I have been asked to expand (1-2t)^(-n/2). So far I have found the log of this function and then differentiated twice in order to find my mean and variance as n and 2n respectively. How do I write my final answer as a power series in t as far as the term in t^2?
Expansion of an MGF as a power series
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statistics
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0possible duplicate of [Expansion of a function with a negative fractional power.](http://math.stackexchange.com/questions/13424/expansion-of-a-function-with-a-negative-fractional-power) – 2010-12-08
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0This is an *exact* duplicate of the cited prior question (the above comment was auto-generated by the vote to close). – 2010-12-08
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0I tried to ask my Q as a part to the previous mentioned Q, but someone said I should ask my own Q! But I am aware it is the exact same Q, I just dont understand how to write my final answer! – 2010-12-08
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0@Sarah: Most likely you were given that advice because in the prior thread since you said you had a *similar* question. But in fact your question is precisely the same as the prior one. – 2010-12-08
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1@Bill: this new question is not the same as the prior one (it is much easier). – 2010-12-08
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0@Shai: How would you tackle this question? I would be grateful for your help! – 2010-12-08
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0I'll answer the question shortly. – 2010-12-08
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0That would be great! Thank you. =) – 2010-12-08
2 Answers
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You have already done most of the work. If the mean $\mu'_1$ and variance $\sigma^2$ are equal to $n$ and $2n$, respectively, then the second moment $\mu'_2$ is given by $\mu'_2 = \sigma^2 + (\mu'_1)^2 = n(n+2)$. Now, the moment-generating function can be expanded as follows: $$ M(t) = 1 + \mu' _1 t + \frac{1}{{2!}}\mu' _2 t^2 + \cdots = 1 + nt + \frac{{n(n + 2)}}{2}t^2 \cdots. $$
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If you want to expand your function in a power series on the argument $t$ you can write its Taylor Polynomials .