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A property-casualty insurance company issues automobile policies on a calendar year basis only. Let $X$ be a random variable representing the number of accident claims reported during calendar year 2005 on policies issued during calendar year 2005. Let $Y$ be a random variable representing the total number of accident claims that will eventually be reported on policies issued during calendar year 2005. The probability that an individual accident claim on a 2005 policy is reported during calendar year 2005 is $d$. Assume that the reporting times of individual claims are mutually independent. Assume also that $Y$ has the negative binomial distribution, with fixed parameters $r$ and $p$, given by $$\mathbb{P}[Y=y]=\binom{r+y-1}{y}p^{r}(1-p)^{y}$$ for $y=0,1,\ldots$. Calculate $\mathbb{P}[Y=y|X=x]$ the probability that the total number of claims reported on 2005 policies is $y$, given that $x$ claims have been reported by the end of the calendar year.

Remark: I know that the solution requires the use of Baye's Theorem and Theorem of Total Probability, and the identity $\binom{y}{x}\binom{r+y-1}{y}=\binom{r+x-1}{x}\binom{(r+x)+(y-x)-1}{y-x}$.

I have not been able to correctly describe $X$ or include $d$ in the analysis. I need your help to understand this better.

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$Y$ represents the total random claim count on year 2005 issued policies. $X$ represents the subset of those claims that were reported in the same calendar year of issue. For a sufficiently large number of claims made, you would expect that $X \approx Y \cdot d$. Another way to say this is that given that $Y$ claims are reported across year 2005 issued policies, the random number $X$ of these claims that were reported in the same year is binomial, since each such claim is independent and identically distributed with probability $d$ of occurring in the same calendar year. Formally, we would write $$X \mid Y \sim \operatorname{Binomial}(Y, d), \\ \Pr[X = x \mid Y = y] = \binom{y}{x} d^x (1-d)^{y-x}, \quad x \in \{0, 1, \ldots, y\}.$$ So now we apply the law of total probability to obtain the marignal (unconditional) distribution of $X$: $$\Pr[X = x] = \sum_{y=0}^{\infty} \Pr[X = x \mid Y = y] \Pr[Y = y].$$ Then, we apply Bayes' theorem: $$\Pr[Y = y \mid X = x] = \frac{\Pr[X = x \mid Y = y]\Pr[Y = y]}{\Pr[X = x]},$$ where the numerator is simply the summand in the previous equation, and the denominator is the value of that sum.

I have left the actual computation of these to you as an exercise, as I regard the salient aspect of your question to be the fact that $X \mid Y$ is binomial.

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