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2 ‫حل هوك‬

‫احمد حسن علي‬

3.1. If the moment-generating function of a random variable

X is

1 2 (  e t )5 , find Pr(X  2 or 3) . 3 3 M (t )  (q  pe t )n 1 2 M (t )  (  e t )5 3 3 2 1 p  ,q  , n  5 3 3 2 3 3 2  5   2   1   5 2   1  40 80 120    0.494 Pr(X  2 or 3)                243 243 243  2   3   3   3 3   3 

3.2. The moment-generating function of a random variable X is

2 1 (  e t )9 . Show that 3 3 x

 5  1   2  Pr(   2  X    2 )        x 1  x   3   3  1 2 p  ,q  , n  9 3 3 9 9*2   np   3, 2  npq  2 3 9   2  3  2* 2  0.172 5

5 x

  2  3  2* 2  5.828 Pr(0.172  X  5.828)  Pr(1  X  5) 3.3. If X is b(n,P), show that

X E    p n 

and

2  X  p (1 p )  E   p    n    n

 X  E ( X ) np E    p n n n  2  X  X 2  E (X 2 )   X   E (X )  2 E    p    E   2  2 p    p 2   2 p  p  2 n   n   n    n   n 2 2 2 2 2   pq  np 2  np 2  np  p 2 npq  n p 2 p 2 p 2 pq  np 2 p p 2           n2 n2 n n  n  p (1  p )  n

1

2 ‫حل هوك‬

‫احمد حسن علي‬

3.4. Let the mutually stochastically independent random variables

X 1 , X 2 , X 3 have the same p.d.f f (x )  3x 2 ,0  x  1 , zero elsewhere. Find the probability that exactly two of these three variables exceed

Pr(x 

1 ) 2



1 1 2

1 . 2

1 7 3x 2 dx  1  8 8 2

1 1  7  147 Pr(exactly two > )  3. .   2 8  8  512 3.5. Let Y be the number of successes in n independent repetitions of a random experiment having the probability of success P = 2 3

. If n = 3, compute

Pr(2 Y ) ; if n = 5, compute Pr(3 Y ) . y

5 y

 5   1   2   5  1   2  240                2  3   3   3  3   3  243

5 y

 5  1   2  80        243  3  3   3 

 5  1   2  Pr(2 Y )        y  2 y   3   3  3

 5  1   2  Pr(3  Y )        y 3  y  3   3  y

3

2

3

3

3

2

2

3.6. Let Y be the number of successes throughout n independent repetitions of a random experiment having probability of success P =

1 4

Determine the smallest value of n so that

Pr(1 Y )  0.70 0

 n  1   3 Pr(1 Y )  1  Pr(1 Y )  1        0  4   4 n

ln 0.30  3 1     0.70  n   4.2  5 ln 0.75  4

2

n

2 ‫حل هوك‬

‫احمد حسن علي‬

3.7. Let the stochastically independent random variables Xl and X2 have binomial distributions with parameters n1  3 , p 1  23 and

n 2  4 , p 2  21 , respectively. Compute Pr(X 1  X 2 ) . Hint. List the four mutually exclusive ways that X 1  X 2 and compute the probability of each. 3  3  2  1 1 1 Pr(X 1 )        , x 1  0,1,2,3  x 1  3   3  x

x

3  4 1  2 1 2 Pr(X 2 )        , x 1  0,1,2,3 x22  2  1 1 Pr(X 1  0)  Pr( X 2  0)  27 16 2 1 Pr(X 1  1)  Pr( X 2 1)  9 4 4 3 Pr( X 1  2)  Pr( X 2  2)  9 8 8 1 Pr( X 1  3)  Pr( X 2  3)  27 4 x

3.8. Let X 1 ,X 2 ,...,X

k 1

x

have a multinomial distribution (a) Find the

moment-generating function of X 2 , X 3 ,..., X k 1 (b) What is the p.d.f. of X 2 , X 3 ,..., X

k 1

? (c) Determine the conditional p.d f.

X 1 given that X 2  x 2 ,..., X k 1  x k 1 (d) What is the conditional expectation E( X 1 | x 2,..., xk 1 )? of

3

2 ‫حل هوك‬

‫احمد حسن علي‬ x



M ( t1 , t2 ,..., tk 1 )  E (e t1x 1 t2x 2 ... t k x k ) 

e t1x1 t2 x 2...  t k 1x k 1

x 1x 2... x k 10 t x 1 t 2 x 2 ...  t k x k

M ( t1 , t2 ,..., tk 1 )  E (e 1

M ( t2 ,..., tk 1 )  E (e t 2x 2... tk x k ) 

)

x p1X 1 p2X 2 ... pkX k x 1 x 2 ...(x k  x k 1  ...  x 1 )

x



x t e 1 p1 x 1x 2... x k 10 x 1 x 2 ...(x k  x k 1  ...  x 1 )



x

  e

x p1 x1 x2 ... x k 1  0 x 1x 2 ...(x k  x k 1  ...  x 1 )



x1

t2

 e x1

p2



x2

t2

p2



x2

...p kX k

...pkX k

M ( t 2 ,..., tk 1 )  ( p1  et 2 p2  ...  pk )x x

x p1X 1 p 2X 2 ...p Xk k    x x x x x ... 1 ) x 1  0 1 2 ...( k k 1

p (x 2 ,...,x

 k 1 )

p (x 2 ,...,x

) p

k 1

X 2

2

...p

Xk k

x x p1X 1 x 2 ...x k 1 x x 1 (x k  x k 1  ... x 1 ) 1 0

x p1X 1 p 2X 2 ...p Xk k x 1x 2...(x k  x k 1  ...  x 1 ) p (x 1 | x 2,...,x k 1 )  x x p 1X 1 p 2X 2... pkX k x 2...x k 1 x x 1 ( x k  x k 1  ...  x 1 ) 1 0 1

X

p 1 x1 ( x k  x k 1  ...  x 1 ) 1  x p1X 1  x 1 0 x 1 (x k  x k 1  ... x 1 ) x

E ( x1 | x 2 , x 3 ,..., x k 1 )  

x 1 0

x

1

x1

p

X1 1

x 1(x k  x k 1  ...  x 1 )  p1X 1  x 1 0 x 1 (x k  x k 1  ...  x 1 ) x

 (x

1

p 1X1

 x k 1  ...  x 1 ) p1X 1  x 1 0 x 1 (x k  x k 1  ...  x 1 )

x 1 0 x

k

3.11. One of the numbers 1, 2, ... , 6 is to be chosen by casting an unbiased die. Let this random experiment be repeated five independent times. Let the random variable x 1 be the number of terminations in the set {x; x = 1,2, 3} and let the random variable x 2 be the number of terminations in the set {x; x = 4, 5}. Compute Pr ( x 1 = 2, x 2 = 1). x

x

(5 x 1  x 2 )

1 2 5! 3 2 1 p (X 1  2, X 2  1,(5  x 1  x 2 ))        x 1 !x 2 !(5  x1  x 2 )! 6   6   6 

2

1

2

5! 5  3  2  1    0.0694       2!1!(5  2 1)!  6   6   6  72

4

2 ‫حل هوك‬

‫احمد حسن علي‬

3.12. Show that the moment-generating function of the negative binomial distribution is

M (t )  p r 1  (1  p )e t 

r

. Find the

mean and the variance of this distribution. Hint. In the summation representing M (t ), make use of the MacLaurin's series for

1   

r

.

 x  r  1  tx r x M (t )  E (e )    e p (1 p ) x x 0    x  r  1 x  r  1 where         x    x    x  r   r  x  p r    1  (1  p )e t   p r      (1  p )e t x 0  x  x 0  x  

tx



x

 p r 1 (1 p )e t 

3.13. Let Xl and X 2 have a trinomial distribution. Differentiate themoment-generating function to show that their covariance is - nPIP2'

5

r...