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Math Help - MGF normal distribution

  1. #1
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    MGF normal distribution

    Hello everyone

    I looking at wikipedia for normal distribution and doing some work on moment generating functions. Please could someone help me with full working from getting from

    e^\frac{\sigma^2{t^2}}{2}

    to formula that say


    Thank you.
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  2. #2
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    Quote Originally Posted by jescavez View Post
    Hello everyone

    I looking at wikipedia for normal distribution and doing some work on moment generating functions. Please could someone help me with full working from getting from

    e^\frac{\sigma^2{t^2}}{2}

    to formula that say


    Thank you.
    Look at the definition and properties of the MGF, you will find that:

    E(x^{2k})=(2k)!\; a_{2k}

    where a_{2k} is the coefficient of t^{2k} in the power series expansion of the MGF. We know the series expansion of e^{\sigma^2t^2/2} so carry on from there.

    RonL
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  3. #3
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    Hi RonL

    Thank you. Am I on the right track with this below then for normal dist MGF:

    1+tx\sigma^2+\frac{t^2\sigma^4x^2}{2!}+...

    I am checking the pages for MGF and normal distribution for this formula below:

    E(x^{2k})=(2k)!\; a_{2k}

    But I did not see it yet, where does this come from please?

    In first point, am I doing power series expansion wrong? Because I do not know where
    2^n will come from for denominator in final equation or how to see that odd moments are 0.

    Thank you again
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  4. #4
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    Quote Originally Posted by jescavez View Post
    Hi RonL

    Thank you. Am I on the right track with this below then for normal dist MGF:

    1+tx\sigma^2+\frac{t^2\sigma^4x^2}{2!}+... Mr F says: Where have these x's come from? See main reply.

    I am checking the pages for MGF and normal distribution for this formula below:

    E(x^{2k})=(2k)!\; a_{2k}

    But I did not see it yet, where does this come from please?

    In first point, am I doing power series expansion wrong? Because I do not know where 2^n will come from for denominator in final equation or how to see that odd moments are 0.

    Thank you again
    The MGF you quoted in your original question is for X ~ N(0, \sigma): m_X(t) = e^{\frac{\sigma^2 t^2}{2}}. So I don't know where you got those x's from in your power series ....??

    By definition, the (2k)th moment is given by E(X^{2k}) = M^{(2k)}(0). That is, you have to differentiate the MGF 2k times and then substitute t = 0.

    The power series is m_X(t) = 1 + \frac{\sigma^2 t^2}{2} + \frac{\left(\frac{\sigma^2 t^2}{2}\right)^2}{2!} + \, .... + \, \frac{\left(\frac{\sigma^2 t^2}{2}\right)^k}{k!} + \, ....

    Note that the kth term of this power series is \frac{ \left( \frac{\sigma^2 t^2}{2} \right)^k}{k!} = \frac{\sigma^{2k} t^{2k}}{2^{2k} k!}.

    If you think about it, the answer you want is got by differentiating this term 2k times .....
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  5. #5
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    Hello Mr F

    Thank you for your help, I tell you where I get my x's!:
    I put in x's because I thinking like this:

    t is variable so i differentiate e^{\frac{\sigma^2 t^2}{2}} with respect to t which I thought would give me t\sigma^2 each time so I would have:

    1 +t\sigma^2e^{\frac{\sigma^2 t^2}{2}}+t^2\sigma^4e^{\frac{\sigma^2 t^2}{2}}+t^3\sigma^6e^{\frac{\sigma^2 t^2}{2}}..

    which I equate against polynomial expression set up and differentiated at same time

    a + bx + cx^2 + dx^3

    I was confuse about this method because I trying to think what point I expand about which no make sense. Is it completely wrong to do this?

    I was thinking that you need e^{\frac{\sigma^2 t^2}{2}} to be e^{\frac{\sigma^2 t^2x}{2}} so that you get {\frac{\sigma^2 t^2}{2}} when you differentiate. Is this property of MGF that I don't understand maybe?
    How does this prove that odd moments are not there?

    Thank you again.
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  6. #6
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    Quote Originally Posted by mr fantastic View Post

    Note that the kth term of this power series is \frac{ \left( \frac{\sigma^2 t^2}{2} \right)^k}{k!} = \frac{\sigma^{2k} t^{2k}}{2^{2k} k!}.
    The kth term of this power series is \frac{ \left( \frac{\sigma^2 t^2}{2} \right)^k}{k!} = \frac{\sigma^{2k} t^{2k}}{2^{k} k!}.
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  7. #7
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    Quote Originally Posted by jescavez View Post
    Hello Mr F

    Thank you for your help, I tell you where I get my x's!:
    I put in x's because I thinking like this:

    t is variable so i differentiate e^{\frac{\sigma^2 t^2}{2}} with respect to t which I thought would give me t\sigma^2 each time so I would have:

    1 +t\sigma^2e^{\frac{\sigma^2 t^2}{2}}+t^2\sigma^4e^{\frac{\sigma^2 t^2}{2}}+t^3\sigma^6e^{\frac{\sigma^2 t^2}{2}}..

    which I equate against polynomial expression set up and differentiated at same time

    a + bx + cx^2 + dx^3

    I was confuse about this method because I trying to think what point I expand about which no make sense. Is it completely wrong to do this?

    I was thinking that you need e^{\frac{\sigma^2 t^2}{2}} to be e^{\frac{\sigma^2 t^2x}{2}} so that you get {\frac{\sigma^2 t^2}{2}} when you differentiate. Is this property of MGF that I don't understand maybe?
    How does this prove that odd moments are not there?

    Thank you again.
    Sorry but you have the wrong idea. The MGF is function of t and so its power series expansion is a function of t. The result is given in my previous reply. The expansion is the standard expansion for an exponential function (it's expanded around the point t = 0). You might need to review this idea a bit more thoroughly.

    The odd moments are zero because if you differentiate an odd number of times, all the terms in the power series expansion contain t. When you put t = 0 the result is therefore zero.
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  8. #8
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    Thank you Mr F. Is there any place online that gives a detailed step by step explanation of MGF like this?

    I see where 2k! come from now also, thanks.
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  9. #9
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    Quote Originally Posted by jescavez View Post
    Thank you Mr F. Is there any place online that gives a detailed step by step explanation of MGF like this?

    I see where 2k! come from now also, thanks.
    What steps do you still need in this question? They can be shown here ....
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  10. #10
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    I guess the part where we say that

    <br />
m_X(t) = e^{\frac{\sigma^2 t^2}{2}}<br />
    has expansion
    <br />
m_X(t) = 1 + \frac{\sigma^2 t^2}{2} + \frac{\left(\frac{\sigma^2 t^2}{2}\right)^2}{2!} + \, .... + \, \frac{\left(\frac{\sigma^2 t^2}{2}\right)^k}{k!} + \, ....<br />
    I no understand that very well. I thought that MGF is proved by Power series expansion of e^{tx} it seems here we just take exponent and leave as it is, why is it unaffacted?

    I also do not know where this assumption come from:
    <br />
E(X^{2k}) = M^{(2k)}(0)<br />
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  11. #11
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    Quote Originally Posted by jescavez View Post
    I guess the part where we say that

    <br />
m_X(t) = e^{\frac{\sigma^2 t^2}{2}}<br />
    has expansion
    <br />
m_X(t) = 1 + \frac{\sigma^2 t^2}{2} + \frac{\left(\frac{\sigma^2 t^2}{2}\right)^2}{2!} + \, .... + \, \frac{\left(\frac{\sigma^2 t^2}{2}\right)^k}{k!} + \, ....<br />
    I no understand that very well. I thought that MGF is proved by Power series expansion of e^{tx} it seems here we just take exponent and leave as it is, why is it unaffacted?

    I also do not know where this assumption come from:
    <br />
E(X^{2k}) = M^{(2k)}(0)<br />
    You know that the series expansion for e^w is 1 + w + \frac{w^2}{2!} + \frac{w^3}{3!} + \, ....... \, + \frac{w^k}{k!} + \, .......

    So to get the series expansion of m_X(t) = e^{\frac{\sigma^2 t^2}{2}} you can just make the substitution w = \frac{\sigma^2 t^2}{2}. Then you will get the result I stated earlier.


    "I thought that MGF is proved by Power series expansion of e^{tx} it seems here we just take exponent and leave as it is, why is it unaffacted?"

    That is done as part of the proof that E(X^{n}) = M^{(n)}(0)<br />
.


    Summary of general results:

    1. The MGF of a random variable X is m_X(t) = E \left( e^{tX} \right).

    2. The nth moment of X , that is, E\left(X^{n}\right) is given by \frac{d^n m_X}{dt^n} evaluated at t = 0, that is, E\left(X^n \right) = m_X^{(n)}(0).


    In your problem, m_X(t) = e^{\frac{\sigma^2 t^2}{2}} and n = 2k. It is convenient to get the (2k)th derivative of m_X(t) by writing it as a power series. Alternatively, you could just differentiate  e^{\frac{\sigma^2 t^2}{2}} 2k times without using the expansion.

    Do not confuse expressing the calculated MGF as a power series in order to easily do the differentiations with using a power series to prove that the nth moment of the MGF is given by the formula in 2. above.

    I hope this clears things up for you.
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  12. #12
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    Hi,

    I'm working on what seems to be essentially the same problem. I'm stuck in one area.

    I understand that the mgf can be written \frac{t^{2k}}{(k!*2^k)} *\sigma^{2k}

    but I don't understand how to get to


    the t^{2k} -> (2k)! is tripping me up.

    Thanks!
    Last edited by mr fantastic; December 28th 2009 at 04:34 PM. Reason: Fixed the sigma
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  13. #13
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    Quote Originally Posted by reflex View Post
    Hi,

    I'm working on what seems to be essentially the same problem. I'm stuck in one area.

    I understand that the mgf can be written \frac{t^{2k}}{(k!*2^k)} *\sigma^{2k}

    but I don't understand how to get to


    the t^{2k} -> (2k)! is tripping me up.

    Thanks!
    You need to differentiate with respect to t the mgf 2k times and then substitute t = 0. Please show your work.
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  14. #14
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    In your problem, m_X(t) = e^{\frac{\sigma^2 t^2}{2}} and n = 2k.

    I understand now, I think. I just have one more question and I'm not sure if it makes sense. I understand it 'works out' if n = 2k, but how do we justify that?
    Last edited by reflex; January 26th 2010 at 03:10 AM.
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