p1soln10 - STAT 330 SOLUTIONS Part I 1(a Starting with P x...

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STAT 330 SOLUTIONS Part I 1( a ) Starting with P x =0 θ x = 1 1 θ , | t | < 1 it can be shown that P x =1 x θ x = θ (1 θ ) 2 , | t | < 1 (1) P x =1 x 2 θ x 1 = 1+ θ (1 θ ) 3 , | t | < 1 (2) P x =1 x 3 θ x 1 = 1+4 θ + θ 2 (1 θ ) 4 , | t | < 1 . (3) ( i ) 1 k = P x =1 x θ x = θ (1 θ ) 2 using (1) gives k = (1 θ ) 2 θ and therefore f ( x )=(1 θ ) 2 x θ x 1 ,x =1 , 2 , ... ;0 < θ < 1 . The graph of f ( x ) below is for θ =0 . 3 . 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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( ii ) F ( x )= 0 if x< 1 x P t =1 (1 θ ) 2 t θ t 1 for x =1 , 2 , ... Note that F ( x ) is speci f ed by indicating its value at each jump point. It can also be shown (not necessary) that x P t =1 (1 θ ) 2 t θ t 1 (1 + x x θ ) θ x for x , 2 ,... ( iii ) E ( X P x =1 x (1 θ ) 2 x θ x 1 =(1 θ ) 2 P x =1 x 2 θ x 1 =( 1 θ ) 2 (1 + θ ) (1 θ ) 3 using (2) = (1 + θ ) (1 θ ) E ¡ X 2 ¢ = P x =1 x 2 (1 θ ) 2 x x 1 θ ) 2 P x =1 x 3 θ x 1 1 θ ) 2 (1 + 4 θ + θ 2 ) (1 θ ) 4 using (3) = (1 + 4 θ + θ 2 ) (1 θ ) 2 Var ( X E ( X 2 ) [ E ( X )] 2 = (1 + 4 θ + θ 2 ) (1 θ ) 2 (1 + θ ) 2 (1 θ ) 2 = 2 θ (1 θ ) 2 ( iv ) Using θ =0 . 3 , P (0 . 5 <X 2) = P ( X =1)+ P ( X =2) . 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
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1 . ( b )( i ) 1 k = Z −∞ 1 1+( x/ θ ) 2 dx =2 Z 0 1 x/ θ ) 2 dx because of symmetry θ Z 0 1 1+ y 2 dy let y = 1 θ x, θ dy = dx θ lim b →∞ arctan ( b )=2 θ ³ π 2 ´ = πθ . Thus k = 1 πθ and f ( x )= 1 h x/ θ ) 2 i ,x < , θ > 0 . The graphs for θ =0 . 5 , 1 and 2 are plotted below. The graph for each di f erent value of θ is obtained from the graph for θ =1 by simply relabeling the x and y axes. That is, on the x axis, each point x is relabeled x θ and on the y axis, each point y is relabeled y/ θ .The graph of f ( x ) below is for θ . Note that the graph of f ( x ) is symmetric about the y axis . -5 -4 -3 -2 -1 0 1 2 3 4 5 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 x f(x) θ =1 θ =2 θ =0.5 Figure 1: 3
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( ii ) F ( x )= x Z −∞ 1 πθ h 1+( t/ θ ) 2 i dt = 1 π lim b →−∞ arctan μ t θ | x b ¸ = 1 π arctan ³ x θ ´ + 1 2 ,x < . ( iii ) Consider the integral Z 0 x h x/ θ ) 2 i dx = θ Z 0 t 1+ t 2 dt = θ lim b →∞ 1 2 ln ¡ b 2 ¢ =+ . Since the integral Z 0 x h x/ θ ) 2 i dx does not converge, the integral Z −∞ x h x/ θ ) 2 i dx does not converge absolutely and E ( X ) does not exist. Since E ( X ) does not exist Var ( X ) does not exist. ( iv ) Using θ =1 , 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
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1 . ( c )( i ) 1 k = Z −∞ e | x θ | dx let y = x θ ,t h e n dy = dx = Z −∞ e | y | dy =2 Z 0 e y dy by symmetry Γ (1) = 2 (0!) = 2 . Thus k = 1 2 and f ( x )= 1 2 e | x θ | ,x < , θ < . The graphs for θ = 1 , 0 and 2 are plotted below. The graph for each di f erent value of θ is obtained from the graph for θ =0 by simply shifting the graph for θ to the right θ units if θ is positive and shifting the graph for θ to the left θ units if θ is negative. Note that the graph of f ( x ) is symmetric about the line x = θ .
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p1soln10 - STAT 330 SOLUTIONS Part I 1(a Starting with P x...

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