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I'm trying to solve this question for a continuous surplus process.

The surplus process is $$U_s=U_0+s-B_s$$ where $B_t$ is a Brownian motion representing payouts, $U_0$ is starting capital, $s$ is premiums

How do I calculate the probability of hitting zero (probability of ruin) before hitting a level $X$, where $X>U_0$?

Thoughts: I've been shown to how to calculate the probability of ruin at a specific time, but I'm not sure how to find the probability of 'infinite' ruin and incorporate in the hitting $0$ before $X$ factor.

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Define

$$\tau:=\tau_x := \inf\{t \geq 0; U_t \notin [0,x]\}.$$

Since

$$U_t = U_0+t-B_t \geq U_0-B_t$$

we can conclude from the continuity of the sample paths and the fact that

$$\sigma := \inf\{t \geq 0; B_t = U_0-2x\} < \infty \quad \text{a.s.}$$

that $\tau<\infty$ a.s. Now set

$$M_t := e^{-2(U_t-U_0)}.$$

Applying Itô's formula, it follows easily that $(M_t)_{t \geq 0}$ is a martingale. Hence, by the optional stopping theorem (applied to the bounded stopping time $t \wedge \tau$)

$$\mathbb{E}M_{t \wedge \tau} = 1.$$

By the definition of the stopping time, we have $|U_{t \wedge \tau}| \leq U_0$; hence $|M_{t \wedge \tau}| \leq e^{2U_0}$. This means that we may apply the dominated convergence theorem and obtain

$$\mathbb{E}M_{\tau}=1. \tag{1}$$

On the other hand,

$$\mathbb{E}M_{\tau} = e^{-2(x-U_0)} \cdot \mathbb{P}(U_{\tau}=x) + e^{2U_0} \mathbb{P}(U_{\tau}=0). \tag{2}$$

Using that $\mathbb{P}(U_{\tau}=0) = 1- \mathbb{P}(U_{\tau}=x)$, we find by combining $(1)$ and $(2)$ that

$$\mathbb{P}(U_{\tau}=0) = \frac{e^{-2U_0}-e^{-2x}}{1-e^{-2x}}.$$


Alternative solution For a diffusion process

$$dX_t = \mu_t \, dt + \sigma_t \, dB_t$$

the scale function is defined by

$$s(x) := \int_{x_0}^x \exp \left(- \int_{y_0}^y \frac{2\mu(z)}{\sigma^2(z)} \, dz \right) \, dy$$

for arbitrary $x_0,y_0$. It satisfies (see e.g. Revuz-Yor)

$$\mathbb{P}^x(\tau_a<\tau_b) = \frac{s(b)-s(x)}{s(b)-s(a)} \tag{3}$$

for any $a<x<b$. If we apply this in the given setting, we find (for $x_0 := 0$, $y_0 :=0$)

$$s(x) := e^{-2x}-1$$

Plugging this into $(3)$ yields

$$\mathbb{P}^{U_0}(\tau_0<\tau_x) = \frac{e^{-2x}-e^{-2U_0}}{e^{-2x}-1}.$$

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  • $\begingroup$ Question: if I take the limit of X->infinity on the final answer, I would expect the probability of ruin to approach 1 (intuitively) as the upper limit moves further away but (I'm not sure if I'm doing it right) I am getting 0 when I take the limit here. Not sure why? $\endgroup$ – JIM May 18 '14 at 10:12
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    $\begingroup$ @JIM No, the limit approaches $e^{-2U_0}$ if $x \to \infty$. But yes, intuivitely, one might expect that the probability converges to $1$. That this is not the case is in my oppinion due to the (positive) drift. $\endgroup$ – saz May 18 '14 at 10:54
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    $\begingroup$ @saz, thanks for the nice solutions, as well as correcting my own. As for the result in the limit $X\rightarrow+\infty$, this is understandable by acknowledging that $\tau_0=\inf\{t\ge0\ :\ B_t=t+U_0\}$ is not always finite. In fact, when $X=\infty$, the probability of hitting $0$ before $X$ is the probability that $\tau_0$ is finite. It can be shown (by the same kind of arguments developed here) that if $\tau_{a,b}=\inf\{t\ge0\ :\ B_t=a+bt\}$, where $a,b\ge0$, then $\mathbb P(\tau_{a,b}<+\infty)=e^{-2ab}$. In the case at hand, $\mathbb P(\tau_0<+\infty)=e^{-2U_0}$. $\endgroup$ – Ian May 18 '14 at 14:06
  • $\begingroup$ @Ian Right ... and this is -intuitively- due the positive drift. (See also this question.) $\endgroup$ – saz May 18 '14 at 17:15
  • $\begingroup$ Nice and clean, +1. $\endgroup$ – Did May 20 '14 at 7:52

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