4
$\begingroup$

This question is a follow-up to a discussion on MESE. On the linked post, the OP asked about the following question (which I will paraphrase slightly):

Suppose $\theta$ is a random variable which takes its values in $[0, \pi/2]$ and with a pdf proportional to $\theta(\pi-\theta)$. Let $\triangle ABC$ be a triangle with congruent legs of length $\ell$ and vertex angle $\theta$. Find the pdf for the area of $\triangle ABC$.

On the original question, the OP wanted to know whether this question was fair, reasonable, too difficult, etc. I have a different question: Is my solution to the problem (below) correct?

Solution. First, let $f_\theta = k\theta(\pi-\theta)$ be the pdf for $\theta$. Then $$1 = \int_0^{\pi/2} f(\theta) \, d\theta = k \int_0^{\pi/2} \theta(\pi-\theta) \, d\theta = k\frac{\pi^3}{12},$$ so $k = \frac{12}{\pi^3}$. Then the cdf for $\theta$ is $$F_\theta(x) = \frac{6x^2}{\pi^2} - \frac{4x^3}{\pi^3}$$

Now, the area of the triangle is related to the angle by the equation $A = \frac{\ell^2}2 \sin\theta$, or equivalently $\theta = \arcsin\left(\frac{2A}{\ell^2}\right)$. Therefore, if we let $F_A$ be the cdf of $A$, then: $$\begin{aligned} F_A(x) &= P(A \le x) \\ &=P\left(\theta \le \arcsin \left( \frac{2x}{\ell^2} \right) \right) \\ &=F_\theta\left(\arcsin\left(\frac{2x}{\ell^2} \right) \right) \\ &=\frac{6 \left(\arcsin\left(\frac{2x}{\ell^2} \right) \right)^2}{\pi^2} - \frac{4 \left(\arcsin\left(\frac{2x}{\ell^2} \right) \right)^3}{\pi^3} \\ \end{aligned}$$ and the pdf of $A$ is therefore $$f_A(x) = \frac{d}{dx} F_A(x)$$ which -- assuming I haven't made any errors in differentiation -- works out to be

$$f_A(x) = \frac{24 \arcsin\left(\frac{2x}{l^2}\right) \left(\pi - \arcsin\left(\frac{2x}{l^2}\right) \right)}{\pi^3 \sqrt{l^4 - 4x^2}}$$

Are there any errors in the above reasoning, or in the work leading to the solution? Can this result be simplified any further? Is there an easier approach to this problem?

$\endgroup$
1
  • 2
    $\begingroup$ @isaiah I think that is not the case for problems tagged [solution-verification], but I could be wrong. $\endgroup$
    – mweiss
    Jun 7 '21 at 20:29
0
$\begingroup$

Thank you for the reference to my test question, mweiss. There are several approaches one can take with the question. I have a very long solution (3 pages) for the benefit of the students that breaks it down in exhaustive detail. Here is what I can tell you so far (before others make any comments) and before I tell you if the answer is correct.

A proportional probability density function to $x (\pi -x)$ (I notice you used theta, which is nice) at each point within $x\in{(0,\displaystyle\frac{\pi}{2}})$

is represented by:

$f(x)=\begin{cases} kx(\pi-x) & \text{if}& x\in [0,\pi/2]\\c & \text{if}& x\not\in [0,\pi/2]\end{cases}$

To find k, impose the following condition:

$1=\displaystyle\int_{-\infty}^{\infty}f(x)dx=\displaystyle\int_{0}^{\pi/2}x(\pi-x)dx$

The area is given by:

$A(x)=\dfrac{1}{2}L^2sin(x)$ or your equivalent (you list $\theta = \arcsin\left(\frac{2A}{\ell^2}\right)$ and that is fine too (I used that form in one of my alternative solutions).

Now, here you can also use the "Change of Variable" Theorem used in Statistics. Are you familiar with it?

We can transform the variable to:

$g(a)=f(A^{-1}(a))\dfrac{1}{A'(A^{-1}(a))}$

$a\in [0,L^2/2]$

I am awful when it comes to Latex and I hardly know it, so I hope it comes out correctly.

Another way you can do it: Sorry, had to do it with an image because I do not feel completely confident in Latex.

solution

$\endgroup$
7
  • $\begingroup$ This is clearly identical to my solution. $\endgroup$
    – mweiss
    Jun 7 '21 at 19:20
  • $\begingroup$ Yes, just slightly different notation. You can also use the idea of a sample space $\omega$ and then use the variable $\eta$ as follows: $\eta = g o \xi = \frac {L^2sin \xi}{2}$ and then work from there. $\endgroup$
    – Wasp
    Jun 7 '21 at 19:36
  • $\begingroup$ I have a much longer solution that takes up to 3 pages, but I dare not use latex for such long solutions since I do not know it well. $\endgroup$
    – Wasp
    Jun 7 '21 at 19:37
  • 2
    $\begingroup$ This does not, however, include an explicit formula for the actual answer... computing $A’(A^{-1})$ is the hard part! $\endgroup$
    – mweiss
    Jun 7 '21 at 20:33
  • $\begingroup$ Oh indeed! I might just post it as an attachment to the answer, but I have to rearrange some things first because I wrote the solution for myself so it's sort of all over the place. I'll try to get it done tonight if possible. $\endgroup$
    – Wasp
    Jun 7 '21 at 20:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.