We define a simple function as one which is a linear

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We define a simple function as one which is a linear combination of indicator func- tions: For some collection A 1 , A 2 , . . . , A n ∈ F , g ( ω ) = n X k =1 b k I A k ( ω ) This gives us a piecewise-constant function on Ω . It also defines a random variable. Note that the collection need not be disjoint. However, we can shuffle things around to write the function as g ( ω ) = n * X k =1 b * k I A * k ( ω )
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ECE 6010: Lecture 2 – More on Random Variables 3 where the A * k s are disjoint, and where the b * k s are unique. Note that A * k = { ω Ω : g ( ω ) = b * k } Now note that E [ I A ] = P ( A ) Based on this, and the disjointness of the A * k , we can write E [ g ] = n * X i =1 b * k P ( A * k ) There are many instances where indicator functions are used to get a “handle” on the prob- ability of an event. Now we will get a bit more technical, dealing with some issues related to the existence of expectations. We have seen how to define expectations for simple functions (which are random variables). But what about more general random variables? Let X 0 be a random variable. We define E [ X ] = sup g simple X E [ g ] where the sup means “least upper bound”, and the limit is taken over all simple functions g satisfying g X . It can be shown that this limit always exists, though it may be infinite. There is thus no question of convergence or anything like that. Generalizing further, let X be an arbitrary r.v. Since the previous result holds for non- negative random variables, let us split X : X = X + - X - where X + is the positive part and X - is the negative part: X + = max( X ( ω ) , 0) 0 X - = - min( X ( ω ) , 0) 0 Now X + and X - have well-defined expectations. We take E [ X ] = E [ X + ] - E [ X - ] which is defined in every case except when E [ X + ] and E [ X - ] are both infinite (in which case the difference is undefined). Let us examine the expectation in light of the Riemann-Stieltjes integral. We define E [ X ] = Z -∞ xdF X ( x ) = lim a →-∞ ,b →∞ Z b a xdF X ( x ) This is a stronger sense of the limit than, for example Z -∞ xdF X ( x ) = lim a →∞ Z a - a xdF X ( x ) For example, sin( x ) has an integral in the latter sense (which is equal to 0), but not in the former sense. Now we will consider an example of a density where the expectation does not exist. Example 1 Cauchy density : f X ( x ) = 1 π (1 + x 2 ) It is straightforward to show that this satisfies the requirements for a p.d.f. It looks a lot like a Gaussian, but has “heavy tails.” It can be shown that a Cauchy r.v. can be obtained as a ratio of Gaussians: X = Y/Z . Now let us attempt to compute E [ X ] = lim a →-∞ ,b →∞ 1 π Z b a x 1 + x 2 dx
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ECE 6010: Lecture 2 – More on Random Variables 4 If a = - b , then the integral is zero, no matter what. If you fix a or b , taking the limit of the other one, the result is . That is, E [ X + ] exists and E [ X - ] exists (although both are ), but they can’t be subtracted. 2 Properties of Expectations 1. If X = c then E [ X ] = c .
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  • Fall '08
  • Stites,M
  • Probability theory, lim

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