7  Continuous random variables


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1 71 Continuous rndom vribles S. Lll, Stnford Continuous rndom vribles Continuous rndom vribles The cumultive distribution function The uniform rndom vrible Gussin rndom vribles The Gussin cdf Collecting dt Induced probbility density functions Exmple: liner trnsformtions Exmple: noninvertible trnsformtions Simultion of rndom vribles
2 72 Continuous rndom vribles S. Lll, Stnford Continuous rndom vribles A continuous rndom vrible x : Ω R is specified by its (induced) cumultive distribution function (cdf) F x (z) = Prob(x z) F() 1 Then we hve Prob(x [, b]) = F x (b) F x ( )
3 73 Continuous rndom vribles S. Lll, Stnford Properties of the cumultive distribution function F() 1 F x () 0 for ll F x is nondecresing function F x () F x (b) if b F x is right continuous, i.e,. lim z + F x (z) = F x ()
4 74 Continuous rndom vribles S. Lll, Stnford Properties of the cumultive distribution function If F x is differentible, then the induced probbility density function (pdf) is p x (z) = df x (z) dz then Prob(x [, b]) = b p x (z) dz Notice tht p x (z) is not probbility; it my be greter thn 1. We use nottion x p x to men x is rndom vrible with pdf p x
5 75 Continuous rndom vribles S. Lll, Stnford Properties of the cumultive distribution function If x is discrete rndom vrible, then F is just stircse function 1 F() The corresponding probbility density function is sum of δ functions.
6 76 Continuous rndom vribles S. Lll, Stnford The uniform rndom vrible The uniform rndom vrible x U[, b] hs pdf p x (z) = 1 b if z b 0 otherwise
7 77 Continuous rndom vribles S. Lll, Stnford Gussin rndom vribles The rndom vrible x is Gussin if it hs pdf p(x) = 1 σ (x µ) 2 2π e 2σ 2 write this s x N(µ, σ 2 ) the men or expected vlue of x is E(x) = xp(x) dx = µ the vrince of x is E ( (x µ) 2) = (x µ) 2 p(x) dx = σ 2
8 78 Continuous rndom vribles S. Lll, Stnford Gussin rndom vribles pdf for x N(0, 1) is p() p is symmetric bout the men decys very fst; but p(x) > 0 for ll x
9 79 Continuous rndom vribles S. Lll, Stnford Computing probbilities for Gussin rndom vribles The error function is erf(x) = 2 π x 0 e t2 dt The Gussin CDF is F N () = ( ) µ 2 erf σ 2 When µ = 0 nd σ = 1, F N ()
10 710 Continuous rndom vribles S. Lll, Stnford Computing probbilities for Gussin rndom vribles so for x N(0, σ 2 ) we hve for 0 ( ) Prob(x [, ]) = erf σ 2 Some prticulr vlues: Prob(x [ σ, σ]) 0.68 Prob(x [ 2σ, 2σ]) Prob(x [ 3σ, 3σ])
11 711 Continuous rndom vribles S. Lll, Stnford Collecting dt For discrete rndom vribles, we cn collect dt nd count the frequencies of outcomes This converges to the true pmf. The nlogous procedure for continuous rndom vribles uses the cumultive distribution function. Suppose S = {z 1,..., z n } re n smples of relvlued rndom vrible. Let F() be the frction of smples less thn or equl to, given by F() = {z S z } n F is piecewise constnt function, clled the empiricl cdf
12 712 Continuous rndom vribles S. Lll, Stnford Exmple: collecting dt Suppose x N(0, 1). The plots below show 25 nd 250 dt points, respectively
13 713 Continuous rndom vribles S. Lll, Stnford Induced probbility density Suppose we hve x : Ω R is rndom vrible with induced pdf p x : R R. y is function of x, given by y = g(x) Wht is the induced pdf of y? The key ide is tht we need to chnge vribles for integrtion of probbilities. Recll the following. If f nd h re continuous, then h(b) h() f(x) dx = b f ( h(y) ) h (y) dy
14 714 Continuous rndom vribles S. Lll, Stnford Induced probbility density Assume g is continuous, nd g is strictly incresing; i.e., if < b then g() < g(b) This implies tht g is invertible, i.e., for every y there is unique x such tht y = g(x). We would like to find the pdf of y is p y, which stisfies for b, Prob(y [, b]) = b p y (y) dy We lso know tht this probbility is Prob(y [, b]) = Prob(g(x) [, b]) = Prob ( x [g 1 (), g 1 (b)] ) since g is incresing = g 1 (b) g 1 () p x (x) dx
15 715 Continuous rndom vribles S. Lll, Stnford Induced probbility density We hve b p y (y) dy = g 1 (b) g 1 () p x (x) dx Now we cn pply the chnge of vribles x = h(y) to the integrl on the right hnd side, where h = g 1. We hve h 1 (y) = g ( g 1 (y) ) becuse g(h(y)) = y, so d dy g(h(y)) = 1, i.e., g (h(y))h (y) = 1 Therefore, by the chnge of vribles formul b p y (y) dy = b p x( g 1 (y) ) g ( g 1 (y) ) dy
16 716 Continuous rndom vribles S. Lll, Stnford Induced probbility density Since this holds for ll nd b, we hve the following. If y = g(x), nd g is strictly incresing with g continuous, then the pdf of y is p y (y) = px( g 1 (y) ) g ( g 1 (y) ) More generlly, if g (x) 0 for ll x, then p y (y) = px( g 1 (y) ) g ( g 1 (y) )
17 717 Continuous rndom vribles S. Lll, Stnford Exmple: liner trnsformtions Suppose x : Ω R, nd y = αx + β y b g à1 () g à1 (b) x We hve p y (y) = 1 α px ( ) y β α
18 718 Continuous rndom vribles S. Lll, Stnford Noninvertible trnsformtions Wht hppens when g is not invertible? e.g., when y = x 2 Prob(y [, b]) = Prob(x 2 [, b]) y b = Prob(x [ b, ]) + Prob(x [, b]) = = b b p x (x) dx + p x ( y) 2 y dy + b p x (x) dx b p x ( y) 2 y dy y = x 2 = p y (y) = 1 ( 2 p x ( y) + p x ( ) y) y x
19 719 Continuous rndom vribles S. Lll, Stnford Simultion of rndom vribles We re given F : R [0, 1] We would like to simulte rndom vrible y so tht it hs cumultive distribution function F We hve source of uniform rndom vribles x U[0, 1] To construct y, set y = F 1 (x) Becuse Prob(y ) = Prob(F 1 (x) ) = Prob(x F()) = F() This works when F is invertible nd continuous
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