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Functions of Random Variables. Method of Distribution Functions X 1,…,X n ~ f(x 1,…,x n ) U=g(X 1,…,X n ) – Want to obtain f U (u) Find values in (x 1,…,x.

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Presentation on theme: "Functions of Random Variables. Method of Distribution Functions X 1,…,X n ~ f(x 1,…,x n ) U=g(X 1,…,X n ) – Want to obtain f U (u) Find values in (x 1,…,x."— Presentation transcript:

1 Functions of Random Variables

2 Method of Distribution Functions X 1,…,X n ~ f(x 1,…,x n ) U=g(X 1,…,X n ) – Want to obtain f U (u) Find values in (x 1,…,x n ) space where U=u Find region where U≤u Obtain F U (u)=P(U≤u) by integrating f(x 1,…,x n ) over the region where U≤u f U (u) = dF U (u)/du

3 Example – Uniform X Stores located on a linear city with density f(x)= ≤ x ≤ 10, 0 otherwise Courier incurs a cost of U=16X 2 when she delivers to a store located at X (her office is located at 0)

4 Example – Sum of Exponentials X 1, X 2 independent Exponential(  ) f(x i )=  -1 e -xi/  x i >0,  >0, i=1,2 f(x 1,x 2 )=  -2 e -(x1+x2)/  x 1,x 2 >0 U=X 1 +X 2

5 Method of Transformations X~f X (x) U=h(X) is either increasing or decreasing in X f U (u) = f X (x)|dx/du| where x=h -1 (u) Can be extended to functions of more than one random variable: U 1 =h 1 (X 1,X 2 ), U 2 =h 2 (X 1,X 2 ), X 1 =h 1 -1 (U 1,U 2 ), X 2 =h 2 -1 (U 1,U 2 )

6 Example f X (x) = 2x 0≤ x ≤ 1, 0 otherwise U=10+500X (increasing in x) x=(u-10)/500 f X (x) = 2x = 2(u-10)/500 = (u-10)/250 dx/du = d((u-10)/500)/du = 1/500 f U (u) = [(u-10)/250]|1/500| = (u-10)/ ≤ u ≤ 510, 0 otherwise

7 Method of Conditioning U=h(X 1,X 2 ) Find f(u|x 2 ) by transformations (Fixing X 2 =x 2 ) Obtain the joint density of U, X 2 : f(u,x 2 ) = f(u|x 2 )f(x 2 ) Obtain the marginal distribution of U by integrating joint density over X 2

8 Example (Problem 6.11) X 1 ~Beta(  X 2 ~Beta(  Independent U=X 1 X 2 Fix X 2 =x 2 and get f(u|x 2 )

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10 Method of Moment-Generating Functions X,Y are two random variables CDF’s: F X (x) and F Y (y) MGF’s: M X (t) and M Y (t) exist and equal for |t| 0 Then the CDF’s F X (x) and F Y (y) are equal Three Properties: –Y=aX+b  M Y (t)=E(e tY )=E(e t(aX+b) )=e bt E(e (at)X )=e bt M X (at) –X,Y independent  M X+Y (t)=M X (t)M Y (t) –M X1,X2 (t 1,t 2 ) = E[e t1X1+t2X2 ] =M X1 (t 1 )M X2 (t 2 ) if X 1,X 2 are indep.

11 Sum of Independent Gammas

12 Linear Function of Independent Normals

13 Distribution of Z 2 (Z~N(0,1))

14 Distributions of and S 2 (Normal data)

15 Independence of and S 2 (Normal Data) Independence of T=X 1 +X 2 and D=X 2 -X 1 for Case of n=2

16 Independence of and S 2 (Normal Data) P2 Independence of T=X 1 +X 2 and D=X 2 -X 1 for Case of n=2 Thus T=X1+X2 and D=X2-X1 are independent Normals and & S 2 are independent

17 Distribution of S 2 (P.1)

18 Distribution of S 2 P.2

19 Summary of Results X 1,…X n ≡ random sample from N(  2 )  population In practice, we observe the sample mean and sample variance (not the population values: ,  2 ) We use the sample values (and their distributions) to make inferences about the population values

20 Order Statistics X 1,X 2,...,X n  Independent Continuous RV’s F(x)=P(X≤x)  Cumulative Distribution Function f(x)=dF(x)/dx  Probability Density Function Order Statistics: X (1) ≤ X (2) ≤...≤ X (n) (Continuous  can ignore equalities) X (1) = min(X 1,...,X n ) X (n) = max(X 1,...,X n )

21 Order Statistics

22 Example X 1,...,X 5 ~ iid U(0,1) (iid=independent and identically distributed)

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24 Distributions of Order Statistics Consider case with n=4 X (1) ≤x can be one of the following cases: Exactly one less than x Exactly two are less than x Exactly three are less than x All four are less than x X (3) ≤x can be one of the following cases: Exactly three are less than x All four are less than x Modeled as Binomial, n trials, p=F(x)

25 Case with n=4

26 General Case (Sample of size n)

27 Example – n=5 – Uniform(0,1)

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