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Lecture 11

# Lecture 11 - 1 Functions of Random Variables We have...

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1 Functions of Random Variables We have encountered functions of random variables before when we were interested in finding the expectation of Y = g ( X ). Now, we are interested in finding the distribution of Y . 1.1 Discrete Case In general, P ( Y = y ) = P ( X ∈ { x : g ( x ) = y } ) (1) Example: Suppose that X can take values - 1 , 0 , 1 with probabilities 1 / 4 , 1 / 2 , 1 / 4, and consider the random variable Y = X 2 . The possible values of Y are 0 and 1. Clearly P ( Y = 0) = P ( X = 0) = 1 / 2 and P ( Y = 1) = P ( X ∈ {- 1 , 1 } ) = 1 / 2. The general formula simplifies significantly if g ( x ) is one-to-one, for in this case there is exactly one element in the set { x : g ( x ) = y } for values of y in the domain of Y . In this case, then there exists an inverse function, say x = h ( y ), and consequently P ( Y = y ) = P ( X = h ( y )) because the event { Y = y } is equivalent to the event { X = h ( y ) } . Suppose X is Bernoulli with parameter p and let Y = 2 X - 1. Then h ( y ) = y +1 2 , and P ( Y = y ) = P ( X = y + 1 2 ) for y ∈ {- 1 , 1 } . 1.2 Continuous Case To deal with the continuous case, suppose first that the function g is increasing. Then there is an inverse mapping such that g ( h ( y )) = y , and as a result, the set { x : g ( x ) y } = { x : x h ( y ) } . Consequently, P ( Y y ) = P ( g ( X ) y ) = P ( X h ( y )) . Taking derivative with respect to Y , we obtain the formula f Y ( y ) = f X ( h ( y )) h 0 ( y ) . Example: Suppose that X is uniformly distributed over the interval [0 , 1] and that Y = g ( X ) = a +( b - a ) X with b > a . Then g ( x ) is clearly increasing and h ( y ) = y - a b - a . Since the density f X ( x ) = 1 on 0 x 1, it follows that the density of Y is given by f Y ( y ) = h 0 ( y ) = 1 b - a on a Y b , so Y is uniformly distributed over the interval ( b, a ). 1

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Example: Suppose X is exponential with parameter λ . Find the density of Y = g ( X ) = aX + b where a > 0. Then, h ( y ) = y - b a and f Y ( y ) = λ a exp( - λ a ( y - b )) on y a , so Y has a shifted exponential distribution with parameter λ/a . If Y = g ( X ) and g is decreasing, then the event { x : g ( x ) y } corresponds to the event { x : x h ( y ) } and as a result, P ( Y y ) = P ( X h ( y )) = 1 - P ( X h ( y )) Taking derivatives we find f Y ( y ) = - f X ( h ( y )) h 0 ( y ) .
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