MTH 202 : Probability and Statistics

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1 MTH 202 : Probability and Statistics Lecture 9 - : 27, 28, 29 January, Functions of a Random Variables 4. : Borel measurable functions Similar to continuous functions which lies to the heart of real analysis, which you have encountered in your basic courses on calculus and several variable calculus, it is measurable functions which are the most relevant objects of interest here. Recall that the ɛ δ definition of a continuous function f : R R Figure. ɛ δ would really mean that : given any open interval I R, the set f (I) is an open set (which mean f (I) is a union of countably many open intervals). Definition 4.. : Let (Ω, S) be a σ algebra and (R, B(R)) be the Borel σ-algebra. A function f : Ω R is called a measurable function if for every Borel set B we have f (B) S.

2 2 As we have seen before, we don t need to involve all Borel sets while verifying whether f is Borel measurable or not. Theorem 4..2 : A function f : Ω R is measurable if and only if f ((, x]) S for any x R. Proof : See [ROY] We usually refer the measurable function by f : Ω R if the structure of the related σ-algebras are understood. A Borel measurable function is rather special : Recall that an RV is a Borel measurable function (see definition in chapter 3). But while dealing with a complicated function f : Ω R it would be difficult to tell whether f is an RV or not. In these situations we would rather employ an old fashioned trick, namely substitution of variables which you might have seen in high school while dealing with tricky integrals. However, if you substitute a bad function it may not work at all. At least you need to ensure these are substituted by some nice functions, which are Borel measurable functions f : R R with respect to the Borel measure in its domain. Theorem 4..4 : Let X be an RV defined on a probability space (Ω, S, P ). If f : R R is a Borel measurable function, then f(x) := f X is an RV. Proof : We consider the function f X : Ω R. Recall that it is enough to show (f X) ((, x]) S for any x R. Now if B = (, x] (f X) (B) = X (f (B)) Since f is Borel measurable and B is a Borel set, f (B) is a Borel set as well. Next since X is an RV, X (f (B)) S. Examples 4..5 : Let X be an RV with DF F. Then X, ax + b (where a( 0) and b are constants), X k (where k N 0}), and X α (α > 0) are all RV s. Define X + X if X 0, = 0 if X < 0.

3 3 and X = X if X 0, 0 if X > 0. Then X + and X are also RV s. [See Page-58, sect. 2.5, [RS]]. Notice that except the function X α (α > 0), all can be expressed by X composed with some continuous function. Using natural logarithm, X α can also be seen that way (Why?). This is not a coincidence : Theorem 4..6 : Every continuous function f : R R is Borel measurable. Proof : See [ROY] Hence if f is continuous and X is an RV, the previous results imply that f(x) is an RV. Next we note that while the DF if known for an RV X, the DF of Y = f(x) can always be determined if f is Borel measurable. Exercise 4..7: Let X be an RV of discrete type with PMF ( ) n P (X = r) = p r ( p) n r, r = 0,, 2,..., n r for some p [0, ]. Find the PMFs of the RVs : (i) Y = ax + b for a, b R and a 0, (ii) Y = X 2, and (iii) Y = X. Solution : (i) There is a bijection t : 0,, 2,..., n} b, a + b, 2a + b,..., na + b} = B sending x ax + b. If y B then P (Y = y) = P (X = y b ) and we a have the following PMF of Y ( n ) y b y b y b p n a ( p) a if y B P (Y = y) = a (ii) Here B = 0,, 4,..., n 2 } and the PMF of Y is ( n y ) p y ( p) n y if y B P (Y = y) = (ii) B = 0,, 2,..., n} and the PMF of Y is ( n ) y p y 2 ( p) n y2 if y B P (Y = y) = 2

4 4 For an RV X of continuous type it is not necessary that the RV Y = g(x) would be of continuous type even if g is a continuous function. It would be ensured provided g is a little more nice : Theorem 4..8 : Let X be an RV of continuous type with PDF f. Let y = g(x) be differentiable for all x and is strictly monotone, i.e. either f (x) > 0 for all x or f (x) < 0 for all x. Then the RV Y = g(x) is also of continuous type with PDF given by f(g (y)) d dy g (y) if α < y < β, where α = ming( ), g(+ )} and β = maxg( ), g(+ )}. Proof : See [RS, Theorem 3, Page 60] Note 4..9 : Recall here that g( ) = lim x g(x) and g(+ ) = lim x g(x). Exercise 4..0 : If the density of X is f(x) = (x R) π(x 2 + ) Find the density of Y = tan (X). Solution : The function y = tan (x) is strictly increasing since > 0 for all x R. Hence we can use previous theorem. dy dx = x 2 + Here y = g(x) = tan x. Hence g (y) = tany. Thus f(tany) d dy (tany) = π(tan 2 y + ).sec2 y = π while π 2 < y < π 2 and the density of Y is : π if π 2 < y < π 2,. There are quite a few important distributions we would encounter whose pdf cannot be obtained using the previous theorem due to lack of universal monotonicity. The following result would be useful in some such cases.

5 Theorem 4.. : Let X be an RV of continuous type with PDF f. Let y = g(x) be differentiable for all x, and assume that g (x) is continuous and non-zero at all but a finite number of x R. Then for every y R, (a) there exists a positive integer n = n(y) and real numbers x (y), x 2 (y),..., x n (y) such that g(x k (y)) = y and g (x k (y)) 0, (k =, 2,..., n(y)) or (b) g (x) 0 for each x satisfies g(x) = y and we write n(y) = 0 in such cases. Finally Y is a continuous RV with PDF given by : n k= f(x k(y)) g (x k (y)) if n > 0, 0 if n = 0 We will see this in the following example Exercise 4..2 : Let X be an RV with PDF f(x) = e x2 2, ( < x < ) 2π What is the PDF of Y = X 2. Solution : We have y = g(x) = x 2. Here every point y R with y 0 has exactly two pre-images (with non-zero derivative) except at the point x = 0. In other words, n(y) = 2 if y > 0; n(0) = 0. Also if y > 0, we have x (y) = y, x 2 (y) = y. Hence if y > 0, we have f( y). 2 y + f( y). 2 y = e y 2 2πy Therefore the PDF of Y is given by 2πy e y 2 if y > 0, 0 if y 0. References : [ROY] Real Analysis, H.L. Royden, 3rd Edition, Macmillan Publishing Co. [RS] An Introduction to Probability and Statistics, V.K. Rohatgi and A.K. Saleh, Second Edition, Wiley Students Edition. 5