5
$\begingroup$

A bike has probability of breaking down $p$, on any given day.

In this case, to determine the number of times that a bike breaks down in a year, I have been told that it would be best modelled with a Poisson distribution, with $\lambda = 365\,p$.

I am wondering why it would be incorrect to use a binomial distribution, with $n=365$. After all, isn't Poisson really an approximation of a sum of Bernoulli random variables?

Thanks!

$\endgroup$
  • $\begingroup$ As I understand, if bernoulli we assume if the bike breaks in a given day, it doesn't break again that day. For poisson he might get it fixed and it break yet again in the same day. $\endgroup$ – JMoravitz Dec 11 '14 at 4:18
5
$\begingroup$

Poisson distribution

a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time and/or space if these events occur with a known average rate and independently of the time since the last event.

Binomial distribution

the discrete probability distribution of the number of successes in a sequence of n independent yes/no experiments, each of which yields success with probability p

Emphasis mine

For the Poisson you need a known interval (365 days) and a known failure rate (average failures per day - Note: this can be any number $> 0$). For the Binomial you would need a fixed number of trials (365) and a known failure rate per trial (failure chance on a given day Note: this must be a number $\in [0,1]$).

For the specific question, it is a matter of interpretation and both could be justified here.

The Poisson is more appropriate if it is conceivable that the bike could break on a given day, be repaired and break again (and again etc.). For minor failures this is appropriate.

The Binomial is more appropriate if a failure on a given day takes the bike out for the rest of the day (but not for more than that because it would then reduce the total number of days). That is, a moderate failure.

I know from your earlier question here that this is then combined with a Gamma distributed cost - there is no mention of the time the repair takes. If there were, this would be a fairly typical queuing problem which typically uses Poisson distributions. I must say that it was this that led me towards the Poisson.

$\endgroup$
2
$\begingroup$

It's not incorrect to use a binomial distribution because indeed that is what it would be.

However, it is best modeled as a poison distribution because the calculations are much simpler and the approximation is sufficiently close for large $n$

$$\mathcal{Bin}(n, p) \approx \mathcal {Pois}(np), \mbox{ for }n\to\infty$$

$\endgroup$
0
$\begingroup$

I think the main difference is that the Poisson distribution is used to approximate a very large sample like casuality in the war, fish caught in a big lake, number of traffic accidents, etc. If you review the derivation of Poission distribution, at some step we let $n\rightarrow \infty$ and assume $np\rightarrow \lambda$, which is a constant. At here $n=365$ is a relatively large number, comparing to a typical sample side of $25-100$, so I think using Poisson distribution is justified if $p$ is relatively small.

The wikipedia article on Poission distribution might also helps.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.