Insurance company with claims following a Poisson Process. Calculate the probability that the capital is always positive throughout the first 4 days.

Question

Suppose that claims are made to an insurance company according to a Poisson process with rate $10$ per day. The amount of a claim is a random variable that has an exponential distribution with mean $1,000$ dollars. The insurance company receives payments continuously in time at a constant rate of $11,000$ dollars per day. Starting with zero initial capital, find the probability that the firm’s capital is always positive throughout its first $4$ days. (Book: Simulation by Sheldon Ross, Chapter 7 Problem 11)

I initially simulated this to approximate the answer as around $13.5$% but I want to find an exact form.

I note that the distribution of the total amount $x$ claimed after $d$ days can be represented with a Gamma($10d,1000$) distribution: $$F(x) = \int_0^x \frac{1}{(1000)^{10d}\Gamma(10d)}t^{10d-1}e^{-\frac{t}{1000}}dt$$

For example, since the insurance company receives $11,000$ dollars after $1$ day, $F(11000)\approx 0.6595$, which means that after $1$ day, the probability of having a negative capital is $1-0.6595 = 0.3405$. However, this doesn't account for the firm once having a negative capital within the day, even if at the end of the day, the capital is positive.

I'm having trouble finding how to deal with this part. My thinking is to find $$\epsilon \cdot P(\text{positive capital on days } [0,\epsilon]) + \epsilon \cdot P(\text{positive capital on } [\epsilon, 2\epsilon] \text{ given positive capital on } [0,\epsilon]) + \epsilon \cdot P(\text{positive capital on } [2\epsilon, 3\epsilon] \text{ given positive capital on } [0,2\epsilon]) + ...$$ but I run into a dead end here since finding $P(\text{positive capital on } [\epsilon, 2\epsilon])$ would require knowing the capital I start with after $\epsilon$ days.

Answer 1

This process can be described using a classical result from ruin theory - Cramér–Lundberg model. Let $X_t^x$ be the firm's capital at the time $t$ (in days) provided that initial capital was equal to $x$, and according to this model $$X_t^x = x + c t - \sum_{k=1}^{N(t)} \xi_k,$$ where $c$ is the payments rate ($11000$ in your case), $N(t)$ is the Poisson process with intensity $\lambda$ ($10$ in your case) for claims, and $\xi_k$ are claims themselves, being independent exponentially distributed random variables.

Your question is essentially about the finite-horizon survival probability, which is defined as $\varphi(x,t) = 1 - \mathbb{P}\left(\inf_{0\leq s \leq t} X_s^x < 0\right)$. There is a classic results in ruin theory about an integro-differential equation for $\varphi(x, t)$: $$ \begin{cases} \frac{\partial \varphi(x, t)}{\partial t} - c \frac{\partial \varphi(x, t)}{\partial x} + \lambda \varphi(x, t) = \lambda \int_0^x \varphi(x-y, t) \mathrm{d} F(y), \\ \varphi(x, 0) = 1, \;\lim_{x\to+\infty} \varphi(x, t) = 1, \end{cases} $$ where $\lambda$ is the rate of Poisson process and $F(y)$ is the CDF of $\xi_k$ (see, e.g. Ruin Probabilities: Smoothness, Bounds, Supermartingale Approach).

Denoting $\mu = 1000$ (mean claim), $F(y) = 1 - e^{-y/\mu}$ for $y \geq 0$, so $\mathrm{d}F(y) = \frac{1}{\mu} e^{-y/\mu} \mathrm{d}y$, and $$ \frac{\partial \varphi(x, t)}{\partial t} - c \frac{\partial \varphi(x, t)}{\partial x} + \lambda \varphi(x, t) = \frac{\lambda}{\mu} \int_0^x \varphi(x-y, t) e^{-y/\mu} \mathrm{d} y. $$ This integral can be simplified using the properties of convolution: $$ \int_0^x \varphi(x-y, t) e^{-y/\mu} \mathrm{d} y = \int_0^x \varphi(y, t) e^{-(x-y)/\mu} \mathrm{d} y = e^{-x/\mu} \int_0^x \varphi(y, t) e^{y/\mu} \mathrm{d} y, $$ so the final form of the equation is $$ \begin{cases} \frac{\partial \varphi(x, t)}{\partial t} - c \frac{\partial \varphi(x, t)}{\partial x} + \lambda \varphi(x, t) = \frac{\lambda}{\mu} e^{-x/\mu} \int_0^x \varphi(y, t) e^{y/\mu} \mathrm{d} y, \\ \varphi(x, 0) = 1, \;\lim_{x\to+\infty} \varphi(x, t) = 1, \end{cases} $$ and the value you need is $\varphi(0, 4)$.

Finding exact solution to this type of equations is quite complicated, and I'm not sure it can be solved in this particular case (although it's relatively easy to find an exact solution for infinite horizon case with exponentially distributed claims), so numerical methods may be the only option.

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EDIT. In fact, there exists a rather cumbersome explicit formula for $\psi(u,t) = 1 - \varphi(u,t)$ for exponentially distributed claims (see Statistical Tools for Finance and Insurance, p. 368). Assuming that $c = 1$ and $\beta = \frac{1}{\mu} = 1$ (exponential distribution is defined here to have CDF $1-e^{-\beta x}$ for $x \geq 0$), $$ \psi(u, T)=\lambda e^{-(1-\lambda) u}-\frac{1}{\pi} \int_0^\pi \frac{f_1(x, u, T) f_2(x, u)}{f_3(x)} \mathrm{d} x, $$ where $$ f_1(x, u, T)=\lambda \exp \{2 \sqrt{\lambda} T \cos x-(1+\lambda) T+u(\sqrt{\lambda} \cos x-1)\}, \\ f_2(x, u) = \cos (u \sqrt{\lambda} \sin x)-\cos (u \sqrt{\lambda} \sin x+2 x), \\ f_3(x) = 1+\lambda-2 \sqrt{\lambda} \cos x. $$ If $\beta \neq 1$ and $c \neq 1$, than $\lambda$ should be replaced with $\frac{\lambda}{\beta c} = \frac{\lambda \mu}{c}$, $u$ with $\beta u = u/\mu$, and $T$ with $\beta c T = \frac{c}{\mu} T$. So, the final formula is $$ \psi(u, T)=\lambda e^{-\left(1-\frac{\lambda \mu}{c}\right) u}-\frac{1}{\pi} \int_0^\pi \frac{f_1(x, u/\mu, cT/\mu) f_2(x, u/\mu)}{f_3(x)} \mathrm{d} x, $$ and the integral can be calculated numerically.

For given values of $\lambda = 10$, $\mu = 1000$, $c = 11000$, $\frac{\lambda\mu}{c} = \frac{10}{11}$, so $$ \psi(0,4) = \frac{\lambda\mu}{c} - \frac{1}{\pi} \int_0^\pi \frac{f_1(x, 0, cT/\mu) f_2(x, 0)}{f_3(x)} \mathrm{d} x = \\ = \frac{10}{11} - \frac{1}{\pi} \int_0^\pi \frac{10 e^{-84 + 8\sqrt{110} \cos x}(1-\cos 2x)}{21 - 22 \sqrt{\frac{10}{11}} \cos x} \mathrm{d}x, $$ and approximate value given by Wolfram is $0.863789$, meaning that $\varphi(0,4) \approx 0.136211$, which is quite close to your number obtained from simulation.

Answer 2

Let $C(t)$ be the amount of capital at time $t\in[0,4].$ Then

$$C(t)=11000t-(X_1+\dots + X_{N(t)})$$ where $N(t)\sim \text{Poisson}(10t)$ and $X_j\sim \exp(.001)$ are independent. You need to find $$\mathbb{P}(\forall t\in [0,4]:C(t)>0)$$

But $C(t)> 0$ is the same as $M(t)\geq N(t)$ where $M(t)\sim \text{Poisson}(11t)$ is the process defined by $$M(t)=\sup\Big\{n\Big|X_1 + \dots + X_n < 11000t\Big\}$$

Arrivals for the Poisson process $M(t)$ happen at a rate of $11$ per day and independently from $N(t)$.

That said, we consider a Markov chain $\{X_n\}_{n\geq 0}$ with $X_0=0$, state space $\{...,-1,0,1,...\}$, and transition probabilities $$\mathbb{P}(X_{n}=j+1|X_{n-1}=j)=\frac{11}{21} \\ \mathbb{P}(X_n=j-1|X_{n-1}=j)=\frac{10}{21}$$

A positive jump in this Markov chain signals a "type $M$" arrival in the compound process $N(t)+M(t)\sim \text{Poisson}(21t)$

If $\tau=\inf\{n:X_n<0\}$, then $\tau$ is supported on $\{1,3,5,...\}\cup\{\infty\}$ and satisfies $$\mathbb{P}(\tau=2k-1)=\Big(\frac{10}{21}\Big)^k\Big(\frac{11}{21}\Big)^{k-1}C_{k-1}$$ for $k\geq 1$. Here $C_{k}=\frac{1}{k+1}{ 2k \choose k}$ are the Catalan numbers, while $\{\tau=\infty\}=\bigcap_{n\geq 0}\{X_n\geq 0\}$ which is the event that the Markov chain never dips below $0$; this event has probability $$\mathbb{P}(\tau =\infty)=1-\sum_{k=1}^{\infty}\mathbb{P}(\tau=2k-1)=\frac{1}{11}$$

Let $p_n$ be the probability that $X_k\geq 0$ for all $ k \leq n$. Then $p_0=1$ and for $n\geq 1$ we have $$p_n=1-\mathbb{P}(\tau\leq n)=1-\sum_{k=1}^{\lceil\frac{n}{2}\rceil}\mathbb{P}( \tau=2k-1)$$ This equals to the following expression. $$p_n=1-\sum_{k=1}^{\lceil \frac{n}{2}\rceil}\frac{1}{k}{2(k-1) \choose k-1} \Big(\frac{10}{21}\Big)^{k}\Big(\frac{11}{21}\Big)^{k-1} \text{ for } n\geq 1$$

The number of steps taken in this Markov chain on $t\in[0,4]$ is governed by the random variable $N(4)+M(4)\sim \text{Poisson}(84)$.

Via the total law, probability you seek is $\sum_{n=0}^{\infty}p_n \mathbb{P}(N(4)+M(4)=n)$ which equals the following expression: $$1-\sum_{n=1}^{\infty}e^{-84}\frac{84^n}{n!}\sum_{k=1}^{\lceil \frac{n}{2}\rceil}\frac{1}{k}{2(k-1) \choose k-1} \Big(\frac{10}{21}\Big)^{k}\Big(\frac{11}{21}\Big)^{k-1}$$