This is a past exam question and I just want your opinions on if it's sufficient or not.
I had to prove:
Let the discrete random variable $X$ have a Poisson distribution with parameter $\lambda$. Then $$\sum_{r=0}^{n}rP[x=r] = {\lambda}P[x \leq n-1]$$
So this is what I did:
$$\sum_{r=0}^{n}rP[x=r] = \sum_{r=0}^{n}{r}{\frac{e^{(-{\lambda})}{\lambda}^r}{r!}}$$
The term is 0 when $r=0$
$$=\sum_{r=1}^{n}{r}{\frac{e^{(-{\lambda})}{\lambda}^{r}}{r(r-1)!}}$$
$$= \lambda \sum_{r=1}^{n}\frac{e^{(-\lambda)}{\lambda}^{r-1}}{(r-1)!}$$
Let $z = r-1$
$$= \lambda \sum_{z=0}^{n-1}\frac{e^{(-\lambda)}{\lambda}^{z}}{z!}$$
Observe that $$\frac{e^{(-\lambda)}{\lambda}^0}{0!} + \frac{e^{(-\lambda)}{\lambda}^1}{1!} + \frac{e^{(-\lambda)}{\lambda}^2}{2!} + ... + \frac{e^{(-\lambda)}{\lambda}^{n-2}}{(n-2)!} + \frac{e^{(-\lambda)}{\lambda}^{n-1}}{(n-1)!} = P[x=0] + P[x=1] + P[x=2] + ... + P[x=n-2] + P[x=n-1] = P[x \leq n-1]$$
Thus, $$\lambda \sum_{z=0}^{n-1}\frac{e^{(-\lambda)}{\lambda}^{z}}{z!} = \lambda P[x \leq n-1]$$