# Probability of reaching ruin time with Brownian motion with positive drift

Question

You are about to start a new job. As your new job is quite well paid, and you are generally responsible with money, you are confident that your income will exceed your outgoings and that your net balance will increase over time. Moreover, you are fairly confident that your financial situation will be healthy after a year or two. However, your main concern at this point is that you want to avoid ever having a negative bank balance so that you can avoid being given a poor credit rating or receiving any  financial penalties.

You have decided that it reasonable to model your future financial balance by $$\{X(t)\}$$, a Brownian motion with positive drift: $$X(t) = c + \mu t + \sigma W(t),$$

where $$W(t)$$ is a standard Brownian motion. Here, $$c > 0$$ is a constant that represents your initial bank balance.

Calculate the probability of never having a negative bank balance, assuming that your model is correct. Express your answer in terms of $$c, \mu \ \mathrm{and}\ \sigma^2$$.

My attempt

\begin{aligned} X(t) & \sim N( c + \mu t,\sigma^2 t) \\ \implies \mathbb{P}[X(t) >0] & = \mathbb{P}\left(Z > \frac{0 - (c + \mu t)}{\sqrt{\sigma^2 t}}\right)\\ & = \Phi\left(\frac{c+\mu t}{\sigma \sqrt{t}}\right) \end{aligned}

However, according to the phrasing of the question, I believe my answer should not contain t, so where have I gone wrong and how should I proceed further?

Any intuitive explanations will be highly appreciated!

• The event of never having a negative bank balance, means that $X(t) \ge 0$ for all $t$, i.e you need to calculate $\mathbb{P}(\inf_t X(t) \ge 0)$. Commented Nov 5, 2023 at 3:51
• @ThànhNguyễn I see. But how should I proceed then? Commented Nov 5, 2023 at 8:39

Define the stopping time $$\tau=\inf\{t:X_t= 0\}$$ with the convention that $$\inf\emptyset=\infty$$. You want to compute $$P(\tau=\infty)$$. Here we are assuming $$c,\mu>0$$. Define the process $$(e^{\theta X_t})_{t \geq 0}$$. Denote with $$(\mathscr{F})_{t\geq 0}$$ the natural filtration of $$W=(W_t)_{t \geq 0}$$. We get for $$s: $$E[e^{\theta (X_t-X_s)}|\mathscr{F}_s]=e^{\mu\theta (t-s)}E[e^{\sigma \theta (W_t-W_s)}]=e^{\mu\theta (t-s)+\sigma^2\theta^2(t-s)/2}$$ We choose $$\theta$$ s.t. $$\mu\theta+\sigma^2\theta^2/2=0$$, which is given by $$\theta=-2\mu/\sigma^2<0$$. In this case, $$(e^{\theta X_t})_{t \geq 0}$$ is thus a $$\mathscr{F}_t$$-martingale. We choose the bounded stopping time $$\tau\wedge n$$ and by optional stopping theorem we get $$e^{\theta c}\stackrel{\textrm{OST}}{=}E[e^{\theta X_{\tau\wedge n}}]=P(\tau Now, if $$\tau\geq n$$ then $$e^{\theta X_n}\leq 1$$ a.s., so $$e^{\theta X_n}\mathbf{1}_{\{\tau\geq n\}}\in [0,1]$$ a.s. Furthermore, $$X_n=n\bigg(\frac{c}{n}+\mu+\sigma\frac{W_n}{n}\bigg)=n\bigg(\frac{c}{n}+\mu+\sigma\frac{1}{n}\sum_{k \leq n}(W_k-W_{k-1})\bigg)\stackrel{\textrm{a.s.}}{\to}\infty$$ Since the term in parenthesis goes to $$\mu>0$$ a.s. by law of large numbers. So $$e^{\theta X_n}\mathbf{1}_{\{\tau \geq n\}}\leq e^{\theta X_n}\to 0$$ a.s. So by dominated convergence and continuity of measures (that is, $$P(\tau) we get $$P(\tau<\infty)=e^{\theta c}\implies P(\tau=\infty)=1-e^{\theta c}$$