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How can I show that, for a given integer $n$, that the following identity is true? $$\left[(-1)^n\binom{n-1}{m}-\sum_{k=m}^{n-2}(-1)^k\binom{n}{k+1}\binom{k}{m}\right]=-(-1)^m$$

This came up in a derivation of arbitrary derivatives of the Black-Scholes function. It is part of a larger equation for the $n^{th}$ order derivative $$\sum_{m=0}^{n-2}\frac{H_m(d)}{(\sigma\sqrt{\tau})^m}\left[(-1)^n\binom{n-1}{m}-\sum_{k=m}^{n-2}(-1)^k\binom{n}{k+1}\binom{k}{m}\right]$$ which, in order to make contact with a known result (Carr, Peter. (2000). Deriving Derivatives of Derivative Securities. Journal of Computational Finance. 4. 101 - 128. https://ieeexplore.ieee.org/document/844609) this identity must hold. I've also checked that it's true in sagemath.

Tom Davis
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2 Answers2

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$$ \begin{align} &(-1)^n\binom{n-1}{m}-\sum_{k=m}^{n-2}(-1)^k\binom{n}{k+1}\binom{k}{m}\\ &=\sum_{k=m}^{n-1}(-1)^{k+1}\binom{n}{n-k-1}\binom{k}{k-m}\tag1\\ &=(-1)^{m+1}\sum_{k=m}^{n-1}\binom{n}{n-k-1}\binom{-m-1}{k-m}\tag2\\ &=(-1)^{m+1}\binom{n-m-1}{n-m-1}\tag3\\[9pt] &=(-1)^{m+1}[n\ge m+1]\tag4 \end{align} $$ Explanation:
$(1)$: $(-1)^n\binom{n-1}{m}$ becomes the $m=n-1$ term of the sum
$\phantom{\text{(1):}}$ furthermore, $\binom{k}{m}=\binom{k}{k-m}$
$(2)$: $\binom{k}{k-m}=(-1)^{k-m}\binom{-m-1}{k-m}$ (negative binomial coefficients)
$(3)$: Vandermonde's Identity
$(4)$: Iverson Brackets

robjohn
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  • The first step relies on $n \geq m+1$ already, and the OP's formula would run into troubles for $n = 0$ anyway. So I think $n \geq m+1$ should just be a standing requirement here. – darij grinberg Mar 04 '21 at 12:28
  • Yes, m<n-1 indeed in the application I am looking at. Thank you all, I now need to see if I can simplify the calc. – Tom Davis Mar 04 '21 at 13:01
  • If binomial identities are formulated properly, the limits of the indices can often be left out, or at least simplified. In this case, $\binom{n}{k+1}=0$ if $k\gt n-1$ and $\binom{k}{m}=0$ if $0\le k\lt m$. We do constrain $k\ge0$ since if $k\lt0$ we don't have $\binom{k}{m}=\binom{k}{k-m}$. Thus, if we simply sum over all $k\ge0$, the limits are imposed by the binomial coefficients. – robjohn Mar 04 '21 at 21:57
  • Does Vandermonde's Identity still hold when the sum does not start from $0$? – Tom Davis Mar 05 '21 at 13:41
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    The bottom indices go to $0$ at the ends of the sum. – robjohn Mar 05 '21 at 14:06
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Have an alternative solution that takes a different approach than the one here. Consider mine more opaque and involved, but worth sharing for the fun of it.

We start with Eq. 1.23 in this reference changing the labeling for convenience $$\sum_{k=0}^{n'}(-1)^k \binom {n'} k \binom {k} m x^k=(-1)^mx^m(1-x)^{n'-m} \binom{n'}{m} $$ and apply $\binom n k = \frac{n}{k} \binom {n-1} {k-1}$ as $$\sum_{k=0}^{n'}(-1)^k \frac{k+1}{n'+1}\binom {n'+1} {k+1} \binom {k} m x^k=(-1)^mx^m(1-x)^{n'-m} \binom{n'}{m}. $$ Substitution $n=n'+1$ then yields $$\sum_{k=0}^{n-1}(-1)^k \frac{k+1}{n}\binom {n} {k+1} \binom {k} m x^k=(-1)^mx^m(1-x)^{n-m-1} \binom{n-1}{m}. $$ Integrate both sides, assume that constants are 0, $x=1$ to get $$\frac{1}{n}\sum_{k=0}^{n-1}(-1)^k \binom {n} {k+1} \binom {k} m=\frac{(-1)^m}{n}. $$ Integral on the RHS is tricky in general, but evaluates nicely in the present case.

Now we can split off the $n-1$ term from the sum and multiply by $n$ to obtain $$(-1)^{n-1} \binom {n-1} m + \sum_{k=0}^{n-2}(-1)^k \binom {n} {k+1} \binom {k} m=(-1)^m. $$ Finally, we multiply by $-1$, note that binomial coefficients with $k < m$ in the sum are 0 and get $$(-1)^{n} \binom {n-1} m - \sum_{k=m}^{n-2}(-1)^k \binom {n} {k+1} \binom {k} m=-(-1)^m. $$

J.K.
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