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How is the Poisson Distribution derived? watch

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    I spent about 4/5 hours learning about combinatorics so I could understand the formula for the Binomial distribution and now I completely understand it, but the next day, we were taught about the Poisson distribution and I have no idea how on Earth someone came up with it.

    And don't just tell me that e^{\lambda} = {\lambda}^0 + {\lambda}^1 + \frac{{\lambda}^2}{2} + ..., so 1 = \frac{e^{\lambda}}{e^{\lambda}} = \frac{\lambda^0}{e^{\lambda}}+ \frac{\lambda^1}{e^{\lambda}}+ \frac{\lambda^2}{2! \times e^{\lambda}} + ... + \frac{\lambda^n}{n! \times e^{\lambda}} + ... because that says absolutely nothing about how it relates to probability other than it's neat because they all add up to 1; like probabilities do.

    My guess is that somebody assumed the mean and the variance are equal, and using this assumption -- amongst others -- you can derive the poisson distribution; but I have absolutely no clue how. If it requires calculus, I'm up for that, but any help at all would be appreciated!
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    Argghhh...my teacher explained this to me once, now I've completely forgotten it.
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    (Original post by Callum Scott)
    I spent about 4/5 hours learning about combinatorics so I could understand the formula for the Binomial distribution and now I completely understand it, but the next day, we were taught about the Poisson distribution and I have no idea how on Earth someone came up with it.

    And don't just tell me that e^{\lambda} = {\lambda}^0 + {\lambda}^1 + \frac{{\lambda}^2}{2} + ..., so 1 = \frac{e^{\lambda}}{e^{\lambda}} = \frac{\lambda^0}{e^{\lambda}}+ \frac{\lambda^1}{e^{\lambda}}+ \frac{\lambda^2}{2! \times e^{\lambda}} + ... + \frac{\lambda^n}{n! \times e^{\lambda}} + ... because that says absolutely nothing about how it relates to probability other than it's neat because they all add up to 1; like probabilities do.

    My guess is that somebody assumed the mean and the variance are equal, and using this assumption -- amongst others -- you can derive the poisson distribution; but I have absolutely no clue how. If it requires calculus, I'm up for that, but any help at all would be appreciated!
    look at Q86 in this link
    http://madasmaths.com/archive/maths_..._questions.pdf
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    (Original post by Callum Scott)
    I spent about 4/5 hours learning about combinatorics so I could understand the formula for the Binomial distribution and now I completely understand it, but the next day, we were taught about the Poisson distribution and I have no idea how on Earth someone came up with it.

    And don't just tell me that e^{\lambda} = {\lambda}^0 + {\lambda}^1 + \frac{{\lambda}^2}{2} + ..., so 1 = \frac{e^{\lambda}}{e^{\lambda}} = \frac{\lambda^0}{e^{\lambda}}+ \frac{\lambda^1}{e^{\lambda}}+ \frac{\lambda^2}{2! \times e^{\lambda}} + ... + \frac{\lambda^n}{n! \times e^{\lambda}} + ... because that says absolutely nothing about how it relates to probability other than it's neat because they all add up to 1; like probabilities do.

    My guess is that somebody assumed the mean and the variance are equal, and using this assumption -- amongst others -- you can derive the poisson distribution; but I have absolutely no clue how. If it requires calculus, I'm up for that, but any help at all would be appreciated!
    Here's something I found last year regarding the Poisson:
    http://www.pp.rhul.ac.uk/~cowan/stat...oissonNote.pdf
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    Right so:

    \displaystyle\sum_{n=0}^{\infty} \frac{n\times e^{-\lambda}\lambda^n}{n!} \equiv e^{-\lambda}\displaystyle\sum_{n=0}^  {\infty} \frac{n\times\lambda^n}{n!}


    e^{-\lambda}\displaystyle\sum_{n=0}^  {\infty} \frac{n\times\lambda^n}{n!}

    = e^{-\lambda}\left[ \frac{0\cdot \lambda^0}{0!} + \frac{1\cdot \lambda^1}{1!} + \frac{2\cdot \lambda^2}{2!} + \frac{3\cdot \lambda^3}{3!} + \cdots \right]

    = e^{-\lambda} \left[ \frac{\lambda^1}{0!} + \frac{\lambda^2}{1!} + \frac{\lambda^3}{2!} + \frac{\lambda^4}{3!} + \cdots \right]

    = e^{-\lambda} \left[ \frac{1}{0!} + \frac{\lambda^1}{1!} + \frac{\lambda^2}{2!} + \frac{\lambda^3}{3!} + \cdots \right]\cdot\lambda

    = e^{-\lambda}e^{\lambda}\cdot\lambda

    = \lambda

    OK, I get that this shows that the expected value is lambda, which is cool and all; but how did the formula get derivedso that it shows that the expected value is lambda?!
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    (Original post by Callum Scott)
    Right so:

    \displaystyle\sum_{n=0}^{\infty} \frac{n\times e^{-\lambda}\lambda^n}{n!} \equiv e^{-\lambda}\displaystyle\sum_{n=0}^  {\infty} \frac{n\times\lambda^n}{n!}


    e^{-\lambda}\displaystyle\sum_{n=0}^  {\infty} \frac{n\times\lambda^n}{n!}

    = e^{-\lambda}\left[ \frac{0\cdot \lambda^0}{0!} + \frac{1\cdot \lambda^1}{1!} + \frac{2\cdot \lambda^2}{2!} + \frac{3\cdot \lambda^3}{3!} + \cdots \right]

    = e^{-\lambda} \left[ \frac{\lambda^1}{0!} + \frac{\lambda^2}{1!} + \frac{\lambda^3}{2!} + \frac{\lambda^4}{3!} + \cdots \right]

    = e^{-\lambda} \left[ \frac{1}{0!} + \frac{\lambda^1}{1!} + \frac{\lambda^2}{2!} + \frac{\lambda^3}{3!} + \cdots \right]\cdot\lambda

    = e^{-\lambda}e^{\lambda}\cdot\lambda

    = \lambda

    OK, I get that this shows that the expected value is lambda, which is cool and all; but how did the formula get derivedso that it shows that the expected value is lambda?!
    if you do not understand from this examplehow that proves it, you need to revise the method of how to find the mean and variance of a discrete distribution
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    Is this for s2 in maths a level? Do you actually need to know this? Like the proof? Wtf


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    (Original post by anoymous1111)
    Is this for s2 in maths a level? Do you actually need to know this? Like the proof? Wtf


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    Keep your hair on... it is not examinable.
    (it would have been if I was writing the syllabus/exam...)
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    (Original post by TeeEm)
    it is not examinable.
    what a shame
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    (Original post by TeeEm)
    if you do not understand from this examplehow that proves it, you need to revise the method of how to find the mean and variance of a discrete distribution
    It proves it, but I want the thought process of coming up with it! I want to understand what it actually does, rather than just use it like a tool belt for S2
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    (Original post by Callum Scott)
    It proves it, but I want the thought process of coming up with it! I want to understand what it actually does, rather than just use it like a tool belt for S2
    That's fair enough just worrying that I need to know it! I don't need to know this do I? For s2?


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    (Original post by Callum Scott)
    It proves it, but I want the thought process of coming up with it! I want to understand what it actually does, rather than just use it like a tool belt for S2
    fact one
    all the terms of this infinite series add up to 1
    Thus it may represent a discrete probability distribution

    fact two
    for any discrete random variable E(X) = SUM(xP(X=x)
    For this probability distribution we get EXPECTATION = λ

    fact three

    for any discrete random variable Var(X) = E(X2)- [E(X)]2, where E(X2) = SUM(x2P(X=x)
    For this probability distribution we get VARAINCE = λ

    conclusion
    this infinite series may define a discrete probability distribution with mean and variance λ.

    END OF TRANSMISSION
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    (Original post by anoymous1111)
    That's fair enough just worrying that I need to know it! I don't need to know this do I? For s2?


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    Nope, not at all I don't think
    As I mentioned in the title, I spent ages learning what on earth the N choose X part did in the binomial distribution formula, and I must say, it really does help to understand the formulae. I had real trouble with the variance and s.d. formula and only got 80/100 on S1, my lowest mark :C but since I've learned what it actually does; it's second nature to me!
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    (Original post by TeeEm)
    ...
    okay but where did you find out in the first place that the poisson distribution has mean=variance, and how do you show that this property is unique to the poisson distribution (thus the formula you get must correspond to the poisson)
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    (Original post by Callum Scott)
    I spent about 4/5 hours learning about combinatorics so I could understand the formula for the Binomial distribution and now I completely understand it, but the next day, we were taught about the Poisson distribution and I have no idea how on Earth someone came up with it.

    And don't just tell me that e^{\lambda} = {\lambda}^0 + {\lambda}^1 + \frac{{\lambda}^2}{2} + ..., so 1 = \frac{e^{\lambda}}{e^{\lambda}} = \frac{\lambda^0}{e^{\lambda}}+ \frac{\lambda^1}{e^{\lambda}}+ \frac{\lambda^2}{2! \times e^{\lambda}} + ... + \frac{\lambda^n}{n! \times e^{\lambda}} + ... because that says absolutely nothing about how it relates to probability other than it's neat because they all add up to 1; like probabilities do.

    My guess is that somebody assumed the mean and the variance are equal, and using this assumption -- amongst others -- you can derive the poisson distribution; but I have absolutely no clue how. If it requires calculus, I'm up for that, but any help at all would be appreciated!
    The Poisson distribution was actually first derived as an approximation for the binomial distribution, see question 4 here https://www0.maths.ox.ac.uk/system/f.../13/Sheet3.pdf
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    (Original post by imsoanonymous123)
    okay but where did you find out in the first place that the poisson distribution has mean=variance, and how do you show that this property is unique to the poisson distribution (thus the formula you get must correspond to the poisson)
    I am trying to explain but with all due respect this looks to be above your current understanding, as your question is circular.

    Your questions is equivalent to asking:

    How do you prove that tanx = sinx / cosx


    Somebody far brighter than me and you (Mr Simeon Poisson) produces an infinite series whose terms add up to 1.

    this implies he could use it as a model.

    what is the expectation and variance of such distribution.
    the same number.

    this distribution is called Poisson.
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    (Original post by TeeEm)
    I am trying to explain but with all due respect this looks to be above your current understanding, as your question is circular.

    Your questions is equivalent to asking:

    How do you prove that tanx = sinx / cosx


    Somebody far brighter than me and you (Mr Simeon Poisson) produces an infinite series whose terms add up to 1.

    this implies he could use it as a model.

    what is the expectation and variance of such distribution.
    the same number.

    this distribution is called Poisson.
    okay then let me rephrase my question: what is the significance of having mean = variance in a situation that models the distribution of the events with the properties that poisson have:

    Uniform, Singular, Indepent, Random

    yea im pretty bad at statistics but come on bro, if its in a level it cant possibly be beyond my understanding
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    (Original post by imsoanonymous123)
    okay then let me rephrase my question: what is the significance of having mean = variance in a situation that models the distribution of the events with the properties that poisson have:

    Uniform, Singular, Indepent, Random

    yea im pretty bad at statistics but come on bro, if its in a level it cant possibly be beyond my understanding
    I would sincerely love to help you further, but at the moment I am working myself and we are going around in circles as I am a terrible teacher.

    Hopefully someone else will come along and might find the "right" words to explain it better.
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    (Original post by anoymous1111)
    Is this for s2 in maths a level? Do you actually need to know this? Like the proof? Wtf


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    Will you recall the concept in 2 years time if you don't learn the proof?
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    (Original post by TeeEm)
    I would sincerely love to help you further, but at the moment I am working myself and we are going around in circles as I am a terrible teacher.

    Hopefully someone else will come along and mind find the "right" words to explain it better.
    ok thanks for your time anyway =)
 
 
 
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