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Math Help - Find E(Y^3) of Poisson, without using generating functions

  1. #1
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    Find E(Y^3) of Poisson, without using generating functions

    Question.

    Let Y be a random variable that has a Poisson distribution with parameter \lambda. Without using generating functions:

    (a) show that E(X) = \lambda

    (b) find E(X^3).


    Answer

    (a)

    E(X)=\Sigma_xxf_X(x)=\Sigma_{x=0}^{\infty}\frac{\l  ambda^xe^{-\lambda}x}{x!}={\lambda}e^{-\lambda}\Sigma_{x=0}^{\infty}\frac{\lambda^{x-1}}{(x-1)!}={\lambda}e^{-\lambda}e^{\lambda}={\lambda}e^0=\lambda

    (I used exp expansion for summation)

    (b) and now I am stuck with the powers of x in my summation expression...

    E(X)=\Sigma_xx^3f_X(x)=\Sigma_{x=0}^{\infty}\frac{  \lambda^xe^{-\lambda}x^3}{x!}={\lambda}e^{-\lambda}\Sigma_{x=0}^{\infty}\frac{\lambda^{x-1}x^2}{(x-1)!}

    I tried to substitute x-1=j:

    ={\lambda}e^{-\lambda}\Sigma_{j=1}^{\infty}\frac{\lambda^j(j+1)^  2}{j!}=

    and then possibly expand (j+1)^2 into {j^2+2j+1} but then I have to deal with j^2 again (as with x^2 before), so it does not seem to help.

    Are there any methods to deal with these kinds of summation calculations? I have looked into many books but they all show calculations of n-th moments straight from the moment generating functions, and my examinor seem to prefer his students to calculate from 'first principles' - this is a sample exam question. Unfortunately, there is no answer to this question in the study guide.
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  2. #2
    Senior Member Sambit's Avatar
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    If your examiner wants to work it out by first principle, then yours is the only way. In your final expression, if you expand (j+1)^2 then you'll get one term like j^2 / j! which can be written as j/(j-1)!; so it seems to be more helpful than the case of x^2.
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  3. #3
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    Right. Let me try and finish this off then. One question though - when I substitute x=j+1 in the summation operator, the 'bottom' number is no longer x=0, but j=1. How do I adjust the exponential expansion formula a^x!/x! which starts with x=0?
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  4. #4
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    Protip: For discrete distributions it is often easier to calculate the so-called factorial moments, and use these to get the regular moments. For example, to get the 3rd moment, first calculate

    <br />
\mbox{E}[Y(Y - 1)(Y - 2)] = \mbox{E}Y^3 - 3 \mbox{E}Y^2 + 2\mbox{E}Y.<br />

    Incidentally, it is easy to show that the j'th factorial moment of the Poisson is \lambda^j.
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  5. #5
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    thanks for the 'protip'! It will be handy to check the final answer. I will go on with x=j+1 then, Sambit! It's a struggle but I won't quit )))
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  6. #6
    Senior Member Sambit's Avatar
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    Ok ok. Do whatever you want. I also suggest you to follow the method written by theodds.It indeed is a good idea.
    However, if the bottom number starts from j=1, the resulting summation will be of the form exp(something) - 1.
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  7. #7
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    Quote Originally Posted by Sambit View Post
    Ok ok. Do whatever you want. I also suggest you to follow the method written by theodds.It indeed is a good idea.
    However, if the bottom number starts from j=1, the resulting summation will be of the form exp(something) - 1.
    It's not that I have a choice, the examiner expects me to work from the definition of the function, no shortcuts.

    I have now ended up with a collection of curious looking sums, and I don't know what to do with them. For example,

    \Sigma_{k=-3}^{\infty}=\frac{\lambda^k}{k!}

    If I move from k to say l=k+3 then I get

    \Sigma_{l=0}^{\infty}=\frac{\lambda^(l-3)}{(l-3)!}

    Does it mean that I have to deduct three last members of the exponential expansion sequence?... Or add three more members with factorials of negative numbers? or these three members with factorials of negative numbers in denominators all equal to zero, so are to be ignored?...
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  8. #8
    Moo
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    Hello,

    The sum can't start at k=-3 As well as in your first post, you can't have (k-1)! in the denominator while you start the sum at k=0. Never write a factorial for a negative number !
    To avoid this, just note that for k=0, \frac{\lambda^k k^3}{k!}=0, so you can start your sum at k=1 without changing the value of the sum. Then, and only then, you can simplify \frac{k^3}{k!}=\frac{k^2}{(k-1)!} and do what you did.


    The problem with theodds' method is that it is just using another method than mgf, but very very very very similar to it. That's kind of cheating
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  9. #9
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    I've realised that already, via a different route. Every time I have -1 as a starting point, I can see that the first element is zero, so I change -1 to 0 and this works every time.

    So every time I get \Sigma_{j=-1}^{\infty}\frac{\lambda^j(j+1)^2}{j!} turns to 0 if j=-1, so I then change it to 0 every time - just like you said.
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  10. #10
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    Finally!! I solved it. It is indeed \lambda^3+3{\lambda}^2+\lambda. I now have the proof
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  11. #11
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    Quote Originally Posted by Moo View Post
    Hello,

    The sum can't start at k=-3 As well as in your first post, you can't have (k-1)! in the denominator while you start the sum at k=0. Never write a factorial for a negative number !
    To avoid this, just note that for k=0, \frac{\lambda^k k^3}{k!}=0, so you can start your sum at k=1 without changing the value of the sum. Then, and only then, you can simplify \frac{k^3}{k!}=\frac{k^2}{(k-1)!} and do what you did.


    The problem with theodds' method is that it is just using another method than mgf, but very very very very similar to it. That's kind of cheating
    I disagree, personally. The factorial moments and the raw moments have a completely transparent algebraic relationship to each other. How can an instructor hold it against the student for noticing a useful algebraic identity?
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