Suppose that the joint probability density function of $(X, Y)$ is given by $$f_{X,Y}(x, y) = \left(1 - \alpha(l-2x)(l-2y)\right)I_{(0,1)}(x)I_{(0,1)}(y)\,,$$ where $-1 < \alpha < 1$.
An isosceles triangle is formed as indicated in the sketch.
(b) If $(X, Y)$ has the joint density given above, pick $\alpha$ to maximize the expected area of the triangle.
(c) What is the probability that the triangle falls within the unit square with corners at $(0, 0), (1, 0), (1, 1)$, and $(0, 1)$?
(d) Find the expected length of the perimeter of the triangle.
I am having a hard time understanding how to built the area of the triangle, and the length of its perimeter.
This is a continuation to a problem I already solved myself: $X$, $Y$ independent if and only if $X$, $Y$ uncorrelated..
Please help me understand the area and perimeter of the triangle.
Given the area and perimeter given by @LeeDavidChung, the expected area is
$$\int _0^1\int _0^1x y (1-\alpha (1-2 x) (1-2 y))dydx=\frac{9-\alpha }{36}$$
The maximum ($1/4$) occurs when $\alpha=0$.
The expected value of the perimeter is
$$\int _0^1\int _0^12 \left(\sqrt{x^2+y^2}+x\right) (1-\alpha (1-2 x) (1-2 y))dydx=\frac{1}{15} \left(-7 \sqrt{2} \alpha +6 \alpha +5 (\alpha +2) \sinh ^{-1}(1)+10 \sqrt{2}+15\right)$$
When $\alpha=0$ the expected value of the perimeter is $\frac{1}{3} \left(2 \sqrt{2}+3+2 \sinh ^{-1}(1)\right)$.