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Find the $n^\text{th}$ moment of $p(x=k)=\dfrac{pq^k}{1-q^{N+1}}$, $k=0,1,2,\ldots,N$ , $0\lt p\lt 1$ and $q=1-p$.

My approach is $$M_x(t)=\sum e^{xt}\cdot p(x)=\frac{p}{1-q^{N+1}} \cdot \sum_{x=0}^N e^{tx}q^x$$

Now, $M_x^n(t)=\frac{p}{1-q^{N+1}}\cdot\sum_{x=0}^N x^ne^{tx}q^x$

Thus, $M_x^n(0)=\frac{p}{1-q^{N+1}}\cdot\sum_{x=0}^N x^nq^x$

Can we get any closed form for the expression?

Michael Hardy
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Vikas Sharma
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  • Note that $\sum_{x=0}^N e^{tx}q^x = \sum_{x=0}^N (e^t q)^x$. – Semiclassical Aug 02 '21 at 18:10
  • That said, while computing $M_X(t)$ is tractable I don't think this leads to a useful closed form for the moments. If you take $N\to\infty$ then this reduces to a geometric random variable: The moments of such are known, but are expressed in terms of the polylogarithm function. So the moments of your 'truncated' random variable are going to be even worse. – Semiclassical Aug 02 '21 at 18:28

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The sum $\displaystyle \sum_{x=0}^N e^{tx}q^x = \sum_{x=0}^N \big( e^t q \big)^x$ is a finite geometric series with first term $1$ and common ratio $e^tq.$ There is a closed form for the sum of a finite geometric series.

I would write $\displaystyle \sum_{k=0}^N e^{tk}q^k$ rather than $\displaystyle \sum_{x=0}^N e^{tx}q^x$ for two reasons: You used the letter $k$ for that earlier, and it is not a good idea to use the same symbol, $x,$ to refer both to the random variable and to the index in this sum.

Michael Hardy
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