NotesOct18 - p X then Y = T X is also discrete The pmf of Y...

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Transformations of random variables Statement of the problem: let X be a ran- dom variable with a known distribution, e.g. known cdf, or known pmf, or known pdf. Let Y = T ( X ) be a function, or a transforma- tion, of X . Find the distribution of Y .
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Computing the cdf of Y = T ( X ) If X is discrete with pmf p X , then F Y ( y ) = P ( Y y ) = P ( T ( X ) y ) = X x i : T ( x i ) y p X ( x i ) , -∞ < y < . If X is continuous with pdf f X then F Y ( y ) = P ( Y y ) = P ( T ( X ) y ) = Z x : T ( x ) y f X ( x ) dx, -∞ < y < .
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Example : Suppose that X is uniformly dis- tributed between - 1 and 1. Find the distribu- tion of Y = X 2 .
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This method sometimes works even for trans- formations of several random variables. Example : Let X and Y be two indepen- dent standard exponential random variables, and Z = T ( X, Y ) = X + Y . Find the distribution of Z .
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Computing the pmf or pdf of Y = T ( X ) . If the computation of the cdf of Y = T ( X ) is not practical, then in the discrete case, one can try computing the pmf of Y = T ( X ); in the continuous case one can try com- puting the pdf of
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Unformatted text preview: p X then Y = T ( X ) is also discrete. The pmf of Y : p Y ( y j ) = P ( Y = y j ) = P ( T ( X ) = y j ) = X x i : T ( x i )= y j p X ( x i ) . If the function T is one-to-one , then p Y ( y j ) = p X ± T-1 ( y j ) ² ( T-1 is the inverse map ) Example Let X be a mean λ Poisson random variable. Find the distribution of Y = 2 X . Let X be continuous with pdf f X . • Sometimes Y = T ( X ) is also continuous. • To compute the density of Y adjust by the derivative of the inverse transfor-mation . Monotone transformations : the function T is either increasing or decreasing on the range of X . A monotone function T is automatically one-to-one. The pdf of Y : f Y ( y ) = f X ± T-1 ( y ) ² ³ ³ ³ ³ ³ dT-1 ( y ) dy ³ ³ ³ ³ ³ ....
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