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The $N(t)$ is a poisson process for the number of events to occur with a mean $\lambda(t) = 3$ per day. I am supposed to find the probability of more than 200 events in 60 days. My theoretical answer does not match the simulations in R, what is wrong?

My calculation is

$$ P(N(t) > 200 | T = 60) = 1 - \int_0^{60} \frac{(3 t)^{200}}{200!} \times e^{-3t} dt = 1 - 0.021 \approx 0.98. $$

Simulations of a 100 processes in R gives a number of 6/100 = 0.06. That is 6 events out of 100 being more than 200. What am I not understanding?

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    $\begingroup$ What's the distribution of the number of events in 60 days? $\endgroup$ – Glen_b -Reinstate Monica Nov 2 '16 at 17:00
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    $\begingroup$ the distribution is $\exp^{-\lambda} \times ( \lambda^n / n! )$? $\endgroup$ – G. Debailly Nov 2 '16 at 17:08
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    $\begingroup$ In your calculation you are mixing $\lambda$, N, t in a wrong way. You should think and carefully write what is the probability that you will see say 201 event in 60 days, and what will be the Poisson $\lambda$ for the the number of events in 60 days. $\endgroup$ – sega_sai Nov 2 '16 at 18:02
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    $\begingroup$ @Erik importantly, for which value of $\lambda$? Note also that first term should be written either as $e^{-\lambda}$ or $\exp(-\lambda)$ rather than $\exp^{-\lambda}$ (that makes no sense ... you're taking a power of a function but without any argument?) $\endgroup$ – Glen_b -Reinstate Monica Nov 2 '16 at 20:59
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    $\begingroup$ You still are not thinking about it in the right way. Don't just spam formulas at the problem and hope for success. What's the Poisson distribution about? What's it used for? Do you know a definition of Poisson processes or anything about sums of independent Poissons? $\endgroup$ – Glen_b -Reinstate Monica Nov 2 '16 at 22:06
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Considering that $\lambda = 3$ and mean intensity over $T = $60 days makes $\lambda * T = 3\times 60 = 180$. Using the Poisson density function $p(i) = P \left \{X = i \right \} = e^{-\lambda} \times \lambda^i / i!$ gives the probability

$$ P(N(T) > 200 | T = 60) = 1 - \sum_{n =1}^{200}\frac{180^n}{e^{180}n!} \approx 0.065. $$

This answer seem sensible given my simulations returning a probability of approximately 0.06.

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    $\begingroup$ Note that $n$ is discrete -- it only takes integer values. If that's intended to be a Riemann integral (as the $\text{d}n$ would suggest) your notation is wrong. You'll need a sum. You might approximate the integral using a continuous function but you can't just stick a discrete pmf into a Riemann integral and have it work. $\endgroup$ – Glen_b -Reinstate Monica Nov 7 '16 at 21:59
  • $\begingroup$ @Glen_b yes you are right. Answer is edited. Thanks again. I would very much like to up vote your comments, but lack reputation :) $\endgroup$ – G. Debailly Nov 9 '16 at 17:27
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    $\begingroup$ A very good way to perform a reality check on such calculations is to approximate the distribution. Because a Poisson distribution with expectation $\mu$, with $\mu$ large, is approximately Normal, we can compute that the value midway between $200$ and the next value $201$, namely $200.5$, is $20.5$ greater than $\mu=3\times 60$. The Poisson SD is $\sqrt{\mu}$, which therefore is between $13$ and $14$ (since $13^2=168$ and $14^2=196$). Thus $20.5$ is close to $1.5$ SDs. The right tail area of a Normal distribution for $1.5$ is $0.067$. $\endgroup$ – whuber Nov 9 '16 at 18:22
  • $\begingroup$ @Erik There's no need to upvote the comment; what matters is that there's a good answer to the question. $\endgroup$ – Glen_b -Reinstate Monica Nov 9 '16 at 23:01
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You want to have between 201 and infinity number of events in 60 days.

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    $\begingroup$ Hi Zahava. It's great that you have tried to give guidance and hints rather than a complete answer for a homework-type question. That's much appreciated. However, it's very short for an answer; it would help if you could supply a little more detail (for example on the reasoning behind that answer, such as a little more detail related to the "number of events in 60 days", or about where the original attempt has gone astray if you can disentangle that.). $\endgroup$ – Glen_b -Reinstate Monica Nov 2 '16 at 21:06
  • $\begingroup$ I appreciate any effort to help :) I think I understand what Sahava means, will have to check further tomorrow. $\endgroup$ – G. Debailly Nov 2 '16 at 22:03

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