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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?

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    $\begingroup$ Please add the self-study tag and read ts [tag wiki](stats.stackexchange.com/tags/self-study/info), modifying your question if necessary to follow the guidelines there. $\endgroup$ – Glen_b -Reinstate Monica Mar 7 '15 at 7:52
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    $\begingroup$ I do not believe the problem as stated has a unique solution. Something more is needed to connect the total number $N$ of claims with the number $L$ of large claims. For instance, $$L|N\sim\mathcal{B}(N,p)\,.$$ $\endgroup$ – Xi'an Mar 7 '15 at 14:18
  • $\begingroup$ i think your are right, but answer given is as: $$N-r|r\sim\mathcal{Poisson}(\lambda(1-p))$$ How about if we assume that small claims follows Binomial distribution then how can we simplify it using definition of conditional probability. $\endgroup$ – user53932 Mar 7 '15 at 15:28
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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.

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