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**Unformatted text preview: **MIT OpenCourseWare http://ocw.mit.edu 14.30 Introduction to Statistical Methods in Economics Spring 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms . 1 14.30 Introduction to Statistical Methods in Economics Lecture Notes 10 Konrad Menzel March 12, 2009 Functions of 2 or more Random Variables Lets recap what we have already learned about joint distributions of 2 or more random variables, say X 1 ,X 2 ,...,X n if X 1 ,...,X n are discrete, their joint p.d.f. is given by f X 1 ,...,X n ( x 1 ,...,x n ) = P ( X 1 = x 1 ,...,X n = x n ) if X 1 ,...,X n are continuous, their joint p.d.f. is a positive function f X 1 ,...,X n ( x 1 ,...,x n ) such that P ( X 1 ,...,X n ) D = ... f X 1 ,...,X n ( x 1 ,...,x n ) dx 1 ...dx n D for any D R n . X 1 ,...,X n are independent if P ( X 1 A 1 ,...,X n A n ) = P ( X 1 A 1 ) ...P ( X n A n ) recall that this is equivalent to f X 1 ,...,X n ( x 1 ,...,x n ) = f X 1 ( x 1 ) ...f X n ( x n ) We are now going to look at how we can generalize from the univariate case discussed above to 2 or more dimensions. As for the single-dimensional case well again distinguish three cases: 1. underlying variables X 1 ,...,X n discrete 2. underlying variable X 1 ,...,X n continuous 3. X continuous and u ( X 1 ,...,X n ) is an n-dimensional one-to-one function 1 1.1 Discrete Case Suppose X 1 ,...,X n are discrete with joint p.d.f. f X 1 ,...,X n ( x 1 ,...,x n ), and Y 1 ,...,Y m are given by m functions Y 1 = u 1 ( X 1 ,...,X n ) . . . Y m = u m ( X 1 ,...,X n ) If we let A y := { ( x 1 ,...,x n ) : r 1 ( x 1 ,...,x n ) = y 1 ,...,u m ( x 1 ,...,x n ) = y m } then the joint p.d.f. of Y 1 ,...,Y m is given by f Y 1 ,...,Y m ( y 1 ,...,y m ) = f X 1 ,...,X n ( x 1 ,...,x n ) ( x 1 ,...,x n ) A y Example 1 (Sum of Binomial Random Variables) Suppose X B ( m,p ) and Y B ( n,p ) are independent binomial random variables with p.d.f.s f X ( k ) = m p k (1 p ) m k k f Y ( k ) = n p k (1 p ) n k k If we define Z = X + Y , what is the p.d.f. f Z ( z ) ? Since X is the number of successes in a sequence of m independent trials, and Y the number of successes in n trials, both with the same success probability, a first guess would be that then Z should just be the number of successes in m + n trials with success probability p , i.e. Z B ( m + n,p ) . This turns out to be true, but well first have to check this formally: P ( Z = z ) = P ( { X = 0 ,Y = z } or { X = 1 ,Y = z 1 } ... or { X = z,Y = 0 } ) z...

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