Transformations and Expectations

Transcription

1 Transformations and Expectations 1 Distributions of Functions of a Random Variable If is a random variable with cdf F (x), then any function of, say g(), is also a random variable. Sine Y = g() is a function of, we can describe the probabilistic behavior of Y in terms of that of. That is, for any set A, P (Y A) = P (g() A), Showing that the distribution of Y depends on the function F and g. Formally, if we write y = g(x), the function g(x) defines a mapping from the original sample space of,, to a new sample space, Y, the sample space of the random variable Y. That is, g(x) : Y. Conveniently, we can write = {x : f (x) > 0} and Y = {y : y = g(x) for some x C}. (1) The pdf of is positive only on the set and is 0 elsewhere. Such a set is called the support set or support of a distribution. We associate with g an inverse mapping, denoted by g 1, which is a mapping from subsets of Y to subsets of, and is defined by g 1 (A) = {x : g(x) A}. It is possible for A to be a point set, say A = {y}. Then g 1 ({y}) = {x : g(x) = y}. In this case, we often write g 1 (y) instead of g 1 ({y}). The probability distribution of Y can be defined as follows. For any set A Y, P (Y A) = P (g() A) = P ({x : g(x) A}) = P ( g 1 (A)). 1

2 It is straightforward to show that this probability function satisfies the Kolmogorov Axioms. If is a discrete random variable, then is countable. The sample space for Y = g() is Y = {y : y = g(x), x }, which is also a countable set. variable. The pmf for Y is f Y (y) = P (Y = y) = = x g 1 (y) f (x), x g 1 (y) Thus, Y is also a discrete random P ( = x) for y Y and f Y (y) = 0 for y / Y. In this case, finding the pmf of Y involves simply identifying g 1 (y), for each y Y, and summing the appropriate probabilities. Example 1.1 (Binomial transformation) A discrete random variable has a binomial distribution if its pmf is of the form f (x) = P ( = x) = where n is a positive integer and 0 p 1. ( ) n p x (1 p) n x, x = 0, 1,..., n, x g(x) = n x. Thus, g 1 (y) is the single point x = n y, and f Y (y) = f (x) = f (n y) x g 1 (y) Consider the random variable Y = g(), where ( ) n = p n y (1 p) n (n y) n y ( ) n = (1 p) y p n y. y Thus, we see that Y also has a binomial distribution, but with parameters n and 1 p. If and Y are continuous random variables, the cdf of Y = g() is F Y (y) = P (Y y) = P (g() y) = P ({x : g(x) y}) = {x :g(x) y} f (x)dx. Sometimes there may be difficulty in identifying {x : g(x) y} and carrying out the integration of F (x) over this region. Example 1.2 (Uniform transformation) Suppose has a uniform distribution on the interval (0, 2π), that is, 1/(2π) 0 < x < 2π f (x) = 0 otherwise. 2

3 Consider Y = sin 2 (). Then P (Y y) = P ( x 1 ) + P (x 2 x 3 ) + P ( x 4 ) = 2P ( x 1 ) + 2P (x 2 π), where x 1 and x 2 are the two solutions to sin 2 (x) = y, 0 < x < π. Thus, even though this example dealt with a seemingly simple situation, the cdf of Y was not simple. It is easiest to deal with functions g(x) that are monotone, that is, those that satisfy either u > v g(u) > g(v) (increasing) or u < v g(u) > g(v) (decreasing). If g is monotone, then g 1 is single-valued; that is, g 1 (y) = x if and only if y = g(x). If g is increasing, this implies that {x : g(x) y} = {x : x g 1 (y)}. If g is decreasing, this implies that {x : g(x) y} = {x : x g 1 (y)}. If g(x) is increasing, we can write F Y (y) = {x :x g 1 (y)} If g(x) is decreasing, we have F Y (y) = f (x)dx = g 1 (y) g 1 (y) f (x)dx = 1 F (g 1 (y)). f (x)dx = F (g 1 (y)). The continuity of is used to obtain the second equality. following theorem. We summarize these results in the Theorem 1.1 Let have cdf F (x), let Y = g(), and let and Y be defined as in (1). a. If g is an increasing function on, F Y (y) = F (g 1 (y)) for y Y. b. If g is a decreasing function on and is a continuous random variable, F Y (y) = 1 F (g 1 (y)) for y Y. 3

4 Example 1.3 (Uniform-exponential relationship-i) Suppose f (x) = 1 if 0 < x < 1 and 0 otherwise, the uniform(0,1) distribution. It is straightforward to check that F (x) = x, 0 < x < 1. We now make the transformation Y = g() = log(). Since d dx g(x) = 1 < 0, for 0 < x < 1, x g(x) is a decreasing function. Therefore, for y > 0, F Y (y) = 1 F (g 1 (y)) = 1 F (e y ) = 1 e y. Of course, F Y (y) = 0 for y 0. If the pdf of Y is continuous, it can be obtained by differentiating the cdf. Theorem 1.2 Let have pdf f (x) and Y = g(), where g is a monotone function. Let and Y be define by (1). Suppose that f (x) is continuous on and that g 1 (y) has a continuous derivative on Y. Then the pdf of Y is given by f (g 1 (y)) d dy f Y (y) = g 1 (y) y Y 0 otherwise. Proof: From Theorem 1.1 we have, by the chain rule, f Y (y) = d dy F f (g 1 (y)) d Y (y) = f (g 1 (y)) d dy g 1 (y) dy g 1 (y) if g is increasing if g is decreasing. Example 1.4 (Inverted gamma pdf) Let f (x) be the gamma pdf f(x) = 1 (n 1)!β n xn 1 e x/β, 0 < x <, where β is a positive constant and n is a positive integer. If we let y = g(x) = 1/x, then g 1 (y) = 1/y and d dy g 1 (y) = 1/y 2. Applying the above theorem, for 0 < y <, we get f Y (y) = f (g 1 (y)) d dy g 1 (y) 1 (1 ) n 1e 1/(βy) = 1 (n 1)!β n y y 2 1 (1 ) n+1e = 1/(βy) (n 1)!β n, y a special case of a pdf known as the inverted gamma pdf. 4

5 Theorem 1.3 Let have pdf F (x), let Y = g(), and define the sample space as in (1). Suppose there exists a partition, A 0, A 1,..., A k, of such that P ( A 0 ) = 0 and f (x) is continuous on each A i. Further, suppose there exist functions g 1 (x),..., g k (x), defined on A 1,..., A k, respectively, satisfying i. g(x) = g i (x), for x A i, ii. g i (x) is monotone on A i, iii. the set Y = {y : y = g i (x) for some x A i } is the same for each i = 1,..., k, and iv. g 1 i (y) has a continuous derivative on Y, for each i = 1,..., k. Then k i=1 f Y (y) = f (gi 1 (y)) d dy g 1 i (y) y Y 0 otherwise. Example 1.5 (Normal-Chi squared relationship) Let have the standard normal distribution f (x) = 1 2π e x2 /2, < x <. Consider Y = 2. The function g(x) = x 2 is monotone on (, 0) and (0, ). The set Y = (0, ). Applying Theorem 1.3, we take The pdf of Y is A 1 = (, 0), g 1 (x) = x 2, g 1 1 (y) = y; A 2 = (0, ), g 2 (x) = x 2, g 1 2 (y) = y. A 0 = {0}; f Y (y) = 1 e ( y) 2 /2 1 2π 2 y + 1 e ( y) 2 /2 1 2π 2 y = 1 2π 1 y e y/2, 0 < y <. So Y is a chi-squared random variable with 1 degree of freedom. by Let F 1 denote the inverse of the cdf F. If F is strictly increasing, then F 1 is well defined F 1 (y) = x F (x) = y. (2) 5

6 However, if F is constant on some interval, then F 1 is not well defined by (2). The problem is avoided by defining F 1 (y) for 0 < y < 1 by F 1 (y) = inf{x : F (x) y}. (3) At the end point of the range of y, F 1 (1) = if F (x) < 1 for all x and, for any F, F 1 (0) =. Theorem 1.4 (Probability integral transformation) Let have continuous cdf F (x) and define the random variable Y as Y = F (). Then Y is uniformly distributed on (0, 1), that is, P (Y y) = y, 0 < y < 1. Proof: For Y = F () we have, for 0 < y < 1, P (Y y) = P (F () y) = P (F 1 [F ()] F 1 (y)) = P ( F 1 (y)) = F (F 1 (y)) = y. At the endpoints we have P (Y y) = 1 for y 1 and P (Y y) = 0 for y 0, showing that Y has a uniform distribution. The reasoning behind the equality P (F 1 [F ()] F 1 1 (y)) = P ( F (y)) is somewhat subtle and deserves additional attention. If F is strictly increasing, then it is true that F 1 (F (x)) = x. However, if F is flat, it may be that F 1 (F (x)) x. Then F 1 (F (x)) = x 1, since P ( x) = P ( x 1 ) for any x [x 1, x 2 ]. The flat cdf denotes a region of 0 probability P (x 1 < x) = F (x) F (x 1 ) = 0. 2 Expected values Definition 2.1 The expected value or mean of a random variable g(), denoted by Eg(), is Eg() = g(x)f (x)dx if is continuous x g(x)f (x) = x g(x)p ( = x) if is discrete, provided that the integral or sum exists. If E g() =, we say that Eg() does not exist. 6