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Car crashes at a traffic stop arrive follows a Poisson Process with rate lambda = 2/hr.

(a) Expected amount of time until the 2nd car crash arrives

(b) P(See at most 2 car crashes during rush hour) , rush hour = 7am-8am and 4pm-5pm

(c) P( first car crash comes after waiting 2 hours)

(d) If car crashes arrive at another traffic stop according to Poisson Process with rate 3/hr, find expected time we wait until a car crash occurs at either traffic stop

(e) Car crashes at our original traffic stop explode independently with prob. 0.3, find the probability we see our 1st explode car crash after 5 hours

Attempt:

for (a) , Let T2 = amount of time until 2nd car crash arrives, T2~ exp (2), so E(T2) = 1/2

(d) E(time of car crash occurs at either traffic stop ) = 1/(2+3) = 1/5

Don't know how to start on the others, anything helps! Thanks so much!

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    $\begingroup$ The expected time until the second crash is $1$ hour. $\endgroup$ – André Nicolas Aug 19 '15 at 5:00
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    $\begingroup$ That you don't know how to handle (b) makes me wonder if somehow you missed the most basic thing you usually learn about the Poisson distribution before learning anything else: $\Pr(X=x) = \dfrac{\lambda^x e^{-\lambda}}{x!}$. Just use that after figuring out what the appropriate $\lambda$ is for rush hour, based on the amount time during rush hour. ${}\qquad{}$ $\endgroup$ – Michael Hardy Aug 19 '15 at 6:28
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(a) The expected time until the second crash is $1$ hour. For let $X_1$ be the time until the first crash, and $X_2$ the time between the first and the second. Then we want $E(X_1+X_2)$, which is $E(X_1)+E(X_2)$. Each of these is $\frac{1}{2}$.

(b) The number $X$ of crashes in a $2$ hour period has Poisson distribution with parameter $4$. The probability that $X\le 2$ is $\Pr(X=0)+\Pr(X=1)+\Pr(X=2)$. Now use the standard formula.

(c) The probability that the waiting time is $\gt 2$ is the probability that an exponentially distributed random variable with parameter $2$ is greater than $2$. This is $e^{-4}$. Alternately, the number $Y$ of crashes in $2$ hours has Poisson distribution with parameter $4$. The probability that $Y=0$ is $e^{-4}$.

(d) Your answer is correct. An explanation should be added.

(e) Hint: The number of explosions in an hour has Poisson distribution with parameter $(0.3)(2)$.

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  • $\begingroup$ Thanks for your explanation! For (e), with your hint, I let Y = number of exploded car in an hour, with poisson distribution (0.6), I got P(see our 1st exploded car after 5 hrs) = P(no car exploded in 5 hours) =[ P(Y=0)]^5, am I doing this right? $\endgroup$ – Bubble Aug 19 '15 at 7:53
  • $\begingroup$ You are welcome. I would slightly prefer the number of explosions in $5$ hours has Poisson distribution parameter $\nu=(0.3)(2)(5)$. So the probability of $0$ explosions is $e^{-\nu}$. This is the same answer as yours. Amazingly dangerous intersection! $\endgroup$ – André Nicolas Aug 19 '15 at 8:07
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(a) No

  • Linearity of expectation says: $\mathsf E(T_{2^{nd}}) = \mathsf E(T_{1^{st}})+\mathsf E(T_{2^{nd}}-T_{1^{st}})$

  • Memorilessness of waiting time says: $ \mathsf E(T_{2^{nd}}-T_{1^{st}}) = \mathsf E(T_{1^{st}})$

(d) Yes

  • If the events are arriving at one stop at $2/{\rm hr}$ and at the other at $3/{\rm hr}$ then events are arriving at both(either) stops at $5/{\rm hr}$. This is itself a Poisson process. $T_{a\vee b} \sim \mathcal {Exp}(\lambda_{a\vee b})$

$$\lambda_{a\vee b} = \lambda_a+\lambda_b = 5\big/{\rm hr} \\ \text{Find: } \mathsf E(T_{a\vee b})$$

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