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The pair { X , Y } ci | H . X exponential ( u / 3 ) , given H = u ; Y exponential ( u / 5 ) , given H = u ; and H uniform [ 1 , 2 ] . Determine a general formula for P ( X > r , Y > s ) , then evaluate for r = 3 , s = 10 .

P ( X > r , Y > s | H = u ) = e - u r / 3 e - u s / 5 = e - a u , a = r 3 + s 5
P ( X > r , Y > s ) = e - a u f H ( u ) d u = 1 2 e - a u d u = 1 a [ e - a - e - 2 a ]
For r = 3 , s = 10 , a = 3 , P ( X > 3 , Y > 10 ) = 1 3 ( e - 3 - e - 6 ) = 0 . 0158
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A small random sample of size n = 12 is taken to determine the proportion of the student body which favors a proposal to expand the student Honor Council by adding twoadditional members “at large.” Prior information indicates that this proportion is about 0.6 = 3/5. From a Bayesian point of view, the populationproportion is taken to be the value of a random variable H . It seems reasonable to assume a prior distribution H beta ( 4 , 3 ) , giving a maximum of the density at ( 4 - 1 ) / ( 4 + 3 - 2 ) = 3 / 5 . Seven of the twelve interviewed favor the proposition. What is the best mean-square estimate of the proportion, given this result? What is the conditionaldistribution of H , given this result?

H Beta ( r , s ) , r = 4 , s = 3 , n = 12 , k = 7

E [ H | S = k ] = k + r n + r + s = 7 + 4 12 + 4 + 3 = 11 19
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Let { X i : 1 i n } be a random sample, given H . Set W = ( X 1 , X 2 , , X n ) . Suppose X conditionally geometric ( u ) , given H = u; i.e., suppose P ( X = k | H = u ) = u ( 1 - u ) k for all k 0 . If H uniform
on [0, 1], determine the best mean square estimator for H , given W .

E [ H | W = k ] = E [ H I { k } ( W ) ] E [ I { k } ( W ) ] = E { H E [ I { k } ( W ) | H ] } E { E [ I { k } ( W ) | H ] }
= u P ( W = k | H = u ) f H ( u ) d u P ( W = k | H = u ) f H ( u ) d u , k = ( k 1 , k 2 , , k n )
P ( W = k | H = u ) = i = 1 n u ( 1 - u ) k i = u n ( 1 - u ) k * k * = i = 1 n k i
E [ H | W = k ] = 0 1 u n + 1 ( 1 - u ) k * d u 0 1 u n ( 1 - u ) k * d u = Γ ( n + 2 ) Γ ( k * + 1 ) Γ ( n + 1 + k * + 2 ) · Γ ( n + k * + 2 ) Γ ( n + 1 ) Γ ( k * + 1 ) =
n + 1 n + k * + 2
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Let { X i : 1 i n } be a random sample, given H . Set W = ( X 1 , X 2 , , X n ) . Suppose X conditionally Poisson ( u ) , given H = u; i.e., suppose P ( X = k | H = u ) = e - u u k / k ! . If H gamma ( m , λ ) , determine the best mean square estimator for H , given W .

E [ H | W = k ] = u P ( W = k | H = u ) f H ( u ) d u P ( W = k | H = u ) f H ( u ) d u
P ( W = k | H = u ) = i = 1 n e - u u k i k i ! = e - n u u k * A k * = i = 1 n k i
f H ( u ) = λ m u m - 1 e - λ u Γ ( m )
E [ H | W = k ] = 0 u k * + m e - ( λ + n ) u d u 0 u k * + m - 1 e - ( λ + n ) u d u = Γ ( m + k * + 1 ) ( λ + n ) k * + m + 1 · ( λ + n ) k * + m Γ ( m + k * ) = m + k * λ + n
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Suppose { N , H } is independent and { N , Y } ci | H . Use properties of conditional expectation and conditional independence to show that

E [ g ( N ) h ( Y ) | H ] = E [ g ( N ) ] E [ h ( Y ) | H ] a . s .

E [ g ( N ) h ( H ) | H ] = E [ g ( N ) | H ] E [ h ( Y ) | H ] a . s . by (CI6) and

E [ g ( N ) | H ] = E [ g ( N ) ] a . s . by (CE5) .

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Consider the composite demand D introduced in the section on Random Sums in "Random Selecton"

D = n = 0 I { k } ( N ) X n where X n = k = 0 n Y k , Y 0 = 0

Suppose { N , H } is independent, { N , Y i } ci | H for all i , and E [ Y i | H ] = e ( H ) , invariant with i . Show that E [ D | H ] = E [ N ] E [ Y | H ] a . s . .

E [ D | H ] = n = 1 E [ I { n } ( N ) X n | H ] a . s .
E [ I { n } ( N ) X n | H ] = k = 1 n E [ I { n } ( N ) Y k | H ] = k = 1 n P ( N = n ) E [ Y | H ] = P ( N = n ) n E [ Y | H ] a . s .
E [ D | H ] = n = 1 n P ( N = n ) E [ Y | H ] = E [ N ] E [ Y | H ] a . s .
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The transition matrix P for a homogeneous Markov chain is as follows (in m-file npr16_07.m ):

P = 0 . 23 0 . 32 0 . 02 0 . 22 0 . 21 0 . 29 0 . 41 0 . 10 0 . 08 0 . 12 0 . 22 0 . 07 0 . 31 0 . 14 0 . 26 0 . 32 0 . 15 0 . 05 0 . 33 0 . 15 0 . 08 0 . 23 0 . 31 0 . 09 0 . 29
  1. Obtain the absolute values of the eigenvalues, then consider increasing powers of P to observe the convergence to the long run distribution.
  2. Take an arbitrary initial distribution p 0 (as a row matrix). The product p 0 * P k is the distribution for stage k . Note what happens as k becomes large enough to give convergence to the long run transition matrix. Does the endresult change with change of initial distribution p 0 ?
ev = abs(eig(P))' ev = 1.0000 0.0814 0.0814 0.3572 0.2429a = ev(4).^[2 4 8 16 24] a = 0.1276 0.0163 0.0003 0.0000 0.0000% By P^16 the rows agree to four places p0 = [0.5 0 0 0.3 0.2]; % An arbitrarily chosen p0 p4 = p0*P^4p4 = 0.2297 0.2622 0.1444 0.1644 0.1992 p8 = p0*P^8p8 = 0.2290 0.2611 0.1462 0.1638 0.2000 p16 = p0*P^16p16 = 0.2289 0.2611 0.1462 0.1638 0.2000 p0a = [0 0 0 0 1]; % A second choice of p0 p16a = p0a*P^16p16a = 0.2289 0.2611 0.1462 0.1638 0.2000
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Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
Aislinn Reply
cm
tijani
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A mouse of mass 200 g falls 100 m down a vertical mine shaft and lands at the bottom with a speed of 8.0 m/s. During its fall, how much work is done on the mouse by air resistance
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Can you compute that for me. Ty
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emma Reply
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what is inorganic
emma
Chemistry is a branch of science that deals with the study of matter,it composition,it structure and the changes it undergoes
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chemistry could also be understood like the sexual attraction/repulsion of the male and female elements. the reaction varies depending on the energy differences of each given gender. + masculine -female.
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A ball is thrown straight up.it passes a 2.0m high window 7.50 m off the ground on it path up and takes 1.30 s to go past the window.what was the ball initial velocity
Krampah Reply
2. A sled plus passenger with total mass 50 kg is pulled 20 m across the snow (0.20) at constant velocity by a force directed 25° above the horizontal. Calculate (a) the work of the applied force, (b) the work of friction, and (c) the total work.
Sahid Reply
you have been hired as an espert witness in a court case involving an automobile accident. the accident involved car A of mass 1500kg which crashed into stationary car B of mass 1100kg. the driver of car A applied his brakes 15 m before he skidded and crashed into car B. after the collision, car A s
Samuel Reply
can someone explain to me, an ignorant high school student, why the trend of the graph doesn't follow the fact that the higher frequency a sound wave is, the more power it is, hence, making me think the phons output would follow this general trend?
Joseph Reply
Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
Joseph
Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
Joseph
"Generation of electrical energy from sound energy | IEEE Conference Publication | IEEE Xplore" ***ieeexplore.ieee.org/document/7150687?reload=true
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Magreth
progressive wave
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Mujahid
A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?
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Source:  OpenStax, Applied probability. OpenStax CNX. Aug 31, 2009 Download for free at http://cnx.org/content/col10708/1.6
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