Conditional Poisson Distribution 
The number of claims, $N$, in a year on a portfolio of policies
  follows a Poisson distribution with parameter $λ$. Large claims have a
  probability $p$ and small claims $(1-p)$, independently of each other.
  Suppose we observe $r$ large claims. Show that the conditional
  distribution of $N-r$ given $r$ is Poisson and find its mean.

I know we can use the definition of a conditional probability and then invoke independence condition on big claims and small claims. But how we find out which PDF does Big claims and small claims follows?
 A: Let $X$ and $Y$ denote the number of large claims and small claims respectively
in an year. Then, $N = X+Y$ is the total number of claims and is known to
be a Poisson$(\lambda)$ random variable. Now, given that $N=n$, the
conditions stated in the problem tell us that the conditional
distribution of $X$ is binomial with parameters $(n,p)$ and that of $Y$
is binomial with parameters $(n,1-p)$. Note that conditioned on $N=n$, $X$ and $Y$ are very much dependent random variables since $Y=n-X$. 
But, unconditionally,
$X$ and $Y$ are independent Poisson$(\lambda p)$ and Poisson$(\lambda(1-p))$
random variables.
To see why all this is so, consider that for $0 \leq r \leq n$,
$$P\{X=r, Y=s \mid N=n\}
= \begin{cases} \displaystyle\binom{n}{r}p^r(1-p)^{s}, & \text{if} ~s = n-r,\\
0, & \text{if} ~ s \neq n-r.
\end{cases}$$
Consequently, for any $r, s \geq 0$,
$$\begin{align}
P\{X = r, Y = s\} &= \sum_{n=0}^\infty P\{X=r, Y=s, N=n\}\\
&= \sum_{n=0}^\infty P\{X=r, Y=s \mid N=n\}P\{N =n\}\\
&=\binom{r+s}{r}p^r(1-p)^{s}e^{-\lambda}\frac{\lambda^{r+s}}{(r+s)!}
&\scriptstyle{\text{only the $n=r+s$ term is nonzero in the sum}}\\
&= \frac{(r+s)!}{r!s!}(\lambda p)^r(\lambda(1-p))^{s}
e^{-\lambda p - \lambda(1-p)}\frac{1}{(r+s)!}\\
&= e^{-\lambda p}\frac{(\lambda p)^r}{r!}\cdot 
e^{-\lambda(1-p)}\frac{(\lambda(1-p))^s}{s!}\\
&= P\{X=r\}P\{Y = s\}
\end{align}$$
showing that $X$ and $Y$ are independent Poisson$(\lambda p)$ and
Poisson$(\lambda(1-p))$ random variables respectively. Consequently, conditioned
on $X = r$, $Y$ continues to be a Poisson$(\lambda(1-p))$ random variable.
