Stochastic processes #2

Printable View

Mar 30th 2010, 09:47 PM
aaronrj

Stochastic processes #2

On the average, Mr. Z drinks and drives once in 4 years. He knows that

According to the laws of his state, the third time when he is caught drinking and driving results in the loss of his driver's license.

Poisson process is the correct model for such ``rare events'' as drinking and driving.

What is the probability that Mr. Z will keep his driver's license for at least 10 years?
Apr 2nd 2010, 11:23 PM
Anonymous1

Let $N(t)=$ the number of times Mr. Z gets pulled over by year $t.$

Then $N(t)$ is a poisson process with rate $1/4.$

Therefore $N(t)=poi(t\times 1/4)$

Now, $P(N(1)<3)= P(N(1)=1)+P(N(1)=2)$

$\Rightarrow P(N(1)<3)= e^{-1/4}\frac{(1/4)^1}{1!} + e^{-1/4} \frac{(1/4)^2}{2!}$

By independent increments,

$\Rightarrow P(N(10)<3)= e^{-1/4\times 10}\frac{(1/4\times 10)^1}{1!} + e^{-1/4\times 10} \frac{(1/4\times 10)^2}{2!}$

I've been studying for my stochastic process test on monday (Rock).

All times are GMT -8. The time now is 09:20 AM.

Copyright © 2005-2017 Math Help Forum. All rights reserved.