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Material Type: Notes; Class: Introduction to Probability Theory I; Subject: Statistics; University: California State University-East Bay; Term: Unknown 1989;
Typology: Study notes
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Geometrical Example.
Suppose X ~ UNIF(0, 1) = BETA(1, 1) and that Y = X^2 = g ( X ).
We know that f (^) X ( x ) = I (0,1)( x ).
(Here, we use the indicator function:
IA ( x ) = 1 for x ∈ A and IA ( x ) = 0 otherwise.)
The support of the random variable X is the unit interval (0, 1). It is crucial in transforming random variables to begin by finding the support of the transformed random variable. Here the support of Y is the same as the support of X.
Now we approximate f (^) Y by seeing what the transformation does to each of the intervals (0, 0.1), (0.1, 0.2), ..., (0.0, 1.0). In the table below, these intervals are in the “ x -Interval” column.
Each of these intervals has length 0.1. The height of f (^) X ( x ) is 1in each interval, so area or probability in within each interval is 0.1:
The “ y -Interval” column of the table shows the transform of each interval. Clearly, the transformed random variable Y must have
Taking into account the (unequal) lengths of the y -intervals, we can see what the (average) height of the f (^) Y must be above each y -interval in order to preserve the probability of the interval under transformation.
The results are shown in the column “ y -Height” and plotted in the rectangles of the graph on the next page.
CDF method.
A more formal approach to finding f (^) Y goes as follows.
The cdf of X is FX ( x ) = x , for 0 < x < 1. Thus the cdf of Y can be found as
FY ( y ) = P( Y ≤ y ) = P( X^2 ≤ y ) = P( X ≤ y 1/2) = y 1/2, for 0 < y < 1.
Differentiating, we have the density function of Y :
FY '( y ) = f (^) Y ( y ) = (1/2) y –1/2 I (0,1)( y ).
Thus, we have shown that Y ~ (^) BETA(1/2, 1). (Not all transforms Y = X k^ of a beta random variable X are beta.)
The density function of Y is plotted in the figure.
This method of finding the distribution of a transformed random variable is called the cdf-method. It is very widely applicable.
Exercises:
PDF method.
For monotone increasing or decreasing functions g , the CDF method can be carried out in general, allowing one to deal only with PDFs.
In our example, within the support of X , the function y = g ( x ) = x^2 is monotone increasing.
For any function y = g ( x ) that is monotone increasing on the support of X , we may carry out the CDF method in a general way:
FY ( y ) = P( Y ≤ y ) = P( g ( X ) ≤ y ) = P( X ≤ g –1^ ( y )) = FX ( g –1^ ( y )).
Whether or not we know the exact functional form of FX , differentiation gives
f (^) Y ( y ) = d FX ( g –1( y )) / dy = f (^) X ( g –1( y )) d g –1^ ( y ) / dy ,
for y in the support of Y, provided that g is a monotone increasing function.
The factor with the derivative of g –1^ results from applying the chain rule of differentiation.
For the example where y = g ( x ) = x^2 , we have
g –1( y ) = x 1/2^ and d g –1( y ) / dx = (1/2) y –1/2.
Geometrical interpretation.
Referring back to the geometrical illustration, the expression
| d g –1( y ) / dy | = | dx / dy |
can be viewed as the appropriate multiple to compensate
for the change in the length of a small interval ∆ x , as g transforms it to ∆ y , in order to ensure that
P( X ∈ ∆ x ) = P( Y ∈ ∆ y ).
Note: If the transforming function is not monotone on the support of X ,
it is often best to use the cdf method.
Advanced Exercises:
Square of standard uniform with larger sample; also with more histogram bins (as specified by breaks parameter).
m <- 500000 x <- runif(m) y <- x^ hist(y, breaks=50, freq=F) yy <- seq(0, 1, by=.001) lines(yy, dbeta(yy, .5, 1))
Square of standard normal with density of CHISQ(1) superimposed (see exercise above).
m <- 100000 x <- rnorm(m) y <- x^ hist(y, freq=F, xlim=c(0,10)) yy <- seq(0, 10, by=.0001) lines(yy, dchisq(yy, 1))
Advanced Exercise:
Copyright © 2002, 2003, 2004 by Bruce E. Trumbo. All rights reserved.
