jasoncondprob

jasoncondprob - Conditional densities, mass functions, and...

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Unformatted text preview: Conditional densities, mass functions, and expectations Jason Swanson April 22, 2007 1 Discrete random variables Suppose that X is a discrete random variable with range { x 1 ,x 2 ,x 3 ,... } , and that Y is also a discrete random variable with range { y 1 ,y 2 ,y 3 ,... } . Their joint probability mass function is p ( x i ,y j ) = P ( X = x i ,Y = y j ) . In order for p ( x i ,y j ) to be a valid joint probability mass function, it is enough that p ( x i ,y j ) 0 for all x i and y j , and that i j p ( x i ,y j ) = 1 . The marginal mass function of X is p X ( x i ) = P ( X = x i ) = j p ( x i ,y j ) , and the marginal mass function of Y is p Y ( y j ) = P ( Y = y j ) = i p ( x i ,y j ) . The conditional mass function of X given Y is p X | Y ( x i | y i ) = P ( X = x i | Y = y j ) = P ( X = x i ,Y = y j ) P ( Y = y j ) = p ( x i ,y j ) p Y ( y j ) . Similarly, the conditional mass function of Y given X is p Y | X ( y j ,x i ) = P ( Y = y j | X = x i ) = P ( Y = y j ,X = x i ) P ( X = x i ) = p ( x i ,y j ) p X ( x i ) . If X and Y are independent, then p X | Y ( x i | y j ) = p X ( x i ) and p Y | X ( y j | x i ) = p Y ( y j ). 1 2 Continuous random variables Suppose X and Y are jointly continuous random variables. This means they have a joint density f ( x,y ), so that P (( X,Y ) A ) = A f ( x,y ) dxdy for all 1 subsets A of the xy-plane. In order for f ( x,y ) to be a valid joint density function, it is enough that f ( x,y ) 0 for all x and y , and that f ( x,y ) dxdy = 1 . The marginal densities of X and Y are f X ( x ) = f ( x,y ) dy, f Y ( y ) = f ( x,y ) dx, and the marginal distributions are F X ( a ) = P ( X a ) = a f X ( x ) dx, F Y ( b ) = P ( Y b ) = b f Y ( y ) dy. Sometimes we are given that Y = y , where y is some constant, and we would like to know the conditional probability that X a . However, P ( X a | Y = y ) = P ( X a,Y = y ) P ( Y = y ) . The reason these are not equal is because the right-hand side is undefined. It is undefined because P ( Y = y ) = 0. We must therefore do something else if we want to define P ( X a | Y = y ). To accomplish this, we define the conditional density of X given Y as f X | Y ( x | y ) = f ( x,y ) f Y ( y ) . This is defined whenever f Y ( y ) > 0. If f Y ( y ) = 0, then the conditional density f X | Y ( x | y ) is undefined. We then define P ( X A | Y = y ) = A f X | Y ( x | y ) dx for all subsets A of the real line. The conditional distribution of X given Y is then F X | Y ( a | y ) = P ( X a | Y = y ) = a f X | Y ( x | y ) dx. 1 As mentioned in the text, this is not really true for all subsets. It is, however, true for all subsets that we care about in this course. We will therefore ignore this technical detail in this and all similar situations....
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jasoncondprob - Conditional densities, mass functions, and...

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