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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 ultimate total claim count. 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 Bayes' Theorem, the 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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3 Answers 3

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Hint: It might assist you to look at a well-known relationship between the geometric distribution and the negative-binomial distribution. (Have a look at their moment generating functions and see if you notice anything about them.)

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  1. Find the conditional distribution of $X$ conditional on $Y$, $\mathbb{P}[X|Y]$, which involves $d$.
  2. Get the joint distribution of $X$ and $Y$, $\mathbb{P}[X,Y] = \mathbb{P}[Y]\mathbb{P}[X|Y]$.
  3. Get marginal distribution of X, $\mathbb{P}[X] = \sum _Y\mathbb{P}[X,Y]$.
  4. Your answer: Conditional distribution of $Y$ given $X$,$\mathbb{P}[Y|X] = \frac {\mathbb{P}[X,Y]}{\mathbb{P}[X]}$.
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Suppose that $r$ policies were issued in 2005. Let $t_1, \cdots, t_r$ denote the times at which accident claims are reported for policies issued in 2005. Additionally, let $z_1, \cdots, z_r$ denote the number of accident claims associated with these policies upon being reported. Assume $t_1, \cdots, t_r$ are independent, $z_1, \cdots, z_r$ are independent, and $t_i,z_i$ are pairwise independent for all $i$. In other words, the reporting time and number of accident claims are independent of one another for a given a policy, and each policy is independent. We shall also assume that $t_i^{\ast}=\mbox{I}\left(t_i = 2005\right) \sim Bernoulli(d)$.

Let $Y$ denote the total number of accident claims associated with all policies issued in 2005. Clearly, from the above set-up, $Y = \sum_{i=1}^r z_i$. Since we assume that $Y \sim \mbox{NB}(r,p)$, it must be the case that $z_i \sim \mbox{Geo}(p)$; that is, the probability mass function of $z_i$ is \begin{eqnarray*} f_{z_i}(z) = p(1-p)^z \quad \mbox{for} \quad z \in \{0, 1, \cdots \}. \end{eqnarray*}

Furthermore, let $X$ denote the number of accident claims associated with all policies issued in 2005 that are reported in 2005. Therefore, $X = \sum_{i=1}^r t_i^{\ast}z_i$. Let $u_i = t_i^{\ast}z_i$. Clearly, $u_1, \cdots, u_r$ are independent and we can write $X = \sum_{i=1}^r u_i$.

Next, we can express $Y$ as $Y=\sum_{i=1}^r t_i^{\ast}z_i + \sum_{i=1}^r \left(1-t_i^{\ast}\right)z_i$. Let $v_i = \left(1-t_i^{\ast}\right)z_i$ and $W=\sum_{i=1}^r v_i$. From the previous independence assumptions, $v_1, \cdots, v_r$ are independent. Therefore, we can write $Y = \sum_{i=1}^r \left(u_i + v_i\right) = X+W$. Although, $u_i$ and $v_j$ are indepedent for $i \ne j$, $u_i$ and $v_i$ are dependent; thus, $X$ and $W$ are dependent. Hence, we will have to use a bivariate discrete transformation to find the joint density of $(u_i,v_i)$ from the joint density of $(t_i^{\ast},z_i)$. This is necessary since by Bayes' theorem, we can write \begin{eqnarray*} P[Y=y|X=x] &=& \frac{P[X=x,Y=y]}{P[X=x]} \\ &=& \frac{P[X=x,W=y-x]}{P[X=x]}. \end{eqnarray*}

Hence, we need only find the joint probability mass function of $(X,W)$ as well as the probability mass function of $X$. To this end, we begin by finding the joint distribution of $(u_i,v_i)$. Note that the transformation from $(t_i^{\ast},z_i)$ to $(u_i,v_i)$ is almost one-to-one since we can solve for $t_i$ and $z_i$ in terms of $u_i$ and $v_i$ as $t_i = u_i/(u_i+v_i)$ and $z_i=u_i+v_i$. Clearly, this transformation breaks down when $(u_i,v_i)=(0,0)$ since $t_i$ will be indeterminate. It is straightforward to show that if $(u_i,v_i)=(0,0)$, then $(t_i^{\ast},z_i)=(0,0)$ or $(t_i^{\ast},z_i)=(1,0)$. When $(u_i,v_i)\ne(0,0)$, it is important to note that either $u_i=0$ or $v_i=0$ since $t_i^{\ast}$ is a Bernoulli random variable. Using the previous transformations, with $u,v \in \mathbb{N}$, if $(u_i,v_i)=(0,v)$, then $(t_i^{\ast},z_i)=(0,v)$ and if $(u_i,v_i)=(u,0)$, then $(t_i^{\ast},z_i)=(1,u)$. Hence, the joint probability mass function of $(u_i,v_i)$ is \begin{eqnarray*} P[u_i=u,v_i=v] = \begin{cases} p & \mbox{if} \,\, (u,v) = (0,0) \\ p(1-d)(1-p)^v & \mbox{if} \,\, u=0,v \in \mathbb{N} \\ pd(1-p)^u & \mbox{if} \,\, v=0,u \in \mathbb{N} \\ 0 & \mbox{otherwise} \end{cases}. \end{eqnarray*}

Since $X$ and $W$ are comprised of $r$ i.i.d. terms, the probability generating function of $X$ and $(X,W)$ may be expressed as the probability generating functions of $u_i$ and $(u_i,v_i)$ raised to the $r$th power. For completeness, we have that the probability generating function of $(X,W)$ is \begin{eqnarray*} G_{X,W}\left(z_1,z_2\right) &=& \mbox{E} \left[z_1^X z_2^W\right] \\ &=& \mbox{E} \left[z_1^{\sum_{i=1}^r u_i} z_2^{\sum_{i=1}^r v_i} \right] \\ &=& \prod_{i=1}^r\mbox{E} \left[z_1^{u_i} z_2^{v_i} \right] \\ &=& \left(\mbox{E} \left[z_1^{u_i} z_2^{v_i} \right]\right)^r \\ &=& \left(G_{u_i,v_i}\left(z_1,z_2\right)\right)^r, \end{eqnarray*} where $G_{u_i,v_i}\left(z_1,z_2\right)$ denotes the probability generating function of $(u_i,v_i)$. Clearly, the probability generating function of $X$, $G_X(z_1) = G_{X,W}\left(z_1,1\right)$.

Now, the probability generating function of $(u_i,v_i)$ is \begin{eqnarray*} G_{u_i,v_i}\left(z_1,z_2\right) &=& \sum_{u=0}^\infty\sum_{v=0}^\infty P[u_i=u,v_i=v] z_1^u z_2^v \\ &=& p + p(1-d) \sum_{v=1}^\infty \left[z_2(1-p)\right]^v + pd \sum_{u=1}^\infty \left[z_1(1-p)\right]^u \\ &=& p(1-d)+p(1-d) \frac{z_2(1-p)}{1-z_2(1-p)} + pd + pd \frac{z_1(1-p)}{1-z_1(1-p)} \\ &=& \frac{p(1-d)}{1-z_2(1-p)} + \frac{pd}{1-z_1(1-p)}. \end{eqnarray*}

By independence, the probability generating function of $(X,W)$ is \begin{eqnarray*} G_{X,W}\left(z_1,z_2\right) &=& \left[\frac{p(1-d)}{1-z_2(1-p)} + \frac{pd}{1-z_1(1-p)}\right]^r. \end{eqnarray*} Using the above equation, we can compute probabilities of $(X,W)=(x,w)$ by using the relation \begin{eqnarray*} P[X=x,W=w] = \frac{ \frac{\partial^{x+w}}{\partial z_1^x \partial z_2^w} G_{X,W}(z_1,z_2)|_{(z_1,z_2)=(0,0)}}{x!w!}. \end{eqnarray*}

Therefore, the answer to the question may be written as \begin{eqnarray*} P[Y=y|X=x] &=& \frac{1}{(y-x)!}\frac{ \frac{\partial^{y}}{\partial z_1^x \partial z_2^{y-x}} G_{X,W}(z_1,z_2)|_{(z_1,z_2)=(0,0)}}{\frac{\partial^{x}}{\partial z_1^x } G_{X,W}(z_1,1)|_{z_1=0}}. \end{eqnarray*}

For instance, we may find $P[Y=0|X=0]$ by evaluating $G_{X,W}(0,0)$ and $G_{X,W}(0,1)$. Specifically, \begin{eqnarray*} P[Y=0|X=0] = \left[\frac{p}{(1-d)+pd}\right]^r. \end{eqnarray*}

This value agrees with that found by simulating. See the following R code

d=.4
p=.3
r=6
B=10^6
cond.check = 0
l = 0
set.seed(555)
for (i in 1:B){
    ti = rbinom(r,1,d)
    zi = rnbinom(r,1,p)
    X = sum(ti*zi)
    Y = sum(zi)
    W = Y-X
    if (X==0){
        cond.check = cond.check + (Y==0)
        l = l+1
    }
}
(cond.check = cond.check/l) #0.005273457
(p/(1-d+p*d))^r  #0.005232781
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