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Math Help - [SOLVED] MGF of Standard Normal

  1. #1
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    [SOLVED] MGF of Standard Normal

    Hi Maths Forum. Suppose that X is N ( μ , 1) and Y = X^2. Find the MGF of Y.

    Can anybody help please with workings?

    Thanks.
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  2. #2
    Moo
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    Hello,

    Brute force... I can't see any simpler way. See the attached pdf.
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  3. #3
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    Quote Originally Posted by Moo View Post
    Hello,

    Brute force... I can't see any simpler way. See the attached pdf.
    The same problem appeared in a recent thread, in a slightly more complicated way (more parameters, hence heavier computation), and I devised a computation-free way in my second post (after having spent some time on the direct way in the first one): when t>0,

    \int e^{-t x^2}e^{-\frac{1}{2}(x-\mu)^2}\frac{dx}{\sqrt{2\pi}}\frac{\sqrt{t}}{\sqrt  {\pi}}

    is the convolution product of the pdf's of a N(0,\frac{1}{2t}) and a N(0,1), evaluated at the point \mu, thus it is the pdf of a N(0,1+\frac{1}{2t}) evaluated at the point \mu (cf. explanation in the quoted post if necessary):

    \int e^{-t x^2}e^{-\frac{1}{2}(x-\mu)^2}\frac{dx}{\sqrt{2\pi}}\frac{\sqrt{t}}{\sqrt  {\pi}} = \frac{1}{\sqrt{2\pi(1+\frac{1}{2t})}}e^{-\frac{\mu^2}{2(1+\frac{1}{2t})}}.

    Thus, E[e^{-tY}]=\int e^{-t x^2}e^{-\frac{1}{2}(x-\mu)^2}\frac{dx}{\sqrt{2\pi}} = \frac{1}{\sqrt{2t+1}}e^{-\frac{t \mu^2}{2t+1}}.

    If you want E[e^{tY}] for positive t (in fact, for 0\leq t<2 otherwise it diverges), this method doesn't work directly. But you can use analytic continuation: t\mapsto E[e^{-t Y}] is analytic on (-2,+\infty) (using the integral formula,...), and it coincides with the above function when t> 0, therefore the coincidence extends to (-2,0], i.e. for t\geq -2. This gives:

    E[e^{tY}]= \frac{1}{\sqrt{1-2t}}e^{\frac{t \mu^2}{1-2t}}

    for any t<2.
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  4. #4
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    Non-Central chi-squared distribution

    Thank you to Moo and Laurent. Your contributions to the Maths web forum are invaluable. It is amazing how you can code and solve the sums at the same time!

    I realised that from the solutions there was a factor in the distributions. I used this to solve Q 2B.

    Can you guys help also with your thoughts/solutions with Q 1B?

    PK.
    Attached Thumbnails Attached Thumbnails [SOLVED] MGF of Standard Normal-set-5.jpg  
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  5. #5
    Moo
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    Laurent's method is more skilled

    For Q 1.B, just remember that the mgf of the sum of independent rv's is the product of the mgf's.
    Since the X_i are independent, the X_i^2 are independent too, and the result should follow easily.
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