,Y x,y dx 1 1 13 dx 4 2 for 3 y 4 f Y y 12 f X,Y x,y dx 1 1 12 dx 4 4 Probability-Berlin Chen 4 Joint CDFs • If X and Y are two (either continuous or discrete) random variables associated with the same experiment , their joint cumulative distribution function (Joint CDF) is defined by F X ,Y x , y P X x , Y y – If X and Y further have a joint PDF f X ,Y ( X and Y are continuous random variables) , then F X ,Y x , y x y f X ,Y s , t dsdt And f X ,Y x , y If F X ,Y x , y 2 xy F X ,Y can be differentiated at the point x, y Probability-Berlin Chen 5 An Illustrative Example • Example 3.12. Verify that if X and Y are described by a uniform PDF on the unit square, then the joint CDF is given by F X ,Y x , y P X x , Y y xy , for 0 x,y 1 Y 2 FX ,Y x, y xy 0,1 1,1 0,0 1,0 X 1 f X ,Y x, y , for all x, y in the unit square Probability-Berlin Chen 6 Expectation of a Function of Random Variables • If X and Y are jointly continuous random variables, and g is some function, then Z g X , Y is also a random variable (can be continuous or discrete) – The expectation of Z can be calculated by E Z E g X , Y g x , y f X ,Y x , y dxdy – If Z is a linear function of X and Y , e.g., Z aX bY , then E Z E aX bY a E X b E Y • Where a and b are scalars We will see in Section 4.1 methods for computing the PDF of (if it has one). Z Probability-Berlin Chen 7 More than Two Random Variables • The joint PDF of three random variables X , Y and Z is defined in analogy with the case of two random variables P X , Y , Z B f X ,Y , Z x , y , z dxdydz X ,Y , Z B – The corresponding marginal probabilities f X ,Y x , y f X ,Y , Z x , y , z dz f X x f X ,Y , Z x , y , z dydz • The expected value rule takes the form E g X , Y , Z g x , y , z f X ,Y , Z x , y , z dxdy dz – If g is linear (of the form aX bY cZ ), then E aX bY cZ a E X b E Y c E Z Probability-Berlin Chen 8 Conditioning PDF Given an Event (1/3) • The conditional PDF of a continuous random variable X , given an event A – If A cannot be described in terms of X , the conditional PDF is defined as a nonnegative function f X A x satisfying P X B A B f X A x dx • Normalization property fX A x dx 1 Probability-Berlin Chen 9 Conditioning PDF Given an Event (2/3) – If A can be described in terms of X ( A is a subset of the real line with P X A 0 ), the conditional PDF is defined as a nonnegative function f X A x satisfying fX f X x , A x P X A 0, if x A otherwise • The conditional PDF is zero outside the conditioning event and for any subset P X B X fX B P X A remains the same shape as f X except that it is scaled along the vertical axis B, X A P X A A B f X x dx P X A A B f X A x dx – Normalization Property f X A x dx A f X A A x dx 1 Probability-Berlin Chen 10 Conditioning PDF Given an Event (3/3) • If A1 , A2 , , An are disjoint events with P Ai 0 for each i , that form a partition of the sample space, then n f X x P A i f X i 1 Ai x – Verification of the above total probability theorem think of as an event , X x B P X x P Ai P X x Ai n i 1 x n f X t dt P Ai x f X i 1 Ai and use the total probability theorem from Chapter 1 t dt Taking the derivative of both sides with respect to x n f X x P Ai f X i 1 Ai x Probability-Berlin Chen 11 Illustrative Examples (1/2) • Example 3.13. The exponential random variable is memoryless. – The time T until a new light bulb burns out is exponential distribution. John turns the light on, leave the room, and when he returns, t time units later, find that the light bulb is still on, which corresponds to the event A={T>t} – Let X be the additional time until the light bulb burns out. What is the conditional PDF of X given A ? X T t , A T t T is exponential e t , t 0 f T t otherwise 0, PT t e t The conditional CDF of X given A is defined by The conditiona l PDF of X given P X x A PT t x T t (where x 0) PT t x T t PT t x PT t e t x e t e x PT t x and T t PT t the event A is also exponentia l with parameter . Probability-Berlin Chen 12 Illustrative Examples (2/2) • Example 3.14. The metro train arrives at the station near your home every quarter hour starting at 6:00 AM. You walk into the station every morning between 7:10 and 7:30 AM, with the time in this interval being a uniform random variable. What is the PDF of the time you have to wait for the first train to arrive? - The arrival time, denoted by X , is a uniform random variable over the interval 7 : 10 to 7 : 30 - Let random varible Y model the waiting time - Let A be a event A 7 : 10 X 7 : 15 (You board the 7 : 15 train) 1/20 - Let B be a event B 7 : 15 X 7 : 30 (You board the 7 : 30 train) - Let Y be uniform conditioned on A - Let Y be uniform conditioned on B For 0 y 5, PY y Total Probability theorem: PY y P A PY A y P B PY B y 1 1 3 1 1 4 5 4 15 10 1 3 1 1 For 5 y 15, PY y 0 4 4 15 20 Probability-Berlin Chen 13 Conditioning one Random Variable on Another • Two continuous random variables X and Y have a joint PDF. For any y with fY y 0 , the conditional PDF of X given that Y y is defined by f X Y x y f X ,Y x , y fY y – Normalization Property f X Y x y dx 1 • The marginal, joint and conditional PDFs are related to each other by the following formulas f X ,Y x , y f Y y f X Y x y , f X x f X ,Y x , y dy . marginalization Probability-Berlin Chen 14 Illustrative Examples (1/2) • Notice that the conditional PDF f X Y x y has the same shape as the joint PDF f X ,Y x , y , because the normalizing factor f Y y does not depend on x f X Y x 3 . 5 f X Y x 3 . 5 f X ,Y x ,3 .5 f Y 3 .5 f X ,Y x , 2 .5 f Y 2 .5 f X Y x 3 .5 f X ,Y x ,1 .5 f Y 1 .5 1/ 4 1 1/ 4 1/ 4 1/ 2 1/ 2 1/ 4 1 1/ 4 cf. example 3.13 Figure 3.16: Visualization of the conditional PDF f X Y x y . Let X , Y have a joint PDF which is uniform on the set S . For each fixed y , we consider the joint PDF along the slice Y y and normalize it so that it integrates to 1 Probability-Berlin Chen 15 Illustrative Examples (2/2) • Example 3.15. Circular Uniform PDF. Ben throws a dart at a circular target of radius r . We assume that he always hits the target, and that all points of impact x , y are equally likely, so that the joint PDF f X ,Y x, y of the random variables x and y is uniform – What is the marginal PDF f Y y 1 , if x , y is in the circle f X ,Y x , y area of the circle 0, otherwise 1 2 2 2 2, x y r πr 0, otherwise fY y f X,Y x,y dx x 2 y 2 r 2 1 πr 2 2 2 x 2 y 2 r 2 1dx 1 πr 2 r y , if y r 2 fX 1 πr 2 Y x y f X,Y x,y fY y dx r2 y2 r2 y 1dx 2 2 πr (Notice here that PDF f Y y is not uniform) 1 πr 2 2 πr 2 r2 y2 1 2 r 2 y For each value y , f X 2 Y , x y if x 2 y2 r2 is uniform Probability-Berlin Chen 16 Conditional Expectation Given an Event • The conditional expectation of a continuous random variable X , given an event A ( P A 0 ), is defined by E X A xf X A x dx – The conditional expectation of a function form Eg X A g x f X – Total Expectation Theorem n EX P Ai E X Ai and i 1 A g X also has the x dx Eg X P Ai Eg X Ai n i 1 A1 , A2 , , An • Where are disjoint events with P Ai 0 for each i , that form a partition of the sample space Probability-Berlin Chen 17 An Illustrative Example • Example 3.17. Mean and Variance of a Piecewise Constant PDF. Suppose that the random variable X has the piecewise constant PDF if 0 x 1, 1 / 3, f X x 2 / 3, 0, if 1 x 2 , otherwise. Define event A1 X lies in the first interval [0,1] event A2 X lies in the second interval [1,2] P A1 01 1 / 3 dx 1 / 3, P A2 12 2 / 3 dx 2 / 3 fX f X x 1, 0 x 1 P X A x 1 A1 0, otherwise Recall that the mean and second moment of a uniform random variable over an interval [ a, b ] is a b / 2 and a 2 ab b 2 / 3 E X A 3 / 2 , E X E X A1 1 / 2 , E X 2 A1 1 / 3 2 2 A2 7 / 3 fX f X x 1, 1 x 2 P X A x 2 A2 0, otherwise E X P A1 E X A1 P A2 E X A2 1 / 3 1 / 2 2 / 3 3 / 2 7 / 6 P A E X E X 2 1 2 A1 P A2 E X 2 A2 1 / 3 1 / 3 2 / 3 7 / 3 15 / 9 var X 15 / 9 7 / 6 2 11 / 36 Probability-Berlin Chen 18 Conditional Expectation Given a Random Variable • The properties of unconditional expectation carry though, with the obvious modifications, to conditional expectation E X Y y xf X Y x y dx E g X Y y g x f X Y x y dx E g X , Y Y y g x , y f X Y x y dx Probability-Berlin Chen 19 Total Probability/Expectation Theorems • Total Probability Theorem – For any event A and a continuous random variable Y P A P A Y y f Y y dy • Total Expectation Theorem – For any continuous random variables X and Y E X E X Y y f Y y dy E g X E g X Y y f Y y dy E g X , Y E g X , Y Y y f Y y dy Probability-Berlin Chen 20 Independence • Two continuous random variables X and Y are independent if f X ,Y x , y f X x f Y y , for all x,y – Since that f X ,Y x , y f Y y f X Y x y f X x f Y X y x • We therefore have f X Y x y f X x , for all x and all y with fY y 0 • Or fY X y x fY y , for all y and all x with f X x 0 Probability-Berlin Chen 21 More Factors about Independence (1/2) • If two continuous random variables X and Y are independent, then – Any two events of the forms X Aand Y B are independent P X A , Y B x A y B f X ,Y x , y dydx x A y B f X x f Y y dydx x A f X x dx y B f Y y dy P X A P Y B – It also implies that F X ,Y x , y P X x , Y y P X x P Y y F X x FY x – The converse statement is also true (See the end-of-chapter problem 28) Probability-Berlin Chen 22 More Factors about Independence (2/2) • If two continuous random variables X and Y are independent, then – – E XY E X E Y var X Y var X var Y – The random variables g X and hY are independent for any functions g and h • Therefore, E g X h Y E g X E h Y Probability-Berlin Chen 23 Recall: the Discrete Bayes’ Rule • Let A1, A2 ,, An be disjoint events that form a partition of the sample space, and assume that P Ai 0, for all i . Then, for any event B such that P B 0 we have P Ai B P Ai P B Ai P B P Ai P B Ai Multiplication rule Total probability theorem nk 1 P Ak P B Ak P Ai P B Ai P A1 P B A1 P An P B An Probability-Berlin Chen 24 Inference and the Continuous Bayes’ Rule • As we have a model of an underlying but unobserved phenomenon, represented by a random variable X with PDF f X , and we make a noisy measurement Y , which is modeled in terms of a conditional PDF f Y X . Once the experimental value of Y is measured, what information does this provide on the unknown value of X ? X f X x Y Measurement fY f X Y x y X y x Inference f X Y x y y x fY y f X t f Y X y t dt f X ,Y x , y f X x fY X Note that f X fY X f X ,Y f Y f X Y Probability-Berlin Chen 25 Inference and the Continuous Bayes’ Rule (2/2) Inference about a Discrete Random Variable • If the unobserved phenomenon is inherently discrete – Let N is a discrete random variable of the form N n that represents the different discrete probabilities for the unobserved phenomenon of interest, and p N be the PMF of N P N n Y y P N n y Y y P N n P y Y y N n P y Y y p N n f Y N y n f Y y y n i f y i p N n f Y p i N N Total probability theorem Y N Probability-Berlin Chen 26 Illustrative Examples (1/2) • Example 3.19. A lightbulb produced by the General Illumination Company is known to have an exponentially distributed lifetime Y . However, the company has been experiencing quality control problems. On any given day, the parameter of the PDF of Y is actually a random variable, uniformly distributed in the interval 1, 3 / 2 . – If we test a lightbulb and record its lifetime ( Y y ), what can we say about the underlying parameter ? f Y y e y , y 0, 0 Conditioned on , Y has a exponential distribution with parameter 2, for 1 3 / 2 f 0, otherwise f Y y f f Y y 3/ 2 1 f t f Y y t dt 2 e y 3/ 2 ty 1 2te dt , for 1 λ 3 / 2 Probability-Berlin Chen 27 Illustrative Examples (2/2) • Example 3.20. Signal Detection. A binary signal S is transmitted, and we are given that P S 1 p and P S 1 1 p . – The received signal is Y S N , where N is a normal noise with zero mean and unit variance , independent of S . S 1 , as a function of the observed value – What is the probability that y of Y ? fY S y s Conditioned on S 2 1 e y s / 2 , for s 1 and -1, and - y 2 s,Y P S 1 Y y has a normal distribution with mean p S 1 f Y S y 1 fY y p S 1 f Y p S 1 f Y S S s and unit variance y 1 y 1 p S 1 fY S y 1 2 1 e y 1 / 2 2 2 2 1 1 e y 1 / 2 1 p e y 1 / 2 2 2 p p e y 2 1 / 2 e y y 2 1 / 2 pe e pe y 1 p e y y 2 1 / 2 pe y pe y 1 p e y Probability-Berlin Chen 28 Inference Based on a Discrete Random Variable • The earlier formula expressing P A Y y in terms of f Y A y can be turned around to yield fY A y P A fY A y P A f Y f Y y P A Y y P A ? f Y y P A Y y f Y t P A Y t dt f Y y P A Y y y dy A f Y y P A Y y dy P A f Y y P A Y y dy ( normalizat ion property : f Y A y dy 1) Probability-Berlin Chen 29 Recitation • SECTION 3.4 Joint PDFs of Multiple Random Variables – Problems 15, 16 • SECTION 3.5 Conditioning – Problems 18, 20, 23, 24 • SECTION 3.6 The Continuous Bayes’ Rule – Problems 34, 35 Probability-Berlin Chen 30

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