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Stat 5101 Lecture Slides: Deck 6 Existence of Integrals and Infinite Sums, Countable Additivity and Monotone Convergence, Existence of Moments, Correlation Charles J. Geyer School of Statistics University of Minnesota This work is licensed


  1. Stat 5101 Lecture Slides: Deck 6 Existence of Integrals and Infinite Sums, Countable Additivity and Monotone Convergence, Existence of Moments, Correlation Charles J. Geyer School of Statistics University of Minnesota This work is licensed under a Creative Commons Attribution- ShareAlike 4.0 International License ( http://creativecommons.org/ licenses/by-sa/4.0/ ). 1

  2. Existence of Integrals Just from the definition of integral as area under the curve, the integral � b a g ( x ) dx always exists when a and b are finite and g is bounded , which means there exists a finite c such that | g ( x ) | ≤ c, a < x < b. In this case � b � b � � � � a g ( x ) dx a | g ( x ) | dx ≤ c ( b − a ) � ≤ � � � � � 2

  3. Existence of Integrals (cont.) It is a theorem of advanced calculus (which we will not prove) that every continuous function g having domain [ a, b ] where a and b are finite is bounded. So if we know g is continuous on [ a, b ], then we know � b a g ( x ) dx exists. It is important that the domain is a closed interval. The function x �→ 1 /x is continuous but unbounded on (0 , 1). So continuous on an open interval is not good enough. 3

  4. Existence of Integrals (cont.) We are worried about non-existence. Clearly � b a g ( x ) dx may fail to exist in one of two cases (I) either a or b is infinite, or (II) g is unbounded (meaning not bounded). 4

  5. Existence of Integrals: Case I So when does � ∞ g ( x ) dx a exist? In probability theory, we require absolute integrability, so � ∞ | g ( x ) | dx a must be finite. 5

  6. Existence of Integrals: Case I (cont.) First we do a very important special case. Suppose a > 0, then � ∞ x α dx < ∞ a if and only if α < − 1. 6

  7. Existence of Integrals: Case I (cont.) Case I of Case I, if α � = − 1, then b � b = b α +1 − a α +1 a x α dx = x α +1 � � � � α + 1 α + 1 a � If α > − 1, then this goes to infinity as b → ∞ . If α < − 1, then this goes to − a α +1 / ( α + 1) as b → ∞ . Case II of Case I, if α = − 1, then � b b � a x α dx = log( x ) � = log( b ) − log( a ) � � � a and this goes to infinity as b → ∞ . 7

  8. Existence of Integrals: Comparison Principle Also obvious from the definition of integral as area under the curve, if | g ( x ) | ≤ | h ( x ) | , a < x < b, then � b � b a | g ( x ) | dx ≤ a | h ( x ) | dx including when either integral is infinite , that is, when the right- hand side is finite, then so is the left-hand side and when the left-hand side is infinite, then so is the right-hand side. 8

  9. Existence of Integrals: Case I (cont.) Suppose a > 0, suppose g is continuous on [ a, ∞ ), and suppose | g ( x ) | lim = c x α x →∞ exists and is finite. If α < − 1, then � ∞ | g ( x ) | dx < ∞ . a Conversely, if c > 0 and α ≥ − 1, then � ∞ | g ( x ) | dx = ∞ . a 9

  10. Existence of Integrals: Case I (cont.) From the definition of limit, we know there exists a finite r such that 2 ≤ | g ( x ) | c ≤ 1 + c, x ≥ r x α and we know that � r a | g ( x ) | dx < ∞ and � ∞ � ∞ � ∞ c x α dx ≤ x α dx | g ( x ) | dx ≤ (1 + c ) 2 r r r Hence the result about g ( x ) follows from the result about x α . 10

  11. Existence of Integrals: Case I (cont.) There exists a constant c such that c f ( x ) = 1 + x 2 + 3( x − 1) 4 , −∞ < x < ∞ is a PDF. Compare f ( x ) | x | − 4 → c 3 as x → −∞ or x → + ∞ . Since − 4 < − 1, it follows that the integral of f is finite. 11

  12. Existence of Integrals: Case I (cont.) In the preceding example we used two important principles. • Constants don’t matter. • In polynomials, only the term of highest degree matters. 12

  13. Existence of Integrals: Case I (cont.) In more detail, if c is a constant � b � b a cg ( x ) dx = c a g ( x ) dx and both sides exist or neither does. And a 0 + a 1 x + a 2 x 2 + · · · + a k x k lim = a k x k x →∞ 13

  14. Existence of Integrals: Case I (cont.) Returning to our example, suppose X has PDF c f ( x ) = 1 + x 2 + 3( x − 1) 4 , −∞ < x < ∞ , then for what positive values of β does E ( | X | β ) exist? Compare | x | β f ( x ) → c | x | β − 4 3 as x → −∞ or x → + ∞ . Since β − 4 < − 1, if and only if β < 3, it follows E ( | X | β ) exists if and only if β < 3 (when β > 0 is assumed). 14

  15. The Cauchy Distribution There exists a constant c such that c f ( x ) = − ∞ < x < ∞ 1 + x 2 is a PDF. Compare f ( x ) | x | − 2 → c as x → −∞ or x → + ∞ . Since − 2 < − 1, it follows that the integral of f is finite. 15

  16. The Cauchy Distribution (cont.) In this case we can actually determine the constant. � t t dx � � 1 + x 2 = atan( x ) = atan( t ) − atan( − t ) , � − t � − t where atan is the arctangent function, which goes from − π/ 2 to π/ 2 as its argument goes from −∞ to ∞ . Thus � ∞ dx � � 1 + x 2 = lim atan( t ) − atan( − t ) = π t →∞ −∞ and c = 1 /π . 16

  17. The Cauchy Distribution (cont.) The distribution with PDF 1 f ( x ) = π (1 + x 2 ) , −∞ < x < ∞ is called the standard Cauchy distribution . The distribution with PDF f µ,σ ( x ) = σ 1 π · σ 2 + ( x − µ ) 2 , −∞ < x < ∞ is called the Cauchy distribution with location parameter µ and scale parameter σ and is abbreviated Cauchy( µ, σ ). 17

  18. The Cauchy Distribution (cont.) The Cauchy( µ, σ ) distributions are a location-scale family. The Cauchy( µ, σ ) distribution is symmetric about µ , so the pa- rameter µ can be called the center of symmetry as well as the location parameter . 18

  19. The Cauchy Distribution (cont.) If X has the Cauchy( µ, σ ) distribution, then for what positive values of β does E ( | X | β ) exist? Compare | x | β f ( x ) → σ | x | β − 2 π as x → −∞ or x → + ∞ . Since β − 2 < − 1, if and only if β < 1, it follows E ( | X | β ) exists if and only if β < 1 (when β > 0 is assumed). Summary: If X has the Cauchy( µ, σ ) distribution, then E ( X k ) exists for no positive integer k . The mean does not exist, neither does the variance. Hence µ cannot be the mean, and σ cannot be the standard deviation. 19

  20. Existence of Integrals: Case I (cont.) The Maclaurin series e x = 1 + x + x 2 2 + · · · + x k k ! + · · · which we also know as the theorem that a Poisson PMF sums to one, shows that e x grows faster than any polynomial as x → ∞ . Similarly for e λx when λ > 0. Hence for any β ∈ R , any α ∈ R , any λ > 0, and any a > 0 x β e − λx x β − α lim = lim = 0 x α e λx x →∞ x →∞ and � ∞ x β e − λx dx < ∞ a 20

  21. Existence of Integrals: Case II Now we turn to case II. The domain of integration is bounded, but the integrand is unbounded. Again we start with the monomial special case. If a > 0, then � a 0 x α dx < ∞ if and only if α > − 1. Note that the magic exponent − 1 is the same, but the inequality is reversed. 21

  22. Existence of Integrals: Case II (cont.) The substitution x = 1 /y reduces this to the other case. � a � 1 /a � ∞ 0 x α dx = 1 /a y − α − 2 dy y − α � − y − 2 � dy = ∞ and we already know the latter is finite if and only if − α − 2 < − 1, which is the same as − α < 1 or α > − 1. 22

  23. Existence of Integrals: Case II (cont.) We can move this theorem to any other point. If a < b , then � b a ( x − a ) α dx < ∞ if and only if α > − 1, and � b a ( b − x ) α dx < ∞ if and only if α > − 1. 23

  24. Existence of Integrals: Case II (cont.) And we can analyze other integrals by comparison. Suppose g is continuous on ( a, b ] and suppose | g ( x ) | lim ( x − a ) α = c x ↓ a exists and is finite. If α > − 1, then � b a | g ( x ) | dx < ∞ . Conversely, if c > 0 and α ≤ − 1, then � b a | g ( x ) | dx = ∞ . 24

  25. Existence of Integrals: Case II (cont.) The case where g is unbounded at b is an obvious modification. Suppose g is continuous on [ a, b ) and suppose | g ( x ) | lim ( b − x ) α = c x ↑ b exists and is finite. If α > − 1, then � b a | g ( x ) | dx < ∞ . Conversely, if c > 0 and α ≤ − 1, then � b a | g ( x ) | dx = ∞ . 25

  26. Existence of Integrals: Case I and II Summary If g ( x ) is continuous on [ a, ∞ ), then � ∞ | g ( x ) | dx < ∞ a only if g ( x ) goes to zero as x → ∞ fast enough, faster than 1 /x . If g ( x ) is continuous on ( a, b ), then � b a | g ( x ) | dx < ∞ only if g ( x ) goes to infinity as x → a or as x → b slow enough, slower than 1 / ( x − a ) or 1 / ( b − x ). 26

  27. Existence of Integrals: Gamma Distribution When does there exist a c such that f ( x ) = cx α − 1 e − λx , 0 < x < ∞ is a PDF? We have already been told this is the PDF of the Gam( α, λ ) distribution when α > 0 and λ > 0, but we haven’t proved it. If λ > 0, then we know from applying our theorem about case I that � ∞ x α − 1 e − λx dx < ∞ a for any real α and any a > 0. 27

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