Hand waving in an Expected value problem

Square Triangle picking is one of the famous problems in Geometric probability. As shown in the mathworld page, a direct attack would entail a sextuple integral and would very very messy to solve by pen and paper.

So it was a surprise when I first a solution to this problem which seems hand wavy. I mean, at first glance, it seems like it shouldn't work and I'm still not convinced why it works. I wanted to check how this approach works on some closely related simpler problems.

We start with the problem of finding the area of the triangle formed by vertex A of a given unit square ABCD and two points selected randomly inside the square.

If we now draw a rectangle with the random points forming a diagonal, then by symmetry they'll be the main diagonal of the rectangle (bottom-left to top-right) 50% of the time and will be the anti-diagonal of the rectangle in the remaining 50% of the time.

Let the two random points be $P_1(X_1,Y_1)$ and $P_2(X_2,Y_2)$. They form the anit-diagonal when $X_1<X_2 \land Y_1>Y_2$ (or the same with the first and the second point interchanged). In this case, we have

$\mathbb{E}(X_1)=\mathbb{E}(Y_2)=1/3$ and $\mathbb{E}(X_2)=\mathbb{E}(Y_1)=2/3$.

Therefore, the average area of the $\triangle AP_1P_2$ is

$\displaystyle \mathbb{E}(\text{Area}(\triangle AP_1P_2)|\text{anti-diagonal case})=\frac{1}{2}\begin{Vmatrix}1&0&0 \\ 1&\frac{1}{3}&\frac{2}{3} \\ 1&\frac{2}{3}&\frac{1}{3}\end{Vmatrix}=\frac{1}{6}$

For the main-diagonal case, we have $X_1<X_2 \land Y_1 < Y_2$. We have,

$\mathbb{E}(X_1)=\mathbb{E}(Y_1)=1/3$ and $\mathbb{E}(X_2)=\mathbb{E}(Y_2)=2/3$

But if we try to find the area with these coordinates, we obviously get 0. Time to get a little creative!

Let's assume we are given $P_2$. Then, by symmetry, $P_1$ is uniform in the rectangle with diagonal $AP_2$. In fact, by the symmetry of a rectangle w.r.t its diagonal, $P_1$ is uniform in the right triangle with hypotenuse $AP_2$ and legs parallel to the sides of the square.

Here is where we use a nice trick. It is very easy to see that the expected area of a triangle formed by one of the sides of a given triangle and a point selected uniformly randomly inside it is one third the area of the given triangle.

Then the expected area we are looking for is simply $(1/3)\mathbb{E}(X_2Y_2/2)$. Because the variables are independent, this works out to be $2/27$.

This finally shows that $\displaystyle \mathbb{E}(\triangle AP_1P_2)=\frac{1}{2}\frac{1}{6}+\frac{1}{2}\frac{2}{27}=\frac{13}{108}$

which is the same answer we get via integration.

We can also use the same approach to answer the following question. Given a point $R(r,0)$, what is the expected area of triangle formed by $R$ and two points selected randomly from an unit square?

Let $R'$ be the point $(r,1)$. With this, we form three different cases - both the points lying in $ARR'D$ (with probability $r^2$), both the points lying in $RBCR'$ (with probability $(1-r)^2$) or one point apiece (with probability $2r(1-r)$).

The first two cases reduced to a scaled version of the first question we dealt with. In the third case, we can reasonably expect

$\mathbb{E}(Y_1)=\mathbb{E}(Y_2)=1/2$, $\mathbb{E}(X_1)=r/2$ and $\mathbb{E}(X_1)=(1+r)/2$

Therefore,

$\displaystyle \mathbb{E}(\text{Area}(\triangle RP_1P_2)|\text{one point apiece})=\frac{1}{2}\begin{Vmatrix}1&0&0 \\ 1&\frac{r}{2}&\frac{1}{2} \\ 1&\frac{1+r}{2}&\frac{1}{2}\end{Vmatrix}=\frac{1}{8}$

Then,

$\displaystyle \mathbb{E}(\text{Area}(\triangle RP_1P_2))=r^2\frac{13r}{108}+2r(1-r)\frac{1}{8}+(1-r)^2\frac{13(1-r)}{108}=\frac{1}{108}(12r^2-12r+13)$

For the special case $R$ being the midpoint of $AB$, the expected area is $5/54$.

We can finally use all this to find the expected area of the triangle formed two randomly selected points and the centre $M$ of the square. The centre divides the square into four smaller squares. Let's say $P_1$ lies in the bottom left square. There are 4 cases depending where $P_2$ is.

(1) On the same square in which case the expected area will be $(1/4)(13/108)$

(2) On adjacent (two ways) squares in which case the expected area will be $(1/2)(1/8)$

(3) On opposite square. This may seem tricky until we realize that we can rotate $P_2$ w.r.t $M$ to get $P_2'$ and see that the area of $MP_1P_2'$ is the same as that of $MP_1P_2$. Therefore, this is no different from the first case.

Then the total expected area is (1/4)(1/4)(13/108) + (2/4)(1/2)(1/8) + (1/4)(1/4)(13/108) = 5/108.

Hope you enjoyed the discussion.

Until then

Yours Aye

Me