s4 - Stat 5101 Lecture Slides Deck 6 Charles J. Geyer...

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Unformatted text preview: Stat 5101 Lecture Slides Deck 6 Charles J. Geyer School of Statistics University of Minnesota 1 Existence of Integrals Just from the definition of integral as area under the curve, the integral Z 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 Z b a g ( x ) dx ≤ Z b a | g ( x ) | dx ≤ c ( b- a ) 2 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 Z b a g ( x ) dx exists. It is important that the domain is a closed interval. The function x 7→ 1 /x is continuous but unbounded on (0 , 1). So continuous on an open interval is not good enough. 3 Existence of Integrals (cont.) We are worried about non-existence. Clearly Z b a g ( x ) dx fails to exist in one of two cases (I) either a or b is infinite, or (II) g is unbounded (meaning not bounded). 4 Existence of Integrals: Case I So when does Z ∞ a g ( x ) dx exist? In probability theory, we require absolute integrability, so Z ∞ a | g ( x ) | dx must be finite. 5 Existence of Integrals: Case I (cont.) First we do a very important special case. Suppose a > 0, then Z ∞ a x α dx < ∞ if and only if α <- 1. 6 Existence of Integrals: Case I (cont.) Case I of Case I, if α 6 =- 1, then Z b a x α dx = x α +1 α + 1 b a = b α +1- a α +1 α + 1 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 Z b a x α dx = log( x ) b a = log( b )- log( a ) and this goes to infinity as b → ∞ . 7 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 Z b a | g ( x ) | dx ≤ Z b 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 Existence of Integrals: Case I (cont.) Suppose a > 0, suppose g is continuous on [ a, ∞ ), and suppose lim x →∞ | g ( x ) | x α = c exists and is finite. If α <- 1, then Z ∞ a | g ( x ) | dx < ∞ . Conversely, if c > 0 and α ≥ - 1, then Z ∞ a | g ( x ) | dx = ∞ . 9 Existence of Integrals: Case I (cont.) From the definition of limit, we know there exists a finite r such that c 2 ≤ | g ( x ) | x α ≤ 1 + c, x ≥ r and we know that Z r a | g ( x ) | dx < ∞ and c 2 Z ∞ r x α dx ≤ Z ∞ r | g ( x ) | dx ≤ (1 + c ) Z ∞ r x α dx Hence the result about g ( x ) follows from the result about x α ....
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This note was uploaded on 09/13/2011 for the course STA 4184 taught by Professor Staff during the Spring '11 term at University of Central Florida.

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s4 - Stat 5101 Lecture Slides Deck 6 Charles J. Geyer...

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