Method: Step 1: Step 2: Find f. Step 3: = Y dy. Solution: 0, ( ) 0, y. Assume
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1 Functions of Rndom Vrible Problem Sttement We know the pdf ( or cdf ) of rndom vrible. Define new rndom vrible Y g( ) ). Find the pdf of Y. Method: Step : Step : Step 3: Plot Y g( ). Find F ( ) b mpping the event Y to proper event in.. Find f Y Y d F = Y d. Liner Function Let Y b. Assume is continuous rndom vrible nd 0. Solution: For 0, FY ( ) PY [ ] b P [ ] ( b F ) For 0, FY ( ) PY [ ] P [ b F ( ) b ]
2 b For 0, FY F fy FY b F b f ( x) dx b b f f ( ) f ( x) dx b b f f 0 0dx b f b For 0, FY F fy FY b F b f ( x) dx b b f b f For n 0, b fy( ) f b
3 3 Moments of Liner Function Y b b b VAR( Y ) Y Y b b VAR( ) Exmple: is Gussin rv with men nd vrince. Y. Find the pdf of Y. b For n 0, fy( ) f f substituting nd b x Since f x e, 4 fy ( ) e e e Y VAR( Y ) 4
4 4 Homework is uniform rv over (0,). Y. Find the pdf of Y. Note. When the pdf is discontinuous, don't tr to use n formul. Find the cdf step b step.
5 5 Polnomil Function Consider Y Find the pdf of Y in terms of the pdf of. Solution: For 0, F ( ) P[ Y ] Y P[ ] F F 0 FY ( ) 0 0 fy FY F F for 0 f ( x) dx f f ( ) f( x) dx f f 0 dx f Y ( ) f, for 0. f
6 6 Homework: chi-squre is N(0,). Y. Find the pdf of Y. Note. When the distribution hs no discontinuit, it is sfe to use the formul. Ans: fy ( ) e 0 Y is clled chi-squre rndom vrible with degree of freedom. The degree of freedom indictes the number of independent Gussin rndom vribles dded fter squring.
7 7 Exponentil Function Y e. Find the pdf of Y in terms of the pdf of. Solution. For 0, Y For 0, F ( ) 0. F ( ) P[ Y ] Y P [ ln ] F fy FY ln (ln ) F ln for 0 f ( x) dx ln f ln ln f ( ) f( x) dx ln f ln f 0 0 dx Ans. fy( ) f(ln ) 0
8 8 Exmple. Log-Norml Distribution is N,. Find the pdf of Y e Ans. fy( ) f(ln ) 0. x f ( ) x e, ln f ( ) Y e 0. Y is referred to s log-norml rndom vrible.
9 9 Normliztion For n rndom vrible, discrete or continuous, with men nd vrince, its normrlized version, 0, hs zero men nd unit vrince. Recll b b VAR( b) VAR( ). 0 0, hs zero men. VAR 0 VAR VAR VAR =, hs unit vrince.
10 0 Exmple. Normlize Uniform RV is U(0,). Normlize. Recll for U, b, b nd ( b ) VAR( ). nd VAR. Therefore 0 is the normlized version of.
11 Exmple. Normlize Gussin Consider N(, ). Normlize. Solution. nd VAR. 0 is the normlized version of. Recll, for Y b, b fy( ) f. In this exmple, nd b. b. x f ( ) x e, f ( ) e 0 e normlized Gusssun pdf
12 Generting Rndom Smples with Computer We wnt to generte rndom number ccording to some pdf Most computers hve built-in routine too generte U (0,). f ( x ). We wnt to convert the rndom number from U (0,) to f ( x). Follow these steps: i) Plot the cdf F ( x). ii) Generte u from U 0,. iii) Findd r tht stisfies u F r F u R obtined from the bove procedure hs pdf () x nd cdf F ( x). ) f () r r Wh does it work? Define new rndom vriblee R F ( U ). Then F () r P R r Rnd R P U F () r F () r hve the sme cdf. - U is uniform rndom vrible in the intervl 0,
13 3 Exmple. Rndom Smples from Exponentil Distribution Generte smples of n exponentil rndom vrible. Solution. Let U be uniform rv within intervl (0,). x For n exponentil rndom vrible with cdf F ( x) e for x0, write R U F ( R) e. ln U Then R Use R s the exponentil rndom smple.
14 4 Probbilit Distribution of Rndom Smples Estimte the cdf from given rndoml generted smples. Suppose we hve three smples: 3,, 8. We ssume the re equll likel. Order the smples in n scending vlue:, 3, 8. The estimted cdf is the lines connecting 3 3 0,0,,, 3,, 8, Exmple. Rndom Smpless from Exponentil Distributio on Exponentil Distribution with men 30. Number of smples: n 0,00,000, nd 0000.
15 5
16 6 Rndom Smples for Discrete Distributions Follow the sme steps s continuous RVs..0 p 3 u cdf p 0.0 p r r r 3
17 7 Bounds on Probbilities Chebshev Inequlit For n rndom vrible with known men nd vrince, VAR( ) P[ ] for n positive rel number Proof. Consider function, u w ( u) 0, otherwise Then w( ) w( x) f ( x) dx Note Therefore w ( ) P u w ( u) for ll u. VAR( ) u VAR( ) P[ ] u
18 8 Exmple 8 Suppose 0 nd VAR( ) 0. Then the Chebshev Inequlit shows P 0 P 0 0 is in the form P 8 8 P 0 0 is in the form P Note. It is not necessril true tht P 0 0 = P 0 0. Exmple Consider ~ U(0,). b b Then /, / Then the Chebshev Inequlit would s P. 4 For / 4, the upper bound is, which is useless. 3 The Chebshev inequlit is frequentl used for proving theorems. However it does not produce tight bound [ref. chernoff bound]. Homework. Chebshev Inequlit for Gussin RV N(, 0.). Using the Chebshev inequlit, find n upper bound to P [ 0].
19 9 Mrkov Inequlit For n nonnegtive rndom vrible, E P 0 Proof. Define n indictor function I I x x 0 otherwise x s Then = P I However x I x for n x0 which ss I
Problem. Statement. variable Y. Method: Step 1: Step 2: y d dy. Find F ( Step 3: Find f = Y. Solution: Assume
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