Extra Topic: DISTRIBUTIONS OF FUNCTIONS OF RANDOM VARIABLES

Transcription

1 Extra Topic: DISTRIBUTIONS OF FUNCTIONS OF RANDOM VARIABLES A little in Montgomery and Runger text in Section 5-5. Previously in Section 5-4 Linear Functions of Random Variables, we saw that we could find the mean and variance of a linear combination of random variables. For example, if Y = 2X 1 + 4X X 3 then E(Y ) = 2E(X 1 ) + 4E(X 2 ) + 10E(X 3 ) and if X 1, X 2, X 3 are independent, then we know V (Y ) = 2 2 V (X 1 ) V (X 2 ) V (X 3 ) But knowing the mean and variance of the random variable Y is not the same as knowing the full probability distribution for Y. Could two random variables have the same mean and variance, but have different distributions? 1

2 ANSWER: Yes. Consider random variables U and Z below such that both have a mean of 0 and a variance of 1, but very different distributions. U Uniform( 3, 3) f U (u) Z Normal(0, 1) Mean and variance are informative, but without knowledge of the shape of the distribution, they don t fully specify a distribution. 2

3 NOTE: If you have a mean & variance for Y AND you know the distribution of Y is normal (i.e. known shape), then your distribution IS fully defined by just µ and σ 2. This is incredibly useful in the case of Y being a linear combination of independent normal random variables, then Y is a normal r.v., as well. (See Section 5-4 Reproductive Property of the Normal Distribution in the book. And see the lecture-notes example on total weight of people on an elevator from the previous section.) In this section of notes, we wish to define a full probability distribution for a random variable Y that is a function of a generic random variable X, or where Y = g(x). 3

4 Transformations (Discrete r.v. s) Suppose we have a random variable X and we know its distribution. We are interested, though, in a random variable Y which is a transformation of X. For example, Y = X 2. We wish to determine the distribution of Y. Example 1: Suppose X is discrete and has the pmf below... x f X (x) Now, let Y = X 2. 4

5 x f X (x) Because the support set for X, {0, 1, 2}, is strictly over the non-negative real number line, the function Y = X 2 is a one-toone mapping from the X-space to the Y -space over the relevant support set. Y=X 2 Y X Because of this one-to-one relationship, we have the pmf of Y as... y f Y (y)

6 We can generalize this case when the transformation or mapping is one-to-one for discrete random variables. For a one-to-one transformation Y = g(x) for discrete X, the pmf of Y or f Y (y) is obtained as, f Y (y) = P (Y = y) = P (g(x) = y) = P (X = g 1 (y)) = f X (g 1 (y)). From the previous example with X {0, 1, 2} and Y = g(x) = X 2 we can obtain the P (Y = 4) as... f Y (4) = P (Y = 4) = P (X 2 = 4) = P (X = + 4) (support of X is = P (X = 2) only non-negatives) = f X (2) = 0.5 6

7 We can use a similar process for obtaining P (Y = 0) and P (Y = 1) by again thinking of the inverse function X = g 1 (y), which is X = + y in this specific case. Or we could write, P (Y = y) = P (X = + y) Y=X 2 Y X 7

8 Example 2: Suppose X has the pmf defined by f X (x) = 0.10x for x in {1,2,3,4}. Now, let Y = X. This is a one-to-one mapping from the X- space to the Y -space over the support set. y f Y (y) We can define the pmf for Y using a formula in this case because the pmf for X was given as a formula: f Y (y) = f X (g 1 (y)) = f X (y 2 ) = 0.10(y 2 ) for all possible values of Y. Or more formally, f Y (y) = 0.10(y 2 ) for y in {1, 2, 3, 2} 8

9 Example 3: (case of NOT one-to-one) Suppose X has the pmf below... x f X (x) Now, let Y = X 2. This is not a one-to-one mapping from the X-space to the Y -space over the support set. What are the possible values for Y? y f Y (y)??? Because X is discrete, we can easily find f Y (y) by looking at which x-values are mapped to the possible y-values. 9

10 Y=X 2 Y Two distinct x-values lead to y= X P (Y = 0) = P (X = 0) = 0.1 P (Y = 1) = P (X = 1) + P (X = 1) = 0.5 P (Y = 4) = P (X = 2) = 0.4 Thus, we have... y f Y (y)

11 In the discrete case when g(x) is not oneto-one, instead of developing an overall rule, we usually obtain f Y (y) in a straightforward manner looking at the mapping itself (as in the above example). In the next set of notes, we ll discuss distributions of functions of continuous random variables. Transformations (Continuous r.v. s) When dealing with continuous random variables, a couple of possible methods are... (1) Distribution function (cdf) technique (2) Change of variable (Jacobian) technique 11