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Unformatted text preview: DISCRETE AND CONTINUOUS PROBABILITY DISTRIBUTIONS Probability mass functions If x ∈ { x 1 , x 2 , x 3 , . . . } is discrete, then a function f ( x i ) giving the probability that x = x i is called a probability mass function. Such a function must have the properties that f ( x i ) ≥ , for all i, and X i f ( x i ) = 1 . Example. Consider x ∈ { , 1 , 2 , 3 , . . . } with f ( x ) = (1 / 2) x +1 . It is certainly true that f ( x i ) ≥ 0 for all i . Also, X i f ( x i ) = Ω 1 2 + 1 4 + 1 8 + 1 16 + · · · æ = 1 . To see this, we may recall that 1 1 − θ = { 1 + θ + θ 2 + θ 3 + · · ·} , whence θ 1 − θ = { θ + θ 2 + θ 3 + + θ 4 + · · ·} . Setting θ = 1 / 2 in the expression above gives θ/ (1 − θ ) = 1 2 / (1 − 1 2 ) = 1 , which is the result that we are seeking. 1 Probability density functions If x is continuous, then a probability density function (p.d.f.) f ( x ) may be defined such that the probability of the event a < x ≤ b is given by P ( a < x ≤ b ) = Z b a f ( x ) dx. Notice that, when b = a , there is P ( x = a ) = Z a a f ( x ) dx = 0 . That is to say, the integral of the continuous function f ( x ) at a point is zero. The value of f ( x ) at a point is described as a probability measure as opposed to a probability. Example. Consider the exponential function f ( x ) = 1 a e − x/a defined over the interval [0 , ∞ ) and with α > 0. There is f ( x ) > 0 and Z ∞ 1 a e − x/a dx = h − e − x/a i ∞ = 1 . Therefore, this function constitutes a valid p.d.f. The exponential distribution provides a model for the lifespan of an electronic component, such as fuse, for which the probability of failing is liable to be independent of how long it has already survived....
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This note was uploaded on 03/02/2012 for the course EC 2019 taught by Professor D.s.g.pollock during the Spring '12 term at Queen Mary, University of London.
 Spring '12
 D.S.G.Pollock

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