# Escape probability for a randomly oriented, single fixed size step from a random coordinate inside a square

I have a square with side length $$s$$.

A (uniformly) random coordinate $$c = (x,y)$$ inside this square is the origin of a (uniformly) randomly oriented vector of fixed length $$l$$.

Vector endpoints can end up both within the square and outside of it, depending on their orientation and $$c$$. The escape probability for a point is not the same everywhere, it depends on their position.

A vector starting closer to the center/halfway along the side of the square has a lower probability than one in the corner to land outside.

Point p here can reach the sides but not the corner, everything closer to the corner can do both and therefore has an even higher probability.

So how does this probability distribution $$P_{escape}(s,l,x,y)$$ look like?

Can it be used to express the average expected escape probability for large amounts of randomly chosen vectors analytically?

$$P_{escape}(s,l) = ?$$

#### Some possibly irrelevant notes:

In the special case where $$l, there exists in the center of the square a region (here in green) where vectors, regardless of orientation, cannot reach the edge. In this special case, there is also a segment of the square edges that have probability 1/2

Also, if $$l = 0$$, then $$P(s,l) = 0$$, or if $$l > \sqrt{2*s^2}$$, then $$P(s,l) = 1$$

• Hint: Assume the bottom left corner of the square is the origin. Consider any point $(x,y)$ and draw a circle with radius $l$. Then any point on the circle be written as $(x + l \cos \theta, y + l \sin \theta)$ for $\theta \in [0, 2 \pi)$. Can you find the range of $\theta$ for which this point lies outside the square? Jun 30, 2023 at 21:32
• @sudeep5221 if you look at the range of $\theta$, the calculus gets quite messy. I've posted an alternative approach as an answer. Jul 2, 2023 at 12:03

To simplify things a little, let's make $$s=2$$ and we'll consider the square $$S$$ with vertices $$(\pm1,\pm1)$$.

Say the start point of the vector is $$P(u,v)$$ and its angle (measured anticlockwise from the positive $$x$$-axis, as usual) is $$\theta$$.

Then the endpoint of the vector is $$Q(u+l \cos \theta,v+l \sin \theta)$$.

It's tempting to pick $$P(u,v)$$ and work out the range of $$\theta$$ that makes $$Q$$ outside $$S$$, but this makes calculations very messy.

It's better to fix $$\theta$$, and consider the ranges of $$u$$ and $$v$$.

Let's restrict $$\theta$$ to be in the range $$\left(0,\frac{\pi}{2}\right)$$ (this will actually cover all cases, by symmetry). This means both $$\cos \theta$$ and $$\sin \theta$$ are positive.

We need to consider the cases $$l \le 2$$ and $$l>2$$ separately.

Case $$l \le 2$$

For the point $$Q$$ to be inside $$S$$, we need both $$-1

and

$$-1

These inequalities define a rectangle, with area $$(2-l \cos \theta)(2-l \sin \theta)$$

So, for a given $$\theta$$, the probability the point $$Q$$ is inside $$S$$ is $$P(\theta) = \frac{(2-l \cos \theta)(2-l \sin \theta)}{4}$$

(the ratio of the areas of the rectangle and the whole square).

Now, to find the overall probability, we need to integrate: \begin{align} P(\text{inside})&=\frac{2}{\pi} \int_0^\frac{\pi}{2} \frac{(2-l \cos \theta)(2-l \sin \theta)}{4} d\theta \\ &= 1+\frac{l(l-8)}{4\pi}\end{align}

So the probability we want (one minus this) is $$\boxed{P(\text{escape}) = \frac{l(8-l)}{4\pi}}$$

Case $$l>2$$

This method needs a slight adjustment for $$l>2$$. Note that now, if $$P=(-1,-1)$$ and $$\cos \theta > \frac{2}{l}$$, $$Q$$ will always be outside $$S$$.

This observation tells us what to do, though; we just change the limits for $$\theta$$ in the integral:

\begin{align} P(\text{inside})&=\frac{2}{\pi} \int_{\cos^{-1} \frac{2}{l}}^{\frac{\pi}{2} - \cos^{-1} \frac{2}{l}} \frac{(2-l \cos \theta)(2-l \sin \theta)}{4} d\theta \\ &= \frac{1}{4\pi} \left(8\sqrt{l^2-4}-16\cos^{-1} \frac{2}{l} + 4\pi - 8 - l^2 \right) \end{align}

So finally, in this case, $$\boxed{P(\text{escape}) = \frac{1}{4\pi} \left(8 + l^2 - 8\sqrt{l^2-4}+16\cos^{-1} \frac{2}{l} \right)}$$

The final position $$(X,Y)$$ is $$(C_x + l\cos\Theta,C_y + l\sin\Theta)$$ where: $$C_x,C_y \sim U(0,s)$$, $$\Theta \sim U(-\pi,\pi)$$, $$l\cos(\Theta)\sim\operatorname{Arcsine}(-l,l)$$.

Let $$L_x=l\cos\Theta$$ and $$L_y = l\sin\Theta$$
$$L_x = \lambda \Rightarrow L_y=\lambda\tan\Theta$$

\begin{align}F_{L_y}(l_y)&=P(L_y\leq l_y)\\&= P(\lambda\tan\Theta\leq l_y)\\&= P\left(\Theta\leq\tan^{-1}\left(\frac{l_y}{\lambda}\right)\right)\\&= F_\Theta\left(\tan^{-1}\left(\frac{l_y}{\lambda}\right)\right)\\&= \frac{1}{2\pi}\left(\tan^{-1}\left(\frac{l_y}{\lambda}\right)+\pi\right) \end{align}

$$\therefore P(L_y|L_x=\lambda)= \frac{1}{2\pi}\frac{\lambda}{\lambda^2+l_y^2}\\$$ \begin{align} P(L_x) &= \frac{1 }{\pi\sqrt{(l_x-(-l))(l-l_x)} } \\&= \frac{1 }{\pi\sqrt{l^2-l_x^2} } \end{align} \begin{align} P(C_y|L_x) &= P(C_y) \end{align} \begin{align} P(C_y+L_y|L_x) &= \int_{-\infty}^\infty{f_{C_y}(y−l_y)f_{L_y}(l_y)dl_y}\\&= \int_{y}^{y-s}{\frac{1}{s}\frac{1}{2\pi}\frac{\lambda}{\lambda^2+l_y^2}dl_y}\\&= \frac{\tan^{-1}\left(\frac{y-s}{\lambda}\right)-\tan^{-1}\left(\frac{y}{\lambda}\right)}{2 \pi s} \end{align} \begin{align} P(C_y+L_y) &= \int_{y}^{y-s}{\frac{1}{s} \frac{1 }{\pi\sqrt{l^2-l_y^2} }dl_y}\\&= \frac{\sin^{-1}\left( \frac{y-s}{l }\right)-\sin^{-1}\left( \frac{y}{l}\right) }{\pi s } \end{align} \begin{align} P(L_x|Y)&= P(L_x|C_y+L_y)\\&= P(C_y+L_y|L_x) \frac{P(L_x)}{P(C_y+L_y) }\\&= \frac{\tan^{-1}\left( \frac{y-s}{l_x }\right)-\tan^{-1}\left( \frac{y}{l_x }\right)}{2 \pi s } \frac{ \frac{1 }{\pi\sqrt{l^2-l_x^2} } }{\frac{\sin^{-1}\left( \frac{y-s}{l }\right)-\sin^{-1}\left( \frac{y}{l}\right) }{\pi s } }\\&= \frac{\tan^{-1}\left( \frac{y-s}{l_x }\right)-\tan^{-1}\left( \frac{y}{l_x }\right)}{2\pi\left(\sin^{-1}\left( \frac{y-s}{l }\right)-\sin^{-1}\left( \frac{y}{l}\right)\right)\sqrt{l^2-l_x^2} } \end{align} $$P(C_x|Y)=P(C_x)$$ \begin{align} \therefore P(X|Y)&= P(C_x+L_x|Y)\\&= \int_{-\infty}^\infty{f_{C_x}(x-l_x)f_{L_x}(l_x)dl_x}\\&= \int_{x}^{x-s}{ \frac{1}{s } \frac{\tan^{-1}\left( \frac{y-s}{l_x }\right)-\tan^{-1}\left( \frac{y}{l_x }\right)}{2\pi\left(\sin^{-1}\left( \frac{y-s}{l }\right)-\sin^{-1}\left( \frac{y}{l}\right)\right)\sqrt{l^2-l_x^2} }dl_x} \end{align}

This is as far as I got,

\begin{align}P(X,Y)&=P(X|Y)P(Y)\\ &=P(C_x+L_x|C_y+L_y)P(C_y+L_y) \end{align}

And, \begin{align}P(0\leq X\leq s, 0\leq Y\leq s)&=\int_0^s{\int_0^sP(X=x,Y=y){dx}dy} \end{align}

This should at least sketch you a solution