Given the distance $d$ between two random points on a segment, find the MLE of its length $L$.

Question

I could solve this problem, but I made a logic jump that I'm not sure how to justify.

We have a segment of length $L$, and two random points with uniform distribution over that segment. That is, $p_{X_{1}X_{2}}(x_{1}, x_{2}) = \frac{1}{L^2}$. The given is the distance between those two points, so $d = |x_{1} - x_{2}|$. I don't know any easy way of dealing with the absolute value, so I took a leap of faith and assumed that $x_{1}$ is the farthest point from $0$, so $d = x_{1} - x_{2}$.

Then I use the delta method:

$$\begin{align} p_{D}(d) &= \iint p_{X_{1}X_{2}}(x_{1}, x_{2})\delta(d - x_{1} + x_{2})dx_{1}dx_{2} \\ &= \int p_{X_{1}X_{2}}(d + x_{2}, x_{2}) dx_{2} \\ &= \int_{-d} ^{L-d}\frac{1}{L^2}\mathbb{1}\{0 < x_{2} < L\}dx_{2} \\ &= \int_{0} ^{L-d}\frac{1}{L^2}dx_{2} \\ &= \frac{1}{L} - \frac{d}{L^2} \end{align}$$ Then: $$\frac{dp_{D}(d)}{dL} = \frac{2d}{\hat{L}^3} - \frac{1}{\hat{L}^2} = 0 \iff \hat{L} = 2d$$

According to my teacher this is right, but I feel like I ignored half of the possible cases by getting rid of the absolute value in $d$.

Why does this yield the same result as if I solved the problem with $d = |x_{1} - x_{2}|$ ?

Thanks.

Answer 1

The function that you found is not a pdf since it is not integrate to 1: $$ \int_0^L \left(\frac1L - \frac{d}{L^2}\right) \textrm{d}d = \frac12. $$ And it differs from the valid pdf by constant multiplier $2$, which does not effect on the points where this function attains maximum. This is the answer to your question.

And then, to get pdf of $|x_1-x_2|$, one can use geometric probability; First find cdf : $$ F_{|x_1-x_2|}(d) = \mathbb P(|x_1-x_2| \leq d). $$ This is exactly probability of meeting. Draw a square of side $L$ and a strip along with diagonal between $x_2=x_1\pm d$. Area of the strip divided by area of the square equals to the above probability. So for $0<d<L$ $$ F_{|x_1-x_2|}(d) = \mathbb P(|x_1-x_2| \leq d)=\frac{L^2-(L-d)^2}{L^2}=\frac{2d}{L}-\frac{d^2}{L^2}. $$ And for $0<d<L$ the pdf is $$ p_{|x_1-x_2|}(d) =(F_{|x_1-x_2|}(d))' =\frac{2}{L}-\frac{2d}{L^2}. $$