The random vector $(X,Y)$ has a uniform distribution on a triangle with vertices $(0,0), (0,1), (1,0)$. Determine the values of $a, b \in \mathbb{R}$ minimizing the value of the
$$f(a,b) = \mathbb{E}(Y- aX - b)^2$$
Hint: minimize first over b, with a fixed a.
So I did it that way: $$f(a,b) = \mathbb{E}(Y- aX - b)^2 = \mathbb{E}(Y^2 +a^2X^2 + b^2 + 2abX - 2aXY - 2bY) =$$ $$ = \mathbb{E}Y^2 + a^2 \mathbb{E}X^2 + b^2 + 2ab \mathbb{E} X - 2a \mathbb{E} XY - 2b \mathbb{E}Y = $$ $$= b^2 + b( 2a \mathbb{E} X - 2 \mathbb{E}Y) + \mathbb{E}Y^2 + a^2 \mathbb{E}X^2 - 2a \mathbb{E} XY$$
Now I know that the distribution is uniform, so: $f_{X,Y}(x,y) = 2 \ \text{ for: } \ x + y \leq 1 \ \text{ and } \ x, y \geq 0$.
Thus:
$f_{X}x = \int_{0}^{1-x} 2 \ dy = 2(1-x) \ \text{ for: } \ 0 \leq y \leq 1 $
and similarly:
$ \ f_{X}x = \int_{0}^{1-y} 2 \ dy = 2(1-y) \ \text{ for: } \ 0 \leq x \leq 1 $
We have that:
$\mathbb{E}X = \int_0^1 2x(1-x) \ dx = x^2 - \left[ \frac{2x^3}{3} \right] \Big|^1_0 = \frac{1}{3}$
$\mathbb{E}Y = \int_0^1 2y(1-y) \ dy = y^2 - \left[ \frac{2y^3}{3} \right] \Big|^1_0 = \frac{1}{3}$
$\mathbb{E}XY = \int_0^1 \int_0^{1-y} 2xy \ dx \ dy = \int_0^1 2y \left[ \frac{x^2}{2} \right] \Big|^{1-y}_0 \ dy = \int_0^1 y(1-y) \ dy = \frac{1}{12}$
$\mathbb{E}X^2 = \int_0^1 2x^2(1-x) \ dx = \frac{1}{6}$
$\mathbb{E}Y^2 = \int_0^1 2y^2(1-y) \ dy = \frac{1}{6}$
Therefore we have that: $$f(a,b) = b^2 + b \frac{2}{3}( a - 1) + \frac{1}{6} + \frac{1}{6}a^2 - \frac{1}{6}a$$
We have a convex parabola with minimum in $b = - \frac{\frac{2}{3}( a - 1)}{2} = \frac{1}{3}( 1 - a)$
Then we have that: $$f(a,b)_{minB} = \frac{1}{9}(1-a)^2 + \frac{2}{9}(1-a)^2 + \frac{1}{6} + \frac{1}{6}a^2 - \frac{1}{6}a = \frac{1}{3}(1-a)^2 + \frac{1}{6} + \frac{1}{6}a^2 - \frac{1}{6}a = $$ $$= \frac{1}{3}a^2 - \frac{2}{3} a + \frac{1}{3} + \frac{1}{6} + \frac{1}{6}a^2 - \frac{1}{6}a = \frac{1}{2}a^2 - \frac{5}{6}a + \frac{1}{2}$$
Similarly, we have a convex parabola with minimum in $a= \frac{\frac{5}{6}}{1} = \frac{5}{6}$
So $b = \frac{1}{3}(1-a ) = \frac{1}{3} \left( 1- \frac{5}{6} \right) = \frac{1}{3} \cdot \frac{1}{6} = \frac{1}{18}$.
Is that correct?