# Can we compute the probability that a Brownian motion hits a quadratic?

Let $A$ be the event that there is some $t$ such that $B_t=1+t^2$, where $B$ is a Brownian mation. Is there any way to compute the probability of $A$, or to approximate it well?

I ask, because we can calculate the probability that $B$ hits a line explicitly. First we change measure with Cameron-Martin, which allows us to reduce to the problem of a Brownian motion hitting a constant. Then we use the reflection principle, which gives the joint distribution of a Brownian motion and its maximum. This approach doesn't seem to generalize to the quadratic.

Thank you.

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## 1 Answer

Some insights which maybe helpful: let us assume that $F_\alpha(t)$ is a cdf for the first hitting time of constant level $\alpha>0$ for Brownian motion and $\tau$ is a first hitting time of moving boundary $1+t^2$.

Consider $X_t = B_t - t^2$, i.e. $$dX_t = -2t\cdot dt+ dB_t$$ with $X_0 = 0$. Let us use Girsanov theorem (generalization of Cameron-Martin one) and put $\theta_s = 2s$, so
$$Z_t = \exp\left\{-\frac23t^3+2\int\limits_0^ts\cdot dB_s\right\}$$ will be a likelihood ratio process leading to the fact that for any finite $T$ the process $(X_t)_{0\leq t\leq T }$ is a Brownian motion under new measure ${\mathsf Q}(\cdot) = \mathsf E[Z_T 1(\cdot)]$. As a results we have: $$\mathsf Q\{\tau\leq T\} = \mathsf Q\left\{\sup\limits_{t\leq T}X_t\geq1\right\} =F_1(T)$$ for any finite $T$ since $X_t$ is a Brownian motion under $\mathsf Q$. If we put $\xi_t = \frac{1}{Z_t} = \exp\left\{-\frac23 t^3-2\int\limits_0^ts \,dX_s\right\}$ then $$\mathsf P\{\tau\leq T\} = \tilde{\mathsf E}\left[\xi_T1\{\tau\leq T\}\right].$$ As a rough upper-bound we can apply Hölder's inequality to get: $$\mathsf P\{\tau\leq T\}\leq \sqrt{\tilde{\mathsf E}[\xi_T^2]}\sqrt{F_1(T)}$$ where $\tilde{\mathsf E}[\xi^2_t] = \exp\{4/3\cdot t^3\}$ so you have $\mathsf P\{\tau\leq T\}\leq F_1(T)\exp\{4/3\cdot T^3\}$ which are pretty rough.

On the other hand you can use different method, namely find $f(a,t)$ such that $$\exp\{aX_t + f(a,t)\}$$ is a martingale for any $a$, then by applying Ito formula you obtain $f(a,t) = at^2-\frac12a^2t$. Then $$\mathsf E[\exp\{aX_{\tau\wedge T} + f(a,\tau\wedge T)\}] = \mathsf E[\exp\{aX_0 + f(a,0)\}] = 1$$ for any $T<\infty$, since $\tau = \infty$ with non-negative probability.

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