14.11.17 1 Stat 100/Sanchez-Intro Probability 1 Jointly Distributed Discrete Random Variables Two variable case This topic is all scrambled throughout Chapter 6 in Ross, and mixed with the discussion of the continuous case. Here you have it all in one place. Stat 100/Sanchez-Intro Probability 2 This lecture: discrete bivariate distributions • 1. Two random variables. Their joint distribution. Distributions that can be obtained from a joint distribution (marginal distributions, conditional distributions) • Independence of two random variables • 2. Functions of two random variables • 3. Covariance as a function of two random variables • 4. Relations between covariances • 5. The multinomial family of multivariate discrete distributions. Stat 100/Sanchez-Intro Probability 3 The Joint probability mass function. Let X and Y be discrete random variables. The joint probability mass function of X and Y is given by: € P x, y( )= P X = x,Y = y( )P x, y( )≥ 0P x, y( )= 1∑∑all x,y⎧ ⎨ ⎪ ⎩ ⎪ Defined for all real numbers x and y. Stat 100/Sanchez-Intro Probability 4 The marginal probability mass functions, marginal expectations and marginal variances The marginal probability mass functions of X and Y respectively, are given by € Marginal for X : P x( )= P x, y( )y∑E(X) = xP(x) Var(X) = x - E(X)( )2P(X )x∑x∑Marginal for Y : P y( )= P x, y( )x∑E(Y ) = yP(y) Var(Y) = y - E(Y)( )2P(Y )y∑y∑Stat 100/Sanchez-Intro Probability 5 The conditional probability mass functions, conditional expectations and variance € P X Y = y( )=P X and Y = y( )P Y = y( )E(X |Y = y) = xP(Xx∑|Y = y) Var(X | Y = y) = x - E(X)( )2x∑P(X |Y = y)P Y X = x( )=P X = x and Y( )P X = x( )E(Y | X = x) = yP(Yy∑| X = x) Var(Y | X = x) = y - E(Y)( )2x∑P(Y | X = x)There are as many conditional distributions of X given Y as values of Y. Similarly, there are as many conditional distributions of Y given X as values of X. Stat 100/Sanchez-Intro Probability 6 Example 1 • A firm that sells word processing systems keeps track of the number of customers who call on any one day and the number of orders placed on any one day. Let X denote the number of calls and Y the number of orders placed, and let P(X,Y) denote the joint probability mass function for X and Y. Records indicate that P(0,0)=0.04, P(1,0)=0.16, P(1,1)=0.1, P(2,0)=0.2, P(2,1)=0.3, P(2,2)=0.2. Thus, for any given day, the probability of, say, two calls and 1 order is 0.3.14.11.17 2 Stat 100/Sanchez-Intro Probability 7 Tabular representation of the Joint pmf, P(X=x, Y=y) Y 0 1 2 0 0.04 0.16 0.2 1 0 0.1 0.3 2 0 0 0.2 X=number of calls Y=# orders X Stat 100/Sanchez-Intro Probability 8 Finding joint joint probabilities of events Example: event (X<2 and Y>0) y 0 1 2 0 0.04 0.16 0.2 1 0 0.1 0.3 2 0 0 0.2 X=number of calls Y=# orders X P(X<2, Y>0)=P(X=0,Y=1)+P(X=0,Y=2)+P(X=1,Y=1)+P(X=1,Y=2)=0.1 Stat 100/Sanchez-Intro Probability 9 y 0 1 2 0 0.04 0.16 0.2 1 0 0.1 0.3 2 0 0 0.2 P(X) 0.04 0.26 0.7 X=number of calls Y=# orders X MARGINAL PMF FOR X,THE NUMBER OF CALLS- Column sum ∑===YYxXPxXP ),()(Stat 100/Sanchez-Intro Probability 10 Marginal Expected Value and Variance of X X P(X) 0 0.04 1 0.26 2 0.7 ( )[ ]5517.03044.0)(3044.066.1)7.0(4)26.0(1)()()()()(66.1)7.0(2)26.0(1)04.0(0)()(22222====−+=−=−===++===∑∑XSdXEXEXPXEXXVarXXPXExxσσµStat 100/Sanchez-Intro Probability 11 Marginal pmf for Y (row sum) y 0 1 2 P(Y) 0 0.04 0.16 0.2 0.4 1 0 0.1 0.3 0.4 2 0 0 0.2 0.2 X=number of calls Y=# orders X ∑===XyYXPyYP ),()(Stat 100/Sanchez-Intro Probability 12 Marginal Expected Value and Variance of Y Y P(Y) 0 0.4 1 0.4 2 0.2 € µ= E (Y) = YP(Y) = 0(0.4) +1(0.4) + 2(0.2) = 0.8y∑σ2= Var(Y) = Y − E (Y)( )y∑2P(Y ) = E(Y2) − E(Y)[ ]2= 1(0.4) + 4(0.2) − 0.82= 0.56σ= Sd(Y) = 0.56 = 0.748314.11.17 3 Stat 100/Sanchez-Intro Probability 13 Independence of the two random variables • The random variables X and Y are independent if P(X,Y)=P(X)P(Y) for all values of x and y. • In our calls-orders example, P(X=0,Y=0)=0.04, P(X=0)=0.04 and P(Y=0)=0.4. So P(X=0)P(Y=0)=0.016 So P(X=0,Y=0)≠P(X=0)P(Y=0) and therefore, X and Y are not independent. All we need is to find one pair for which the condition does not hold. However, if we had found that they were equal, we would have to check the condition for all X and Y. Stat 100/Sanchez-Intro Probability 14 Conditional probability mass functions in our example Stat 100/Sanchez-Intro Probability 15 Conditional expectations (examples) • E(X|Y=0)=(0)(0.1)+1(0.4)+2(0.5)=1.4 E(Y|X=2)=0(0.286)+(1)(0.429)+(2)(0.286)= 1.001 Stat 100/Sanchez-Intro Probability 16 2. Expected Values of functions of two random variables. When we encounter problems that involve more than one random variable, we often combine the variables into a single function. Suppose that the discrete random variables (X,Y) have a joint probability function given by P(x,y). If g(X,Y) is any real-valued function of (X,Y) then € E g X,Y( )[ ]= g(x, y)P(x, y)y∑x∑Stat 100/Sanchez-Intro Probability 17 Examples of functions of two random variables Let X, Y be two random variables P(X,Y) discrete. We can define the following functions of X and Y. € g x, y( )= xg x, y( )= yg x, y( )= x2€ g x, y( )= y2g x, y( )= xyg x, y( )= x + y€ g x, y( )= x − µx( )2g x, y( )= y − µy( )2g x, y( )= x − µx( )y − µy( )Expected values can be found with he general formula : € E g(X,Y )( )= g x, y( )P(x, y)y∑x∑Stat 100/Sanchez-Intro Probability 18 A very special function: g(x,y)=xy 0 1 20XY=(0)(0)=0 [0.04](1)(0)=0 [0.16](2)(0)=0 [0.2]1(0)(1) = 0 [0](1)(1)=1 [0.1](2)(1)=2 [0.3]2(0)(2) = 0 [0](1)(2)=2 [0](2)(2)=4 [0.2]XY14.11.17 4 Stat 100/Sanchez-Intro Probability 19 (cont.) XY P(XY) 0 [0.4]+[0.16]+[0.2]+[0] = 0.4 1 [0.1] = 0.1 2 [0.3]+[0] = 0.3 4 [0.2] = 02 € E XY( )= 0(0.4) +1(0.1) + 2(0.3) + 4(0.2) = 1.5Another special function: g(X,Y) = aX+ bY • E(aX+bY) = aE(X)+bE(Y) • Var(aX+bY) = a2Var(X) +b2 Var(Y) +2abCov(X,Y) • We need to introduce Cov(X,Y), another expectation. Stat 100/Sanchez-Intro Probability 20 Stat 100/Sanchez-Intro Probability 21 3. Covariance between two discrete random variables The covariance measures the strength and direction of the linear association between two random variables. € Cov
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