Math 3338, Homework 4. f(u)du. (Note that f is discontinuous, whereas F is continuous everywhere.)

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1 Math 3338, Homework 4. Define a function f by f(x) = { if x < if x < if x. Calculate an explicit formula for F(x) x f(u)du. (Note that f is discontinuous, whereas F is continuous everywhere.) 2. Define a function f(x) = max( x,). (a) Show that f can be written in case form as if x < +x if x < f(x) = x if x < if x. (b) Calculate an explicit formula for F(x) x f(u)du. (Note that f(x) is continuous but not differentiable at x = or x = ±. On the other hand, F is continuous and differentiable everywhere.) 3. Let f be given by Compute F(x) = x f(u)du. { if x or x = f(x) = 4 x if < x <. 4. Let f be given in exercise, and for ǫ > define f ǫ (x) ǫ f(x/ǫ). (a) If g(x) is continuous at x = a, show that lim ǫ g(x)f ǫ (x a)dx = g(a). (b) Let F ǫ (x) x f ǫ(u)du. Show that for every x R, lim ǫ F ǫ (x) = H(x), where H denotes the Heaviside function; (defined by us so that H() = ). (c) Suppose this time f is as given in exercise 2. Conclude part (a) is still true whereas part (b) is not quite (but is essentially) true for all x. We will call a distribution function F(x) absolutely continuous if it is given by F(x) = x f(u)du for some f which is integrable on (, ).

2 Clearly such an F is continuous everywhere. Also F is nondecreasing f (almost everywhere). Via integration, the function little f defines a distribution function F when () f(x) for all x R and (2) In this case little f is called F s density function. f(x)dx =. An excellent question to ask at this point is, are all continuous probability distributions absolutely continuous? Generally speaking they are not; see e.g. []. However, continuous probability distributions are absolutely continuous if they fail to be differentiable only on a countable set of exceptional points. For such distributions F we can define its density f by d F(x) for nonexceptional x s f(x) = dx otherwise, and note that the values assigned to f at exceptional points are arbitrary. As we ve seen, not every probability distribution is continuous. In this course we will only consider distributions which can be decomposed as follows. F(x) = F ac (x)+f pp (x), where F ac denotes the absolutely continuous part of F, and F pp denotes what is called the pure point or discrete part of F. (This decomposition is not complete for all possible distribution functions. However, it s pretty close. For those interested see [2] for a related discussion of this issue.) The discrete part F pp is a nondecreasing piecewise constant function with jumps at F s points of discontinuity. We will write F pp (x) = n p n H(x a n ), where p n denotes the jump of F(x) at x = a n ; i.e. p n = F(a n ) F(a n + ). The numbers p n are often called probability masses and they must satisfy p n, and F( ) = F ac ( )+ n p n =. From lecture and exercise 4(b) above, we can now define the density function f of a probability distribution F which has the form F ac +F pp f(x) = f ac (x)+ n p n δ(x a n ) [] set and function [2] s decomposition theorem 2

3 where f ac (x) denotes the derivative of F ac (x) and δ(x) denotes the derivative of the Heaviside function H(x). Recall that the δ function is not really a function at all. It is merely a formalism. We will use the probability density to compute what is called the expected value of functions of a random variable X. Suppose X has F X as its distribution. Let f X denote F X s associated density. Then the notation E(g(X)) stands for E(g(X)) g(u)f X (u)du. Note that when g(u) = we must have E() = F X ( ) =. The mean of a random variable X is denoted by µ X and is given by µ X E(X) = The variance of X is denoted by σ 2 X Check that σ 2 X E((X µ X) 2 ) = and is given by uf X (u)dx. (u µ X ) 2 f X (u)du. σ 2 X = E((X µ X) 2 ) = E(X 2 2µ X X +µ 2 X ) = E(X 2 ) 2µ X E(X)+µ 2 XE() = E(X 2 ) µ 2 X. The standard deviation of X is denoted by σ X and is found by taking the square root of the variance σ 2 X, σx = E(X 2 ) µ 2 X. As will be seen later, all of these integrals may or may not be defined. Here s an example. Suppose X has distribution if x < x/3 if x < /2 if x < 2 if x 2. To compute the derivative of the absolutely continuous part, all you have to do if differentiate one case at a time. This yields d()/dx if x < d(x/3)/dx if x < d(/2)/dx if x < 2 d()/dx if x 2. = { /3 if x < otherwise. F X (x) is discontinuous at x = with jump 2 3 = 6 and at x = 2 with jump 2 = 2. Therefore, the derivative of the discrete part is 6 δ(x ) + 2δ(x 2). Putting these 3

4 together we get ({ ) f X (x) = /3 if x < + otherwise. 6 δ(x )+ 2 δ(x 2). Now compute E(g(X))= g(u)f X(u)du with g(x) =, g(x) = X and g(x) = X 2 E() = E(X) = E(X 2 ) = () 3 du+ 6 () u= + 2 () u=2 = = (duh), = 3 = 3 (u) 3 du+ 6 (u) u= + 2 (u) u= = 4 3, (u 2 ) 3 du+ 6 (u2 ) u= + 2 (u2 ) u= = 4 8. The standard deviation is σ X = 4/8 (4/3) 2 = /2. 5. Suppose a random variable X has the following distribution. { e x if x otherwise. (a) Determine F X s density function. (b) Determine X s mean µ X. (c) Determine the variance σ 2 X. 6. Suppose a random variable X has the following distribution. { e x if x otherwise. (a) Determine F X s density function. (b) Determine X s mean µ X. (c) Determine the variance σ 2 X. 7. Suppose X has distribution if x < x/2 if x < /2 (x+)/2 if /2 x < if x. (a) Decompose F X into itsabsolutely continuous part plus itsdiscrete part. (b) Determine its density f X. (c) Compute X s mean µ X. (d) Compute X s variance σx 2. 4

5 8. A random variable is said to be Poisson distributed if e λ λ k k! H(x k), k= for some positive parameter λ. (a) Conclude that F X ( ) =. (b) Determine f X and then compute µ X. (c) Compute X s variance σ 2 X. 9. Consider a random variable X that has { if x < (x )/x if x as its distribution. X has no moment E(X n ) for any n. Why? Answers:. F(x) is equal to if x <, x+ if x < and if x. 2. F(x) is equal to if x <, 2 (x+)2 if x <, 2 (x )2 if x < and if x. 3. F(x) is equal to if x <, ( x)/2 if x <, (+ x)/2 if x < and if x. 4. (c) Change variables, u/ǫ y, to get x x x/ǫ F ǫ (x) = f ǫ (u)du = ǫ f(u/ǫ)du = f(y)dy = F(x/ǫ). From exercise 2 F() = /2 which shows F ǫ () = /2 for all ǫ >. When x <, see that F(x/ǫ) = for all ǫ satisfying < ǫ x. When x >, see that F(x/ǫ) = for all < ǫ x. Therefore, for any x R we have limf ǫ (x) = limf(x/ǫ) = ǫ ǫ 5. (a) f X (x) = e x H(x). (b) µ X =. (c) σ 2 X =. { if x < /2 if x = if x >. 6. (a) f X (x) = e x H(x )+( e )δ(x ). (b) µ X = +e. (c) σ 2 X = 2e e (a) min(max(x/2,),/2) + 2 H(x /2). (b) For x (,) f X(x) = /2+ 2 δ(x /2). Otherwise f X(x) =. (c) µ X = /2. (d) σ 2 X = / Recall that e x = k= xk /k!. (a) F X ( ) = e λ k= λk /k! = e λ e λ =. (b) µ X = λ. (c) σ 2 X = λ. 5