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Define $$\Delta^ra_i= \sum_{j=0}^{r}\binom{r}{j}(-1)^{r-j}a_{i+j}.$$ For consistency also let $$\Delta^0a_i=a_i.$$

My problem is the following: for two sequences $a_i$ and $b_i$, $i=0,\ldots,\nu$, show that $$ \sum_{r=0}^{\nu}a_r\binom{\nu}{r}\Delta^rb_i=\sum_{j=0}^{\nu}b_{i+j}\binom{\nu}{j}(-1)^{\nu-j}\Delta^{\nu-j}a_j. $$

This problem is taken from the book of William Feller-An introduction to Probability Theory and its allocations. Vol II, page 221. (http://gen.lib.rus.ec/book/index.php?md5=5EE1E3564BAA8470668E6A8A28C7E416).

Chamberlain Mbah
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1 Answers1

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I'm sure there are much more elegant ways of doing it, but here goes: \begin{align} \sum_{r=0}^\nu a_r \binom{\nu}{r} \Delta^r b_i &= \sum_{r=0}^\nu a_r \binom{\nu}{r} \sum_{j=0}^r \binom{r}{j}(-1)^{r-j}b_{i+j} \\ &= \sum_{j=0}^\nu b_{i+j} \sum_{r=j}^\nu a_r \binom{\nu}{r} \binom{r}{j}(-1)^{r-j} \\ &= \sum_{j=0}^\nu b_{i+j} \sum_{r=0}^{\nu-j} a_{r+j} \binom{\nu}{r+j} \binom{r+j}{j} (-1)^r \\ &= \sum_{j=0}^\nu b_{i+j} \binom{\nu}{j} \sum_{r=0}^{\nu-j} a_{r+j} \binom{\nu-j}{r}(-1)^r \\ &= \sum_{j=0}^\nu b_{i+j} \binom{\nu}{j} (-1)^{\nu-j} \sum_{r=0}^{\nu-j}\binom{\nu-j}{r}(-1)^{\nu-j-r}a_{r+j} \\ &= \sum_{j=0}^\nu b_{i+j} \binom{\nu}{j} (-1)^{\nu-j} \Delta^{\nu-j}a_j. \end{align} Explanation of each step:

(1) Expand $\Delta^r b_i$ using the definition.

(2) Swap the sums: This changes $\sum_{r=0}^\nu \sum_{j=0}^r$ into $\sum_{j=0}^\nu \sum_{r=j}^\nu$.

(3) Shift the second sum so that it starts from $r=0$: This changes every $r$ into $r+j$.

(4) Use the fact that $\binom{\nu}{r+j} \binom{r+j}{j} = \binom{\nu}{j} \binom{\nu-j}{r}$.

(5) Note that $(-1)^r = (-1)^{-r} = (-1)^{2(\nu-j)}(-1)^{-r} = (-1)^{\nu-j}(-1)^{\nu-j-r}$.

(6) Use the definition to contract the second sum into $\Delta^{\nu-j}a_j$.

ekkilop
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