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Unformatted text preview: PROBABILITY DISTRIBUTIONS: (continued) The change of variables technique. Let x ∼ f ( x ) and let y = y ( x ) be a monotonic transformation of x such that x = x ( y ) exists. Let A be an event defined in terms of x , and let B be the equivalent event defined in terms of y such that if x ∈ A , then y = y ( x ) ∈ B and vice versa. Then, P ( A ) = P ( B ) and we can find the the p.d.f of y denoted by g ( y ). The continuous case. If x,y are continuous random variables, then Z y ∈ B g ( y ) dy = Z x ∈ A f ( x ) dx. If we write x = x ( y ) in the second integral, then the change of variable technique gives Z y ∈ B g ( y ) dy = Z x ∈ B f { x ( y ) } dx dy dy. If y = y ( x ) is a monotonically decreasing transformation, then dx/dy < 0, and f { x ( y ) } > 0, and so f { x ( y ) } dx/dy < 0 cannot represent a p.d.f since g ( y ) ≥ 0 is necessary. The recourse is to change the sign on the yaxis. When dx/dy < 0, it is replaced by its modulus  dx/dy  > 0. In general, g ( y ) = f { x ( y ) } Ø Ø Ø Ø dx dy Ø Ø Ø Ø . 1 The Standard Normal Distribution. Consider the function g ( z ) = e − z 2 / 2 , where −∞ < z < ∞ . There is g ( z ) > 0 for all z and also Z e − z 2 / 2 dz = 1 √ 2 π . It follows that f ( z ) = 1 √ 2 π e − z 2 / 2 constitutes a p.d.f., described as the standard normal and denoted by N ( z ; 0 , 1). In general, the normal distribution is denoted by N ( x ; μ,σ 2 ); so, in this case, there are μ = 0 and σ 2 = 1. The General Normal Distribution. This can be derived via the change of variables technique. Let z ∼ N (0 , 1) = 1 √ 2 π e − z 2 / 2 = f ( z ) , and let y = zσ +...
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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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