Lecture 2 math 2

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Let X1, X2, ..., Xn be independent and uniformly distributed on [0, 1]. Let Yi be the ith largest element in {X1,X2,...,Xn}. So, Y1 = max(X1,X2,...,Xn) y2 is the second largest Xi value, and (The probability of ties is zero, so don’t worry about that.) (a) Find P(Y1 0.5) in terms of n. (b) Find P(Y2 0.5) in terms of n. (c) Find P(Yi 0.5) in terms of n and i. Solution F(x)=(x-a)/(b-a); a<=x<=b. (a): P[Y1>=.5]=P[max(X1,X2,X3,......,Xn)>=.5]=P[X1>=.5]P[X2>=.5]P[X3>=.5]......P[Xn>=.5]=(1- P[X1<=.5])(1-P[X2<=.5])......(1-P[Xn<=.5])=[1-F.5(X)]n=[1-((.5-0)/(1-0))]n=(.5)n. (b):P[Y2>=.5]=P[max(X1,X2,X3,......,Xn-1)>=.5]=P[X1>=.5]P[X2>=.5]P[X3>=.5]......P[X(n- 1)>=.5]=(1-P[X1<=.5])(1-P[X2<=.5])......(1-P[X(n-1)<=.5])=[1-F.5(X)]n-1=[1-((.5-0)/(1-0))]n- 1=(.5)n-1. (c):P[Yi>=.5]=P[max(X1,X2,......,Xn)>=.5]=[1-F.5(X)]n-i=[1-((.5-0)/(1-0))]n-i=(.5)n- i..

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2. Y = XSin-1 (X )

3. Y = Tan-1 (Ln(x))

5. Y’ = Sin-1X .1 - X . 11 - X 2 . 1/2X -1/2 [Sin-1(X )]2

6. Y’ = X1+(Ln(x))2 . 1X

8. Y = eTan-1(X)

9. Y = Ln(1eX)

10.

11. eTan-1X.11 + X2

12. ln eX-12= -12ln eX Y'= -12 1eX . eX = -1/2

13.

14.

15.

16.

17. XX2+1 dx

18. TanX . Sec(X)1+Sec(X) dx

19.

20. 12 LnX2+1+c

21. Ln 1+SecX+c

22.

23. 11-X2 dx

24. 41 +X2 dx

25.

26. Tan-1 (X) + c

27. 4 Tan-1 (X) + c

28. 12 Sin-1 (2X) + c

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