Poisson Distribution Question and Memoryless property

poisson distributionprobability

I have this question assigned to me

Suppose that car accidents in a given town have a Poisson distribution, with an average of 5 per week (i.e. λ = 5). Also, note that, given a Poisson process with mean λ, the waiting time until the first outcome has an exponential distribution with mean 1/λ .

Suppose that no accidents have occurred for two weeks. What is the probability that no accidents will occur for another two weeks? Give the exact value or round your answer to six decimal places.

When I first read this, I interpreted it as asking what is the probability that no accidents will occur for 4 weeks, given there were no accidents in the first two weeks.

I executed the following simple code in R using the dpois function to come up with my answer

#k = 0 because that is the provided number of "successes" we would like to determine"
#lamda = 20 because the rate of accidents is 5 a week over 4 weeks would be 20

dpois(0,20,log = FALSE)

However this does not give the right answer. The correct answer is given by executing

dpois(0,10,log = FALSE)

I believe this has something to do with the memoryless property which is a little lost on me. Is there someone that can help explain what I am missing here?

Best Answer

what is the probability that no accidents will occur for 4 weeks, given there were no accidents in the first two weeks.

Correct.

dpois(0,20,log = FALSE) does not give the right answer

Indeed, because this is not the conditional probability you intended to calculate. This is just the unconditional probability of no accidents over four weeks.

To obtain the conditional probability you need to divide by the probability of the conditioning event, which is ... ?

The correct answer is given by executing dpois(0,10,log = FALSE) I believe this has something to do with the memoryless property which is a little lost on me. Is there someone that can help explain what I am missing here?

Indeed you are right, but from the point of view of Poisson processes we don't always call it the memoryless property, it's rather part of the definition: $P(\text{no accidents in weeks 3 and 4 }|\text{ no accidents in weeks 1 and 2}) = P(\text{no accidents in weeks 3 and 4})$ because the counts of the Poisson process over two disjoint periods of time are independent.

From the point of view of exponential distributions, this is the memoryless property, $P(X>\text{4 weeks} | X\text{> 2 weeks}) = P(X \text{> 2 weeks})$

Related Solutions

[Math] Poisson Distribution and Median

Part (a) is definitely incorrect. The correct answer is

 Pr[A<3] = 0.1247.

You can verify this in R - The Statistical Computing Platform:

 ppois(q=2, lambda=5)

where ppois() gives the cumulative probability, i.e. the case where k=0,1,2.

Part (b) To elaborate on @deinst excellent comment: If the distribution is assumed to have a mean of 5.5 accidents per week, then $\lambda=5.5$, but median = 5. That is, while the mean, and therefore expected value, can be a non-intger, the median, as the middle value will always be an integer.

If, however, $\lambda=5.7$, then median = 6.

Interestingly, a closed form for the median of a Poisson distribution is not simple, so we have bounds and an approximation.

Again, in R, the median of a poisson distribution with rate r can be given by:

 mpois <- function(r) { floor(r + 1/3 - 0.02/r) }

so mpois(5.7) is seen to be 6.

The reference for this approximation can be found in Wikipedia: Poisson Properties, referring to a 1994 paper Choi KP (1994) On the medians of Gamma distributions and an equation of Ramanujan. Proc Amer Math Soc 121 (1) 245–251

[Math] Poisson Distributions: Defining the poisson random variable

A Poisson random variable is frequently used to model the probability of observing a count or frequency; i.e., "how many times did something happen?" It is a discrete random variable--we don't say "we observed exactly 3.4 car accidents in three days," or "there were 4.7 bad apples in a bucket."

An exponential random variable is used to model the probability of a failure time, or event time; i.e., "how long did it take for something to happen?" It is a continuous random variable because we can measure time in very fine increments that do not have to be integers; e.g., "the phone failed after 3.26 years of use."

When a Poisson process is used to model the occurrence of an event of interest, it happens that the frequency distribution is Poisson, but the interarrival times are exponentially distributed; that is, the random number of events that occur in a given period of time follows a Poisson distribution, but the random time that elapses between one observed event and the next is exponentially distributed.

Therefore, the time $T$ that elapses until you observe the first factory accident is exponentially distributed with rate parameter $\lambda = 2$ accidents per week. The survival function of $T$ is $S_T(t) = e^{-\lambda t}$, so we want $$\Pr[T > 2] = S_T(2) = e^{-2(2)} = e^{-4}.$$ The other part of your question is addressed similarly, except you need to convert $2$ days into an equivalent (fractional) number of weeks.

Best Answer

Related Solutions

[Math] Poisson Distribution and Median

[Math] Poisson Distributions: Defining the poisson random variable

Related Question