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Diego Ravignani
Diego Ravignani
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Using the law of rare events

Apart from the mathematical demonstration, the process can be understood intuitively.

In the example in the question, the reasoning is that the there many rabbits (N). Then there is a small probability $p_1$ that a rabbit pass over the trap. These are the premises of the law of rare events, so the number of rabbits passing the trap follows a Poisson distribution with a parameter $r = N \, p_1$.

Note that the problem implicitly assumes large N and small $p_1$. If not the case, the number of the passing rabbits would be a binomial variable instead of a Poisson one.

Let us add the trap efficiency. The probability that a single rabbit pass over the trap and it is caught is the product of the probabilities of these two independent events, $p' = p_1 \, p$. Then, end-to-end, there N rabbits each with a small probability $p'$ of being caught. This is again a Poisson process. The corresponding parameter representing the mean caught rabbits is:

$\mu = N \, p_1 \, p = r \, p $

I adapted this answer from another I made about a convolution of a Poisson with a binomial variable that uses another example.

Using the law of rare events

Apart from the mathematical demonstration, the process can be understood intuitively.

In the example in the question, the reasoning is that the there many rabbits (N). Then there is a small probability $p_1$ that a rabbit pass over the trap. These are the premises of the law of rare events, so the number of rabbits passing the trap follows a Poisson distribution $r = N \, p_1$.

Note that the problem implicitly assumes large N and small $p_1$. If not the case, the number of the passing rabbits would be a binomial variable instead of a Poisson one.

Let us add the trap efficiency. The probability that a single rabbit pass over the trap and it is caught is the product of the probabilities of these two independent events, $p' = p_1 \, p$. Then, end-to-end, there N rabbits each with a small probability $p'$ of being caught. This is again a Poisson process. The corresponding parameter representing the mean caught rabbits is:

$\mu = N \, p_1 \, p = r \, p $

I adapted this answer from another I made about a convolution of a Poisson with a binomial variable that uses another example.

Using the law of rare events

Apart from the mathematical demonstration, the process can be understood intuitively.

In the example in the question, the reasoning is that the there many rabbits (N). Then there is a small probability $p_1$ that a rabbit pass over the trap. These are the premises of the law of rare events, so the number of rabbits passing the trap follows a Poisson distribution with a parameter $r = N \, p_1$.

Note that the problem implicitly assumes large N and small $p_1$. If not the case, the number of the passing rabbits would be a binomial variable instead of a Poisson one.

Let us add the trap efficiency. The probability that a single rabbit pass over the trap and it is caught is the product of the probabilities of these two independent events, $p' = p_1 \, p$. Then, end-to-end, there N rabbits each with a small probability $p'$ of being caught. This is again a Poisson process. The corresponding parameter representing the mean caught rabbits is:

$\mu = N \, p_1 \, p = r \, p $

I adapted this answer from another I made about a convolution of a Poisson with a binomial variable that uses another example.

Source Link
Diego Ravignani
Diego Ravignani
  • 45
  • 7

Using the law of rare events

Apart from the mathematical demonstration, the process can be understood intuitively.

In the example in the question, the reasoning is that the there many rabbits (N). Then there is a small probability $p_1$ that a rabbit pass over the trap. These are the premises of the law of rare events, so the number of rabbits passing the trap follows a Poisson distribution $r = N \, p_1$.

Note that the problem implicitly assumes large N and small $p_1$. If not the case, the number of the passing rabbits would be a binomial variable instead of a Poisson one.

Let us add the trap efficiency. The probability that a single rabbit pass over the trap and it is caught is the product of the probabilities of these two independent events, $p' = p_1 \, p$. Then, end-to-end, there N rabbits each with a small probability $p'$ of being caught. This is again a Poisson process. The corresponding parameter representing the mean caught rabbits is:

$\mu = N \, p_1 \, p = r \, p $

I adapted this answer from another I made about a convolution of a Poisson with a binomial variable that uses another example.