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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1 Functions of Rndom Vrible Problem Sttement We know the pdf ( or cdf ) of rndom r vrible. Define new rndom vrible Y = g. Find the pdf of Y. Method: Step : Step : Step 3: Plot Y = g( ). Find F ( y) by mpping the event Y y to proper event in. Find f Y Y ( y d dy ) F ( = Y y ). Liner Function Let Y = + b. Assume is continuous rndom vrible nd 0. Solution: For > 0, FY ( y ) = PY [ y] y b = P [ ] y b = F For < 0, FY ( y ) y b = PY [ y] = P[ ] y b = F
2 y b For > 0, FY ( y) = F fy ( y) = FY ( y) y yb = F y = y yb f ( x) dx yb yb yb = f f ( ) + f( x) dx y y y yb y b = f f 0+ 0dx y b = f y b For < 0, FY ( y) = F fy ( y) = FY ( y) y yb = F y = f ( x) dx y yb yb yb y =f y b = f For ny 0, y b fy( y) = f
3 3 Moments of Liner Function Men: Y = + b = + b = + b Vrince: VAR( Y ) = Y Y ( b b) = + = = = VAR( ) Exmple: is Gussin rv with men μ nd vrince σ. Y = +. Find the pdf of Y. Solution. y b For ny 0, fy( y) = f y = f substituting = nd b= ( xμ ) Since f x = e σ, πσ y μ ( y+ μ ) 4 fy ( y) = e σ = e σ = e πσ πσ π σ ( y ( μ )) ( σ ) Y = μ + VAR( Y ) = 4σ
4 4 Homework is uniform rv over (0,). Y = +. Find the pdf of Y. Note. When the pdf is discontinuous, don't try to use ny formul. Find the cdf step by step.
5 5 Polynomil Function Consider Y = Find the pdf of Y in terms of the pdf of. Solution: For y 0, F ( y) = P[ Y y] Y = P[ y y] F y F y y 0 FY ( y) = 0 y < 0 fy y = FY y y { } = F y F y for y 0 y = y y y f ( x) dx y = f y y f y ( y) + f( x) dx y y y y y = f y + f y + 0 dx y y y { } fy( y) = f y + f y, for y 0. y
6 6 Homework: chi-squre is N(0,). Y =. Find the pdf of Y. Note. When the distribution hs no discontinuity, it is sfe to use the formul. Answer: y fy ( y) = e y 0 π y 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 y 0, Y For y > 0, F ( y) = 0. F ( y) = P[ Y y] Y = P [ ln y] = F fy y = FY y y ln y (ln y) = F ( ln y) for y 0 y = y f ( x) dx ln y = f ( ln ) ( ln ) y y f ( ) + f( x) dx y y y ln y = f ( ln ) y + f dx y Ans. fy( y) = y f(ln y) y > 0
8 8 Exmple. Log-Norml Distribution is N μσ,. Find the pdf of Y = e Ans. fy( y) = f(ln y) y> 0. y ( xμ) Since f x = e σ, πσ ( ln yμ ) f Y y = e σ y> 0. y πσ Y is referred to s log-norml rndom vrible.
9 9 Normliztion For ny rndom vrible, discrete or continuous, with men μ nd vrince σ, its normrlized version, μ 0, σ hs zero men nd unit vrince. Recll + b = + b VAR( + b) = VAR( ). 0 μ = σ μ = σ σ = 0, hs zero men. VAR 0 μ = VAR σ = VAR σ = VAR σ =, 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( μσ, ). Normlize. Solution. = μ nd VAR = σ. 0 μ = is the normlized version of. σ Recll, for Y = + b, y b fy( y) = f. In this exmple, μ = nd b=. σ σ y b = σy + μ. ( xμ ) Since f x = e σ, πσ f ( y) = σ e 0 πσ y ( σy+ μμ) σ = e < y< π normlized Gusssun pdf
12 Generting Rndom Smples with Computer We wnt to genertee rndom number n ccording to some pdf f ( x). Most computers hve built-in routine r to generte U (0,). 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 iii) Find r tht stisfies u = F ( r) r r = F u 0,. R obtined from the bove procedure hs pdf () x nd cdf F ( x). ) f r Why does it work? Define new rndom vrible R = F - ( U). Then F () r = P R r Rnd R hve the sme cdf. [ = P U F () r = F () r ] U is uniform rndom vri ble in the in ntervl ( 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,). λx For n exponentil rndom vrible with cdf F ( x) = e for x 0, write λr U = F ( R) = e. ln ( U ) Then R = λ Use R s the exponentil rndom smple.
14 4 Probbility Distribution of Rndom Smples Estimte the cdf of non-negtive rv from rndomly generted smples. Supposee we hve three smples: 3,, 8. We ssume they re eqully likely. Order the smples in n scending vlue:, 3, 8. The estimted cdf is the lines connecting 0,0, ( 3),, 3,, 3 ( 8,) Exmple. Rndom Smples from Exponentil Distribution Exponentil Distribution with men 30. λ = 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 Chebyshev Inequlity For ny rndom vrible with known men nd vrince, VAR( ) P[ ε] for ny positive rel number ε ε Proof. Consider function, u ε wε ( u) 0, otherwise Then wε = wε( x) f ( x) dx = P ε u Note Therefore ( u ) wε ( u) for ll u. ε ( ) VAR( ) wε = ε ε Finlly we hve VAR( ) P[ ε ] ε u
18 8 Exmple 8 Suppose = 0 nd VAR( ) = 0. Then the Chebyshev Inequlity shows P 0 = P 0 0 is in the form P ε 8 8 P 0 0 is in the form P ε 0 = Note. It is not necessrily true tht P 0 0 = P 0 0. Exmple Consider ~ U(0,). ( b) + b Then = = /, σ = = / Then the Chebyshev Inequlity would sy P ε. ε 4 For ε = / 4, the upper bound is, which is useless. 3 The Chebyshev inequlity is frequently used for proving theorems. However it does not produce tight bound [ref. chernoff bound]. Homework. Chebyshev Inequlity for Gussin RV N(, 0.). Using the Chebyshev inequlity, find n upper bound to P [ 0].
19 9 Mrkov Inequlity For ny 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 ny x 0 which sys I
Method: Step 1: Step 2: Find f. Step 3: = Y dy. Solution: 0, ( ) 0, y. Assume
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