{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

stat2303_tutorial10

# stat2303_tutorial10 - THE UNIVERSITY OF HONG KONG...

This preview shows pages 1–3. Sign up to view the full content.

Page 1 of 6 THE UNIVERSITY OF HONG KONG DEPARTMENT OF STATISTICS AND ACTUARIAL SCIENCE STAT2303 Probability Modelling (2010-2011) STAT2803 Stochastic Models (2010-2011) Tutorial 10 (Suggested Solutions) 1) Let { 𝑁𝑁 ( 𝑡𝑡 ), 𝑡𝑡 ≥ 0} be a non-homogeneous Poisson process with intensity function 𝜆𝜆 ( 𝑡𝑡 ) , 𝑡𝑡 ≥ 0 . Denote 𝑇𝑇 as the time until the first arrival of the process. a) Find the probability density function of 𝑇𝑇 The probability mass function of 𝑁𝑁 ( 𝑡𝑡 ) is given by 𝑃𝑃 [ 𝑁𝑁 ( 𝑡𝑡 ) = 𝑛𝑛 ] = 𝑒𝑒 −𝑚𝑚 ( 𝑡𝑡 ) [ 𝑚𝑚 ( 𝑡𝑡 )] 𝑛𝑛 𝑛𝑛 ! , 𝑛𝑛 = 0,1,2, … where 𝑚𝑚 ( 𝑡𝑡 ) is the mean value function 𝑚𝑚 ( 𝑡𝑡 ) = ∫ 𝜆𝜆 ( 𝑦𝑦 ) 𝑑𝑑𝑦𝑦 𝑡𝑡 0 , 𝑡𝑡 ≥ 0 The survival function of 𝑇𝑇 is given by 𝑆𝑆 ( 𝑡𝑡 ) = 𝑃𝑃 ( 𝑇𝑇 > 𝑡𝑡 ) = 𝑃𝑃 [ 𝑁𝑁 ( 𝑡𝑡 ) = 0] = 𝑒𝑒 −𝑚𝑚 ( 𝑡𝑡 ) , 𝑡𝑡 ≥ 0 Hence, the probability density function of 𝑇𝑇 is 𝑓𝑓 ( 𝑡𝑡 ) = 𝑑𝑑 [1 − 𝑆𝑆 ( 𝑡𝑡 )] 𝑑𝑑𝑡𝑡 = −𝑆𝑆 ( 𝑡𝑡 ) = 𝑚𝑚 ( 𝑡𝑡 ) 𝑒𝑒 −𝑚𝑚 ( 𝑡𝑡 ) = 𝜆𝜆 ( 𝑡𝑡 ) 𝑒𝑒 −𝑚𝑚 ( 𝑡𝑡 ) , 𝑡𝑡 ≥ 0 b) If 𝜆𝜆 ( 𝑡𝑡 ) = 𝑐𝑐 1+ 𝑡𝑡 , 𝑡𝑡 ≥ 0 , show that 𝐸𝐸 ( 𝑇𝑇 ) < if and only if 𝑐𝑐 > 1 𝑚𝑚 ( 𝑡𝑡 ) = 𝑐𝑐 1 + 𝑦𝑦 𝑑𝑑𝑦𝑦 𝑡𝑡 0 = 𝑐𝑐 [log(1 + 𝑦𝑦 )] 0 𝑡𝑡 = 𝑐𝑐 log(1 + 𝑡𝑡 ) 𝑓𝑓 ( 𝑡𝑡 ) = 𝜆𝜆 ( 𝑡𝑡 ) 𝑒𝑒 −𝑚𝑚 ( 𝑡𝑡 ) = 𝑐𝑐 1 + 𝑡𝑡 𝑒𝑒 −𝑐𝑐 log (1+ 𝑡𝑡 ) = 𝑐𝑐 (1 + 𝑡𝑡 ) 𝑐𝑐 +1 , 𝑡𝑡 ≥ 0 𝐸𝐸 ( 𝑇𝑇 ) = 𝑐𝑐𝑡𝑡 (1 + 𝑡𝑡 ) 𝑐𝑐 +1 𝑑𝑑𝑡𝑡 0 = 𝑐𝑐 � 1 (1 + 𝑡𝑡 ) 𝑐𝑐 1 (1 + 𝑡𝑡 ) 𝑐𝑐 +1 � 𝑑𝑑𝑡𝑡 0 = 𝑐𝑐 �− 1 ( 𝑐𝑐 − 1)(1 + 𝑡𝑡 ) 𝑐𝑐− 1 + 1 𝑐𝑐 (1 + 𝑡𝑡 ) 𝑐𝑐 0 When 𝑡𝑡 → ∞ , the second term is finite for all 𝑐𝑐 > 0 . The first term is finite if and only if 𝑐𝑐 > 1 . Therefore, 𝐸𝐸 ( 𝑇𝑇 ) < if and only if 𝑐𝑐 > 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
STAT2303/2803 Probability Modelling/Stochastic Models Semester 1, 2010-11 Page 2 of 6 2) Let { 𝑁𝑁 ( 𝑡𝑡 ), 𝑡𝑡 ≥ 0} be a compound Poisson process with 𝑋𝑋 ( 𝑡𝑡 ) = 𝑋𝑋 𝑖𝑖 𝑁𝑁 ( 𝑡𝑡 ) 𝑖𝑖 =1 . Suppose that 𝜆𝜆 = 2 and 𝑃𝑃 ( 𝑋𝑋 𝑖𝑖 = 𝑗𝑗 ) = 𝑗𝑗 3 for 𝑗𝑗 = 1,2. Calculate 𝑃𝑃 [ 𝑋𝑋 (2) = 3] Obviously 𝑃𝑃 [ 𝑋𝑋 (2) = 3| 𝑁𝑁 (2) = 0] = 𝑃𝑃 �� 𝑋𝑋 𝑖𝑖 = 3 0 𝑖𝑖 =1 = 0 𝑃𝑃 [ 𝑋𝑋 (2) = 3| 𝑁𝑁 (2) = 1] = 𝑃𝑃 �� 𝑋𝑋 𝑖𝑖 = 3 1 𝑖𝑖 =1 = 𝑃𝑃 (
This is the end of the preview. Sign up to access the rest of the document.