I am not able to understanding a passage on a paper that talk about total variation between the law of numbers of fixed points in a random permutations ($\pi_{N}(x)$) and the Poisson distribution with ${\lambda = 1}$. This paper is "On a Markov construction of couplings" (Persi Diaconis and Laurent Miclo).
Actually I tried to use the recursive formula of derangements, but with no useful results.
Here is what I am trying to understand:
${\sum _{x=0}^{N} {|\pi_{N}(x) - P(x,1)|}}$
= $ {\sum _{x=0}^{N} (\frac{D_{N-x}}{(N-x)!} - {e^{-1}})} \frac{1}{x!}$
= ${\sum _{x=0}^{N} (\frac{D_{x}}{x!} - {e^{-1}})} \frac{1}{N-x!}$
How did it go from $(\frac{D_{N-x}}{(N-x)!} - {e^{-1}}) \frac{1}{x!}$ to $(\frac{D_{x}}{x!} - {e^{-1}}) \frac{1}{N-x!}$
Could you help me? Thank you very much for your help!
For compactness, denote $D(x) = \frac{D_x}{x!} - e^{-1}$. So, the first summation is $\sum_{x=0}^{N} D(N-x)\frac{1}{x!}$ and the resultant summation is $\sum_{x=0}^{N} D(x)\frac{1}{(N-x)!}$.
If we expand the summation, we have \begin{align} \sum_{x=0}^{N} D(N-x) \frac{1}{x!} & = D(N) \frac{1}{0!} + D(N-1) \frac{1}{1!} + \dots + D(0)\frac{1}{N!}\\ & = D(0)\frac{1}{N!} + \dots + D(N-1) \frac{1}{1!} + D(N) \frac{1}{0!} \\ & = \sum_{x=0}^N D(x) \frac{1}{(N-x)!}. \end{align}
They are swapping $x$ in the previous summation to $N-x$ in the new one and vice versa.