By Sheldon M Ross

ISBN-10: 0123814464

ISBN-13: 9780123814463

**Read Online or Download Introduction to Probability Models, Student Solutions Manual (e-only): Introduction to Probability Models 10th Edition PDF**

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**Additional resources for Introduction to Probability Models, Student Solutions Manual (e-only): Introduction to Probability Models 10th Edition**

**Example text**

0 2 1 F¯ e (x) = e−x/2 + e−x 3 3 With μ = (1)1/2 + (2)1/2 = 3/2 equal to the mean interarrival time ∞ F(y) ¯ F¯ e (x) = dy μ x and the earlier formula is seen to be valid. 45. The limiting probabilities for the Markov chain are given as the solution of r1 = r2 r2 = r1 1 + r3 2 r1 + r2 + r3 = 1 1 5 ri μi and so, ∑i ri μi 2 4 3 , P2 = , P 3 = . 9 9 9 47. (a) By conditioning on the next state, we obtain the following: μj = E[time in i] = ∑ E[time in i|next state is j]Pij = ∑ tij Pij i (b) Use the hint.

Condition on X(t1 ) to obtain −∞ h Therefore, =0 P(M|X(t1 ) = y) √ , d = e−σ n = X(0)dn X 2 (t1 )}] = E[X 3 (t1 )] + (t2 − t1 )E[X(t1 )] ∞ h X(t) = X(0)u∑i=1 Xi dn−∑i=1 Xi = E[X(t1 )E[X 2 (t2 ) | X(t1 )]] P(M) = X(s) > x − y} = 2P{X(t2 − t1 ) > x − y} 3. E[X(t1 )X(t2 )X(t3 )] = E[X(t1 ){(t2 − t1 ) + y≥x Var log 2 1 e−y /2t1 dy 2πt1 X(t) X(0) t = 4σ 2 h p(1 − p) h → σ2 t where the preceding used that p → 1/2 as h → 0. 54 55 Answers and Solutions 13. If the outcome is i then our total winnings are xi oi − ∑ xj = oi (1 + oi )−1 − ∑ (1 + oj )−1 j=i 1 − ∑ (1 + ok )−1 j=i Using the formula for the moment generating function of a normal random variable we see that 2 e−c t/2 E[ecB(t) |B(s)] 2 2 = e−c t/2 ecB(s)+(t−s)c /2 k (1 + oi )(1 + oi )−1 − ∑ (1 + oj )−1 = 1 − ∑ (1 + ok )−1 Thus, {Y(t)} is a Martingale.

A) Generate the X(i) sequentially using that given X(1) , …, X(i−1) the conditional distribution of X(i) will have failure rate function λi (t) given by ⎧ t < X(i−1) , ⎪ ⎨ 0, X(0) ≡ 0. λi (t) = ⎪ ⎩ (n − i + 1)λ(t), t > X(i−1) f(i) (t) = and so n! (n − i) 21. Pm+1 {i1 , …, ik−1 , m + 1} ∑ = Pm {i1 , …, ik−1 , j} j≤m j=i1 ,…,ik−1 k 1 m+1k 1 1 1 = (m − (k − 1)) m m+1 m+1 k k 25. See Problem 4. 27. First suppose n = 2. Var(λX1 + (1 − λ)X2 ) = λ2 σ12 + (1 − λ)2 σ22 . × (F(t))n−i f (t) The derivative of the above is 2λσ12 − 2(1 − λ)σ22 and equating to 0 yields n!

### Introduction to Probability Models, Student Solutions Manual (e-only): Introduction to Probability Models 10th Edition by Sheldon M Ross

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