Boundary value problem for Laplace's equation / probability a BMP leaves cone via the ice cream rather than the biscuit

Question

Suppose I start a Brownian motion $X_t$ at position $(0,0,\varepsilon)$ in $\mathbb R^3$. Suppose also I have a cone with its apex at the origin, given by $U=\{(x,y,z):\sqrt{x^2+y^2}< z\tan\alpha\}$ (recall $\alpha$ is the angle of incidence). Let $\tau_1$ be the first time $X_t\in U^c$, and $\tau_2$ the first time $|X_t|=r$. If $\tau=\min(\tau_1,\tau_2)$, then what is the probability that $|X_\tau|=r$ (or equivalently, $\tau=\tau_2$)? In other words, how likely is $X_t$ to remain in the cone for a significant amount of time? An explicit formula would be nice, but I am equally interested in a lower bound for this probability.

Bonus: Is there anything interesting to be said if $\varepsilon=0$? I don't have much intuition for Brownian motion yet, but if I had to guess, I would presume that any sensible formulation of this question would just result in the probability being zero?

Edit: The projection onto the $z$-axis is a 1-dimensional BMP and thus is a.s. negative in every interval $[0,t]$, so the answer to the bonus question is indeed that there is nothing interesting to be said.

Second Edit: This has been mentioned in a comment, but I wanted to reiterate that this can be viewed as a boundary value problem for Laplace's equation: $u(w)=\mathbb P(\tau=\tau_2\left|X_0=w\right.)$ is the unique bounded function which satisfies Laplace's equation in $U\cap B(0,r)$ and $u(w):=\chi_{\{|w|=r\}}$ on the boundary (except possibly on a null set wrt surface measure). If a separable solution exists, this should greatly simplify the question, but I have not been able to find it. And indeed, maybe the behaviour of $u$ should (in spherical coordinates) resemble something like $R^{\alpha-\theta}$, in which case the solution won't be separable.

Answer 1

Without loss of generality we assume $r=1$. The probability you are looking for is given by $f(X_0)$, where solving for $\Delta f=0$ with $f=0$ on the cone and $f=1$ on the 1-sphere. In particular, if we find a solution $\tilde f$ with values in $[0,1]$ of $\Delta \tilde f=0$ with $\tilde f=0$ on the cone and $0\leq \tilde f \leq 1$ on the 1-sphere then we will have $f\geq \tilde f$.

Let us look for such a function $\tilde f$ that can be expressed as $f = g(r) h(\theta)$, where $(r, \theta, \phi)$ are polar coordinates: the Laplacian is $$ \Delta \tilde f = \frac{\partial^2 f}{\partial r^2} + \frac{2}{r} \frac{\partial f}{\partial r} f + \frac{1}{r^2 \sin \theta}\left(\cos \theta \frac{\partial f}{\partial \theta} + \sin \theta \frac{\partial^2 f}{\partial \theta^2}\right) $$ so that $r^2\Delta \tilde f = 0$ becomes $$ r^2 g'' h + 2r g'h + \frac{1}{\tan \theta} gh' + h'' = 0 . $$ Let us look for a solution where $g(r) = r^\beta$ (necessarily $\beta>0$): this becomes $$ \beta (\beta+1) h + \frac{h'}{\tan \theta} + h'' = 0 . $$ It turns out (thanks to a formal solver) that the solutions to this differential equation can be expressed in terms of Legendre functions. Let us check it: since $\theta \in [0,\pi/2]$ and $\cos$ is a bijection on this interval, write $h(\theta) = y(\cos(\theta))$: $y$ satisfies the ODE $$ (1-u^2) y''(u) - 2u y'(u) + \beta(\beta+1) y(u) = 0 $$ which is exactly Legendre's equation for $\lambda = \beta$ and $\mu=0$. Hence, any solution is of the form $y(u) = c_1 P_\beta(u) + c_2 Q_\beta(u)$ for some $c_1, c_2\in\mathbb R$. Because we want $\tilde f$ to take values in $[0,1]$, we need to have $\tilde f(1,0,0) = y(1) \in [0,1]$. Since $Q_\beta(1) = +\infty$ this means we must have $c_2 = 0$; and because $P_\beta$ is maximal (on $[0,1]$) at $1$ and equals $1$ there we can take $c_1=1$.

Of course this $\tilde f$ is only a solution if it is zero on the cone, i.e. if $y(\cos\alpha)=0$ and $y>0$ on $(\cos\alpha, 1]$. We can check that the last zero of $P_\beta$ on $[-1,1]$ is an increasing function of $\beta$ and goes to $1$ as $\beta\to\infty$ and equals zero when $\beta=1$, hence for every $\alpha \in (0,\pi/2]$ we can find $\beta \in [1,\infty)$ such that $$ \tilde f = r^{\beta} P_\beta(\cos \theta) \leq f . $$ This means that your probability of exiting the cone by the ice cream rather than the biscuit is at least $\varepsilon^\beta$. Taking a cone with a slightly larger angle of incidence and bounding $f$ from above you could show that it is at most $C(\beta') \varepsilon^{\beta'}$ for every $\beta'<\beta$, with an explicit constant $C(\beta')$ that goes to $+\infty$ as $\beta'\to\beta$.

In fact I would conjecture that $f(\varepsilon, 0, 0) \sim c \varepsilon^\beta$ as $\varepsilon \to 0$ for some $c\geq 1$ (perhaps even $c=1$).

Update: in fact, every $\beta\geq 1$ such that $P_\beta(\cos\alpha) =0$ gives a solution $f_\beta = r^\beta P_\beta(\cos(\theta))$ to $\Delta \tilde f_\beta=0$ that is zero on the cone. This solution is bounded but might not be positive. My knowledge of Legendre functions is lacking, but it seems reasonable to expect that for any fixed $\alpha$ the set of such $\beta$ forms an increasing sequence $(\beta_n)_{n\geq 1}$, where $\beta_1$ is the $\beta$ we found above. We could then completely describe our original $f$ if we manage to approximate the indicator of the interval $(\cos\alpha, 1]$ with the functions $(P_{\beta_n})$ restricted to $[\cos\alpha, 1]$, à la Fourier. This could lead to a proof of my conjectured asymptotic.