A point is randomly chosen from a $1\times1$ square. Find the probability that the distance between the point and the nearest diagonal is less than z

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I have to solve the following problem:

If a point $(x_0 , y_0)$ is randomly chosen from a square, whose sides are equal to $1$, find the probability that the distance between the point and the nearest diagonal is less than z.

I have done the following:

We know that

$f(x,y)=\left\{\begin{matrix}
0, (x,y) \notin [0,1]\times [0,1] \\
1, (x,y) \in [0,1]\times [0,1]
\end{matrix}\right.$

We want to calculate $P$("the distance between the point and the nearest diagonal is less than z")$=P(B_z)$

As we deduce that $z \in [0, \frac{\sqrt{2}}{4})$, we've got the following cases:

When $z=0$, as we are in an apex or in one of the diagonals, the distance is $0$.
When $z \in (0, \frac{\sqrt{2}}{4})$, the distance to the nearest diagonal would be $\sqrt{(x-x_0)^2+(y-y_0)^2}\le z$

But what do I have to do now? I've tried to calculate the area of $B_z$ this way, but I don't know if I have to do it like this:

$Area(B_z)=\int_0^z \int_0^{\sqrt{z-(x-x_0)^2}+y_0}1 dydx$

Am I doing it ok? Or do I have to calculate something else?

Best Answer

This diagram shows the areas within distance $\frac{0.1}{\sqrt2}$ of the diagonals. Instead of calculating the area of the somewhat complicated X shape, you can calculate the area of the four right isosceles triangles that form the complement and subtract it from the total area $1$.

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Whenever $A$ lies more than $\alpha$ from the edge this is $ 10^{-4}\pi\alpha^2$. For sufficiently small $\alpha$ this is a reasonable approximation of the required probability.

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$$10^{-4}\big(\pi\alpha^2 - 400\int_0^{\alpha} \operatorname {chordSlice}(x,\alpha)\operatorname d x +4\int_0^{\alpha}\!\!\!\int_0^{\alpha}\operatorname {twoChordSlice}(x,y,\alpha)\operatorname d x\operatorname d y \Big) $$

[Math] Find the probability that the sum of two randomly chosen positive numbers…. followed by some conditions.

Comment: You have a path to the answer; this is just a way to confirm you are on the right track without actually working the problem.

I assume you have independent random variables $X$ and $Y,$ each distributed $\mathsf{Unif}(0, 1).$ And you seek $P(X+Y \le 1, XY \le 2/9),$ where as usual the comma means intersection.

Here is a simulation in R statistical software of 100,000 independent $X$'s and $Y$'s that illustrates the problem and your approach. [This method of getting the answer by simulation is sometimes called 'Monte Carlo integration'. It is mainly used for problems that are too messy to handle with ordinary Riemann integration (often in more than 2 dimensions). You should use ordinary integration for your answer.]

set.seed(731);  m = 10^5;  x = runif(m);  y = runif(m)
A = (x + y <= 1);  B = (x*y <= 2/9) # A and B are logical vectors of TRUEs and FALSEs.
mean(A);  mean(B);  mean(A & B)     # mean of a logical vector is its proportion of TRUEs.
[1] 0.50086  # aprx P(A) = 1/2
[1] 0.55795  # aprx P(B)
[1] 0.48821  # aprx answer P(A & B)

Because of the order in which I have plotted the 100,000 points below, it is the percentage of yellow ones that you want, as in your diagram. (About 0.488.)

plot(x, y, pch=".")
  points(x[B], y[B], pch=".", col="red")        # Read [ ] as "such that"
  points(x[A], y[A], pch=".", col="green2")
  points(x[A&B], y[A&B], pch=".", col="yellow")

For an exact solution it may be easiest to subtract the area of the green 'nibble' from 1/2. (As @copperhat has suggested.)

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