Looking for clarity on the mathematical induction of $\sum_{x=r}^\infty (x-r)f(x) = -(r-\lambda)(1-F(r))+\lambda f(r) $ I got this equation that I want to proof by induction.
The equation
$$\sum_{x=r}^\infty (x-r)f(x) = -(r-\lambda)(1-F(r))+\lambda f(r) $$
where $f(x)$ and $F(x)$ are respectively the pmf and the CDF of the Poisson distribution.
So $$f(x) = \frac{e^{-\lambda}\lambda^x}{x!}$$ and $$F(x) = \sum_{i=0}^x \frac{e^{-\lambda}\lambda^i}{i!}$$
I want to proof this result via mathematical induction.
The base step
$r=0$
$$\sum_{x=0}^\infty x f(x) = -(0-\lambda)(1-F(0))+\lambda f(0) = \lambda(1-F(0)+f(0)) = \lambda$$
The inductive step
$$\sum_{x=r+1}^\infty (x-r+1)f(x)$$
$$=\sum_{x=r}^\infty (x-r+1)f(x) - (x-r+1)f(r)$$
$$=\sum_{x=r}^\infty (x-r+1)f(x) - f(r)$$
$$=\sum_{x=r}^\infty (x-r)f(x) - f(r) + (1-F(r-1))$$
$$=\sum_{x=r}^\infty (x-r)f(x) - f(r) + 1-F(r)+f(r)$$
$$=\sum_{x=r}^\infty (x-r)f(x) + 1-F(r)$$
$$=-(r-\lambda)(1-F(r))+\lambda f(r)+1-F(r)$$
$$=-(r-\lambda-1)(1-F(r))+\lambda f(r)$$ or
$$=(-r+1+\lambda)(1-F(r))+\lambda f(r)$$
The question
I would think that I would end up with $(-r+1+\lambda)(1-F(r+1))+\lambda f(r+1)$, but this isn't equal to $\sum_{x=r+1}^\infty (x-r+1)f(x)$ and $(-r+1+\lambda)(1-F(r))+\lambda f(r)$ is.
So my question is; are my calculations complete to finish the proof(and if not, what am I missing) and why is $$\sum_{x=r+1}^\infty (x-r+1)f(x)=(-r+1-\lambda)(1-F(r+1))+\lambda f(r+1)$$ not the outcome of this inductive step?
 A: Your inductive step begins by figuring out
$$\sum_{x=r+1}^\infty (\color{red}{x-r+1})f(x),$$
but that is not part of anything you set out to show. Shouldn't you be trying to figure out
$$\sum_{x=r+1}^\infty \left(\color{red}{x-(r+1)}\right)f(x)\,?$$
A: Forgot: for the people interested in the full derivation of the inductive step:
$$\sum_{x=r+1}^\infty (x-(r+1)) f(x) $$
$$=\sum_{x=r+1}^\infty x-r-1 f(x) $$
$$=\sum_{x=r}^\infty x-r-1 f(x) + f(r) $$
$$=\sum_{x=r}^\infty x-r f(x) - \sum_{x=r}^\infty f(x)+ f(r) $$
$$=\sum_{x=r}^\infty x-r f(x) - (1-F(r-1)) + f(r) $$
$$=-(r-\lambda)(1-F(r))+\lambda f(r) - (1-F(r-1)) + f(r) $$
$$=-(r-\lambda)(1-F(r+1)+f(r+1))+\lambda f(r) - (1-F(r-1)) + f(r) $$
$$=-(r-\lambda)(1-F(r+1))-rf(r+1)+\lambda f(r+1) +\lambda f(r) - (1-F(r-1)) + f(r) $$
Now because
$$\lambda f(r) = \frac{e^{-\lambda}\lambda^{r+1}}{r!} = (r+1)f(r+1) = (r+1)\frac{e^{-\lambda}\lambda^{r+1}}{(r+1)!} = \frac{e^{-\lambda}\lambda^{r+1}}{r!} $$, we can proceed with
$$=-(r-\lambda)(1-F(r+1))-\lambda f(r+1) + f(r+1) - (1-F(r-1)) + f(r) $$
And because $$(1-F(r+1) = 1-F(r+1)+f(r+1)+f(r)$$, we can proceed with:
$$=-(r-\lambda)(1-F(r+1))-\lambda f(r+1) - 1+F(r+1) $$
which leads to
$$=-((r+1)-\lambda)(1-F(r+1))+\lambda f(r+1)$$
