Chapter 2 Transformations and Expectations. , and define f

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1 Revew for the prevous lecture Defnton: support set of a ranom varable, the monotone functon; Theorem: How to obtan a cf, pf (or pmf) of functons of a ranom varable; Eamples: several eamples Chapter Transformatons an Epectatons Chapter. Dstrbutons of Functons of a Ranom Varable Theorem..8: Let have a pf f ( ) an Y = g( ), an efne f ( ), let Y = g(), an efne the sample space = { : f > }. Suppose there ests a partton of, A, A,, Ak, such that P ( A ) = an f ( ) s contnuous on each A. Further, suppose there est functons g ( ),, ( ) gk, efne on A,, Ak, respectvely, satsfyng. g ( ) = g, for A.. g( ) s monotone on A. the set Y= { y: y= g for some A} s the same for each =,,, k, an v. g has a contnuous ervatve on Y, for each =,...,k, then.

2 f Y k f ( ( )) ( ) g y g y y Y = = y otherwse. Note: If any of the contons of Theorem..8 s not satsfe, t wll be very ffcult to fn the strbuton of Y = g( ). Eample..9 (Normal-ch square relatonshp): Let have a stanar normal strbuton,.e., / f = e, < <. π Show that Y = has pf fy e π y y / =, y < <. In ths case, the pf of Y s the pf of a ch-square ranom varable wth egree of freeom. Theorem.. (Probablty ntegral transform): Let have contnuous cf F an efne the ranom varable Y as Y = F ( ). Then Y s unformly strbute on (, ),.e., PY ( y) = y, < y <. Proof: Defne F = nf{ : F y} for < y <, then F s ncreasng (why?);

3 PY ( y) = PF ( ( ) y) = PF F F y F ( [ ( )] ( )) ( s ncreasng) = P F y ( ( )) = F F y F ( ( )) (efnton of ) = y. (contnuty of F ). Note: We requre that s a contnuous ranom varable. If s not contnuous, many propertes of n the proof o not hol anymore. F use Eample: Suppose that a cf n eample.5., what s Soluton: f < < /8 f < F = 4/8 f < 7/8 f < 3 f 3 < F. F 3 f 7/8< y f 4/8<y 7/8 = f /8<y 4/8 f <y /8 f y= Note: Theorem.. has a very mportant applcaton. If one s ntereste n generatng a ranom varable from a populaton wth cf F ( ), one only nees to generate a unform ranom number u, between an, an solve for n the equaton F = u. 3

4 Eample: Let be a contnuous ranom varable wth cf F = e. In ths case, we say that has an eponental strbuton. Then f U = F ( ), then then u = F = e =log( u). Secton. Epecte Values Defnton..: The epecte value or mean of a ranom varable g( ), enote by Eg( ), s Eg( ) = gf ( ) ( ) f s contnous gf ( ) = gp ( ) ( = ) f s screte, prove that the nterval or sum ests. If E g( ) =, we say that Eg( ) oes et. Eample.. (Eponental mean): Suppose has an eponental (λ ) strbuton, that s, t has pf gven by Fn the mean of. Soluton: f e λ / λ =, <, λ >. 4

5 / λ E = f = e λ / λ / λ / λ e e λe = + = = λ (ntegraton wth parts) Theorem (Integraton wth parts): If the functon u( ) an v( ) have ervatves, an v ( ) u ( ) an u ( ) v ( ) est, then we have u ( ) v ( ) = uv ( ) ( ) v ( ) u ( ). Eample..3 (Bnomal mean): If has a bnomal strbuton, ts pmf s gven by n n P ( = ) = p( p), =,,, n, p, where n s a postve nteger an for every fe par n an p the pmf sums to. Fn the mean or epecte value of a bnomal r.v.. Soluton: 5

6 n n n n E = P( = ) = p ( p) = = n n n = n p ( p) np p ( p) = = = n n y ( n) y = np p ( p) = np. y= y n n ( n) Eample..4 (Cauchy mean): A Cauchy r.v. has pf Fn E. Soluton: f = π +, < <. Thus M log( + ) M log( + M ) = = +. M log( + M ) M E = lmm lm π = =. + π Queston: Does the ntegral est? π + 6

7 Theorem..5: Let be a ranom varable an let a, b, an c be constants. Then for any functons g ( ) an g ( ) whose epectatons est, we have E( ag ( ) + bg ( ) + c) = aeg ( ) + beg ( ) + c. a. b. If g for all, then Eg ( ). c. If g g for all, then Eg( ) Eg( ).. If a g b for all, then. a Eg ( ) b. Proof: (screte case) a. b. E( ag ( ) + bg ( ) + c) = ( ag + bg + c) P( = ) = ag P( = ) + bg P( = ) + cp( = ) = aeg ( ) + beg ( ) + c. Eg ( ) = g P( = ). c. If g g for all, then g g for all, thus. Smlar as the proof of c. E( g g ) = Eg Eg. Eample..6 (Mnmzng stance): Fn the value b that mnmzes Soluton : E( ) b. 7

8 E( b) = E( E + E b) = E E + E E E b + E E b ( ) ( )( ) ( ), notce that E( E)( E b) = ( E b)( E E( E)) =, then we have E( b) = E( E) + E( E b), therefore, mn E ( b b ) E ( E ) Soluton : = when b E E( b) = E( be + b ) = E + b be = E + ( b E ) + ( E ). =. Eample..7 (Unform-eponental relatonshp II): Let have a unform (,) strbuton,.e., f ( ) = for < <. Defne g( ) = log. Fn Eg( ). Soluton : Usng the ensty of. Eg( ) = log = log + log =. Soluton : Usng the ensty of g( ) : ( ) ( ) f g( ) y = e = e, for < <. Ths s the specal case of eponental strbuton wth λ =. 8