Introdução aos Processos Estocásticos_Poisson

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Universidade Federal de Minas Gerais 
 
 
 
 
Introdução aos Processos 
Estocásticos 
 
 
 
 
Listas de Exercícios – Poisson 
 
 
 
 
 
 
Nome: Anderson Luiz Tomé Pereira Barbosa 
Matricula: 2020680992 
Departamento: PPGEE-M 
Professor: Eduardo Mazoni Andrade Maçal Mendes 
Universidade Federal de Minas Gerais 
 
Exercício 03 – Find the probability of 6 arrivals of a Poisson random process in 
the time interval [7,12] if 𝜆 = 1. Next determine the average number of arrivals 
for the same time interval. 
O mesmo será para o intervalo {0,5} (Estacionariedade) 
𝑃[𝑛(𝑠) = 6] = 𝑒
(1(5))
6
 = 𝑒 21,70138889 = [0,146228081] 
𝐸|𝑛(𝑡)| = 𝜆𝑡 = 5 
 
Exercício 04 – For a Poisson random process with an arrival rate of 2 arrivals 
per second,find the probability of exactly 2 arrivals in 5 successive time intervals 
of length 1 second each. 
 
Devido a Estacionariedade e independência dos incrementos temos: 
𝑃 = [𝑛(1) = 2] = 𝑒
[(2)(1)]
2!
= (2𝑒 ) = 32𝑒 = [ 0,01452797755] 
 
Exercício 05 – What is the probability of a single arrival for a Poisson random 
process with arrival rate A in the time interval [𝑡, 𝑡 + Δ𝑡] if Δ𝑡 → 0 ? 
 
𝑃[𝑛(∆𝑡 + 𝑡) − 𝑛(𝑡) = 1] = 𝜆∆𝑡 𝑙𝑜𝑔𝑜 𝑥 ∗ → 0 𝑐𝑜𝑚 ∆𝑡 → 0 . 
Onde X* e um número qualquer 
 
Exercício 06 – Telephone calls come into a service center at an average rate 
of one per 5 seconds. What is the probability that there will be more than 12 
calls in the first one minute? 
 
𝑃[𝑛(60) > 12] = 1 − 𝑃[𝑛(60) ≤ 12] 
= 1 − 𝑒
(𝜆𝑡)
𝑘!
 
Universidade Federal de Minas Gerais 
 
 
= 1 − 𝑒
( )
1
5
(60)
𝑘!
 = 1 − 𝑒
(12)
𝑘!
 = 0.424053 
 
 
Exercício 08 – Two independent Poisson random processes both have an arrival 
rate of A. What is the expected time of the first arrival observed from either of the 
two random processes? Explain your results. Hint: Let this time be denoted by T 
and note that 𝑇 = min (𝑇( ), 𝑇( )), where 𝑇( ) is the first arrival time of the ith. 
random process. Then, note that 𝑃[𝑇 > 𝑡] = 𝑃 𝑇( ) > 𝑡; 𝑇( ) > 𝑡 . 
 
𝑃[𝑡 > 𝑡] = 𝑃 min 𝑇
( )
, 𝑇
( )
> 𝑡 = 𝑃 𝑇
( )
> 𝑡 , 𝑇
( )
> 𝑡 
 
= 𝑃 𝑇
( )
𝑃 𝑇
( ) Independencia * 
 
= 𝑃 (𝑇 > 𝑡) 
= 𝜆 𝑒 𝑑𝑢 = [−𝑒 − 𝑑𝑢| ] = 𝑒 = 1 − 𝐹𝑡(𝑡) ∗ 
*CDF 
𝑃 (𝑡) = 
𝑑
𝑑𝑡
1 − 𝑒 = 2𝜆𝑒 
𝐸|𝑡| = 
1
2𝜆
= 
1
2
 𝐸|𝑇
( )
| 
 
 
Exercício 12 – 
 
𝑃(|𝑥 > 𝑘 + 𝑘 ||𝑥 > 𝑘 |) = 
𝑃(|𝑥 > 𝑘 |, |𝑥 > 𝑘 + 𝑘 |)
|𝑥 > 𝑘 |
 = 
𝑃|𝑥 > 𝑘 + 𝑘 |
𝑃|𝑥 > 𝑘 |
 
 
Universidade Federal de Minas Gerais 
 
Mas 
𝑃(𝑥 > 𝑘) = (1 − 𝑝) = 𝑃
(1 − 𝑝)
1 − (1 − 𝑝)
 = (1 − 𝑝) 
 
= 
(1 − 𝑝)
(1 − 𝑝)
= (1 − 𝑝) 
= 𝑃[𝑥 > 𝑘 + 𝐾 |𝑋 > 𝑘 ] = (1 − 𝑝) 
= 𝑃[𝑥∆𝑡 > ∆𝑡𝑘 + ∆𝑡𝐾 |𝑋∆𝑡 > ∆𝑡𝑘 ] = (1 − 𝑝) 
= 𝑃[𝑥∆𝑡 > 𝑧 + 𝑧 |𝑋∆𝑡 > 𝑧 ] = (1 − 𝜆∆𝑡) 
= 𝑃[𝑍 > 𝑧 + 𝑧 |𝑧 > 𝑧 ] → 𝑒 = 𝑒 = 𝑃|𝑍 > 𝑧 |