p1soln09

# p1soln09 - STAT 330 SOLUTIONS Part I 1(a Starting with x=0...

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STAT 330 SOLUTIONS Part I 1( a ) Starting with X x =0 t x = 1 1 t , | t | < 1 it can be shown that X x =1 xt x = t (1 t ) 2 , | t | < 1 (1) X x =1 x 2 t x 1 = 1+ t (1 t ) 3 , | t | < 1 (2) X x =1 x 3 t x 1 = 1+4 t + t 2 (1 t ) 4 , | t | < 1 . (3) ( α ) 1 k = X x =1 x (0 . 3) x = 0 . 3 (1 0 . 3) 2 using (1) gives k = (0 . 7) 2 0 . 3 and therefore f ( x )=(0 . 49) x (0 . 3) x 1 ,x =1 , 2 ,... 1 2 3 4 5 6 7 8 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 graph of f(x) x f ( x ) 1

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( β ) F ( x )= 0 if x< 1 x P t =1 (0 . 49) t (0 . 3) t 1 for x =1 , 2 ,... Note that F ( x ) is speci f ed by indicating its value at each step point. ( γ ) E ( X X x =1 x (0 . 49) x (0 . 3) x 1 =(0 . 49) X x =1 x 2 (0 . 3) x 1 =( 0 . 7) 2 (1 + 0 . 3) (0 . 7) 3 using (2) = 13 7 E ¡ X 2 ¢ = X x =1 x 2 (0 . 49) x (0 . 3) x 1 . 49) X x =1 x 3 (0 . 3) x 1 0 . 7) 2 (1 + 4(0 . 3) + (0 . 3) 2 ) (0 . 7) 4 using (3) = 229 49 Var ( X E ( X 2 ) [ E ( X )] 2 = 229 49 μ 13 7 2 = 60 49 ( δ ) P (0 . 5 <X 2) = P ( X =1)+ P ( X =2) =0 . 49 + (0 . 49)(2)(0 . 3) = 0 . 784 P ( X> 0 . 5 | X 2) = P ( 0 . 5 ,X 2) P ( X 2) = P ( X P ( X P ( X 2) 2
1 . ( c )( α ) 1 k = Z −∞ f ( x ) dx = Z −∞ 1 1+ x 2 dx =2 Z 0 1 x 2 dx because of symmetry = 2 lim b →∞ arctan ( b )=2 ³ π 2 ´ = π and thus k =1 / π . -4 -3 -2 -1 0 1 2 3 4 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 x f ( x ) graph of f(x) Figure 1: ( β ) F ( x )= x Z −∞ 1 t 2 dt = 1 π lim b →−∞ arctan ( t ) | x b ¸ = 1 π arctan ( x )+ 1 2 , −∞ <x< . ( γ ) Consider Z 0 x x 2 dx = lim b →∞ 1 2 ln ¡ b 2 ¢ =+ . Since the integral R 0 x 1+ x 2 dx does not converge, the integral 1 π R −∞ x 1+ x 2 dx does not converge absolutely and E ( X ) does not exist. Since E ( X ) does not exist Var ( X ) does not exist. 3

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( δ ) P (0 . 5 <X 2) = F (2) F (0 . 5) = 1 π [arctan (2) arctan (0 . 5)] 0 . 2048 P ( X> 0 . 5 | X 2) = P ( 0 . 5 ,X 2) P ( X 2) = P (0 . 5 2) P ( X 2) = F (2) F (0 . 5) F (2) = arctan (2) arctan (0 . 5) arctan (2) + π 2 0 . 2048 0 . 8524 0 . 2403 . 4
1 . ( d )( α ) 1 k = Z −∞ f ( x ) dx = Z −∞ e | x | dx =2 Z 0 e x dx Γ (1)=2(0 !)=2 and thus k = 1 2 -5 -4 -3 -2 -1 0 1 2 3 4 5 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 graph of f(x) x f ( x ) ( β ) F ( x )= x R −∞ 1 2 e t dt if x 0 0 R −∞ 1 2 e t dt + x R 0 1 2 e t dt if x> 0 = ( 1 2 e x if x 0 1 2 + ¡ 1 2 e t | x 0 ¢ =1 1 2 e x if 0 ( γ ) Since the improper integral R 0 xe x dx = Γ (2) = 1! = 1 converges, the integral 1 2 R −∞ xe | x | dx converges absolutely and by the symmetry of f ( x ) we have E ( X )=0 . E ¡ X 2 ¢ = 1 2 Z −∞ x 2 e | x | dx = μ 1 2 (2) Z 0 x 2 e x dx = Γ (3) = 2! = 2 Var ( X E ( X 2 ) [ E ( X )] 2 0 2 5

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( δ ) P (0 . 5 <X 2) = F (2) F (0 . 5) = 1 2 ( e 0 . 5 e 2 ) 0 . 2356 P ( X> 0 . 5 | X 2) = P ( 0 . 5 ,X 2) P ( X 2) = P (0 . 5 2) P ( X 2) = F (2) F (0 . 5) F (2) = 1 2 ( e 0 . 5 e 2 ) 1 1 2 e 2 0 . 2527 1 . ( e )( α ) 1 k = Z −∞ f ( x ) dx = 1 Z 0 (1 x ) 5 dx = 1 6 (1 x ) 6 | 1 0 = 1 6 and thus k =6 . 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 1 2 3 4 5 6 graph of f(x) x f ( x ) ( β ) Since F ( x )=6 x Z 0 (1 u ) 5 du = (1 u ) 6 | x 0 =1 (1 x ) 6 , 0 <x< 1 therefore F ( x )= 0 if x 0 1 (1 x ) 6 if 0 1 1 if x 1 6
E ( X 2 )=6 1 Z 0 x 2 (1 x ) 5 dx let u =1 x, x u, du = dx = 6 0 Z 1 (1 u ) 2 u 5 du =6 1 Z 0 ¡ u 5 2 u 6 + u 7 ¢ du μ 1 6 u 6 2 7 u 7 + 1 8 u 8 | 1 0 12 7 + 3 4 = 1 28 Var ( X )= E ( X 2 ) [ E ( X )] 2 = 1 28 μ 1 7 2 = 3 196 ( δ ) P (0 . 5 <X 2) = F (2) F (0 . 5) h 1 (1 0 . 5) 6

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## This note was uploaded on 02/22/2010 for the course STAT 330 taught by Professor Paulasmith during the Fall '08 term at Waterloo.

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p1soln09 - STAT 330 SOLUTIONS Part I 1(a Starting with x=0...

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