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Statistics 510: Notes 19 Reading: Sections 6.3, 6.4, 6.5, 6.7 I. Sums of Independent Random Variables (Chapter 6.3) It is often important to be able to calculate the distribution of X Y from the distribution of X and Y when X and Y are independent. At the end of last class, we derived the results: FX Y (a) P{ X Y a} FX (a y) fY ( y)dy and d f X Y ( a ) FX (a y ) fY ( y ) dy da f X (a y ) fY ( y )dy Example 1: Sum of two independent uniform random variables. If X and Y are two independent random variables, both uniformly distributed on (0,1), calculate the pdf of X Y . II. Conditional Distributions (Chapters 6.4-6.5) (1) The Discrete Case: If X and Y are jointly distributed discrete random variables, the conditional probability that X xi given that Y y j is, if pY ( y j ) 0 , then the conditional probability mass function of X|Y is p X |Y ( xi | y j ) P( X xi | Y y j ) P( X xi , Y y j ) P(Y y j ) p X ,Y ( xi , y j ) pY ( y j ) This is just the conditional probability of the event X xi given that Y y j . If X and Y are independent random variables, then the conditional probability mass function is the same as the unconditional one. This follows because if X is independent of Y, then p X |Y ( x | y ) P( X x | Y y ) P ( X x, Y y ) P(Y y ) P( X x) P(Y y ) P(Y y ) P( X x) Example 3: In Notes 17, we considered the situation that a fair coin is tossed three times independently. Let X denote the number of heads on the first toss and Y denote the total number of heads. The joint pmf is given in the following table: y x 0 1 2 3 0 1/8 2/8 1/8 0 1 0 1/8 2/8 1/8 What is the conditional probability mass function of X given Y? Are X and Y independent? (2) Continuous Case If X and Y have a joint probability density function f ( x, y ) , then the conditional pdf of X, given that Y=y is defined for all values of y such that fY ( y) 0 , by f ( x, y) f X |Y ( x | y) X ,Y f ( y) . Y To motivate this definition, multiply the left-hand side by dx and the right hand side by (dxdy ) / dy to obtain f ( x, y )dxdy f X |Y ( x | y )dx X ,Y fY ( y )dy P{x X x dx, y Y y dy} P{ y Y y dy} P{x X x dx | y Y y dy} In other words, for small values of dx and dy , f X |Y ( x | y ) represents the conditional probability that X is between x and x dx given that Y is between y and y dy . The use of conditional densities allows us to define conditional probabilities of events associated with one random variable when we are given the value of a second random variable. That is, if X and Y are jointly continuous, then for any set A, P{X A | Y y} f X |Y ( x | y)dx . A In particular, by letting A (, a] , we can define the conditional cdf of X given that Y y by a FX |Y (a | y) P( X a | Y y) f X |Y ( x | y)dx . Note that we have been able to give workable expressions for conditional probabilities even though the event on which we are conditioning (namely the event Y y ) has probability zero. Example 4: Suppose X and Y are two independent random variables, both uniformly distributed on (0,1). Let T1 min{ X , Y }, T2 max{ X , Y } (these are called the order statistics of the sample – Section 6.6). What is the conditional distribution of T2 given that T1 t ? Are T1 and T2 independent? III. Joint Probability Distribution of Functions of Random Variables Let X1 and X 2 be jointly continuous random variables with joint pdf f X1 , X 2 . It is sometimes of interest to obtain the joint distribution of random variables Y1 and Y2 , which arise as functions of X1 and X 2 . Specifically, suppose that Y1 g1 ( X1 , X 2 ) and Y2 g2 ( X1 , X 2 ) for some functions g1 and g 2 . Assume that the functions g1 and g 2 satisfy the following conditions: 1. The equations y1 g1 ( x1 , x2 ) and y2 g2 ( x1 , x2 ) can be uniquely solved for x1 and x2 with solutions given by, say, x1 h1 ( y1 , y2 ), x2 h2 ( y1 , y2 ) . 2. The functions g1 and g 2 have continuous partial derivatives at all points ( x1 , x2 ) and are such that the following 2x2 determinant g1 g1 x1 x2 g g g g J ( x1 , x2 ) 1 2 1 2 0 g 2 g 2 x1 x2 x2 x1 x1 x2 at all points ( x1 , x2 ) . Under these two conditions, it can be shown that the random variables Y1 and Y2 are jointly continuous with joint density function given by fY1 ,Y2 ( y1 , y2 ) f X1 , X 2 ( x1 , x2 ) | J ( x1, x2 ) |1 (1.1) A proof of equation (1.1) proceeds along the following lines: P{Y1 y1 , Y2 y2 } f X1 , X 2 ( x1 , x2 )dx1dx2 ( x1 , x2 ): g1 ( x1 , x2 ) y1 g 2 ( x1 , x2 ) y2 The joint density function can now be obtained by differentiating the above equation with respect to y1 and y2 . That the result of this differentiation will be equal to the right hand side of equation (1.1) is an advanced calculus result. Example 5: Let ( X , Y ) denote a random point in the plane and assume that the rectangular coordinates X and Y are independent standard normal random variables. We are interested in the joint distribution of R, , the polar coordinate representation of the point. 2 2 1 ( R X Y , tan (Y / X ) )