Let $X$ be Poisson distributed where $P(X=n) = p_n(\lambda) = \frac{\lambda^n}{n!}e^{-\lambda}$ for $n \geq 0$. Show $$P(x \leq n) = 1 - \int_0^\lambda P_n(x) dx.$$
This is the argument given:
$\frac{d}{d\lambda} p_n = p_{n-1} - p_n$ where $p_{-1} = 0$. Hence $$(\frac{d}{d \lambda})P(X \leq n) = p_n(\lambda).$$
Can someone please explain this argument better, because I am understanding nothing at all.
I think the correct formula should be
$$\tag{1}(\frac{d}{d \lambda})P(X \leq n) = -p_n(\lambda).$$
(with a minus sign in the RHS). The proof of (1) is as follows:
Let us apply the classical rule $d(fg)/d\lambda=(df/d\lambda).g+f.(dg/d\lambda)$ to the formula:
$$p_n(\lambda) = \frac{\lambda^n}{n!}e^{-\lambda}$$
We get:
$$\frac{d}{d\lambda} p_n = n\dfrac{\lambda^{n-1}}{n!}e^{-\lambda} + \dfrac{\lambda^{n}}{n!}(-e^{-\lambda})=\dfrac{\lambda^{n-1}}{(n-1)!}e^{-\lambda} - \dfrac{\lambda^{n}}{n!}e^{-\lambda}=p_{n-1} - p_n \ \ \ \text{(property} \ P_n)$$
Thus, adding the equations of properties $P_0, P_1, \cdots P_n$, one obtains (because of concellations in the RHS):
$$\frac{d}{d\lambda}(p_0+p_1+\cdots+p_n)=p_0-p_n=-p_n$$
proving (1) because the LHS is nothing but $p_0+p_1+\cdots+p_n=P(X \leq n)$.
Edit : (following a good remark of @Henry) How convention
$$\tag{2}p_{-1}=0$$
can be explained ? Let us have a look at case $n=0$ :
$$\frac{d}{d\lambda} p_0 = \frac{d}{d\lambda} e^{-\lambda}=-e^{-\lambda}=-p_0=p_{-1}-p_0$$
if one takes convention (2).
Edit (following another remark of @Henry) : Let us establish the first relationship written under the following form (with a capital $X$ and a more neutral variable "$u$" in the integral):
$$\tag{3}P(X \leq n) = 1 - \int_0^\lambda P_n(u) du.$$
Here is a proof; clearly, from (1), one can deduce :
$$\tag{4}P(X \leq n) = K - \int_0^\lambda P_n(u) du$$
for a certain constant $K$ (this kind of additive constant appears whenever a primitive function is taken on both sides of an equality). The question is now, comparing (3) and (4) : why do we have $K=1$ ?
In fact, there is an explicit expression for the integral in (4) (see (https://en.wikipedia.org/wiki/List_of_integrals_of_exponential_functions)):
$$\tag{5}\int_0^\lambda P_n(u) du=\int_0^\lambda \frac{u^n}{n!}e^{-u} du=\left[\left(\sum_{k=0}^n \frac{u^k}{k!}\right)e^{-u}\right]_{u=0}^{u=\lambda}=\left(\sum_{k=0}^n \frac{\lambda^k}{k!}\right)e^{-\lambda}-1$$
Plugging this result into (4), one gets
$$\tag{6}P(X\leq n)=K+\left(\sum_{k=0}^n \frac{\lambda^k}{k!}\right)e^{-\lambda}-1$$
When $n \rightarrow \infty$, for any fixed value of $\lambda$,
the LHS of (6) tends to $1$,
the RHS of (6) tends to $K+e^{\lambda}e^{-\lambda}-1=K+1-1=K$ (due to the series expansion of "exp" function).
explaining why $K=1.$