Let $X$ and $Y$ have joint density $f(x,y)=\frac{6x}{11}$ over the trapezoid with vertices $(0,0),(2,0),(2,1)$ and $(1,1)$

Let $X$ and $Y$ have joint density $f(x,y)=\frac{6x}{11}$ over the trapezoid with vertices $(0,0),(2,0),(2,1)$ and $(1,1)$.

(a) Find $P(X>1)$

(b) Find $E(Y^2)$

My attempt

(a) $P(X>1)=\int_{0}^{1}\int_{1}^{2}f(x,y)dxdy=\int_{0}^{1}\int_{1}^{2}\frac{6x}{11}dxdy=\frac{9}{11}$

(b) $f_Y(y)=\int_{0}^{2}f(x,y)dx=\int_{0}^{2} \frac{6x}{11}dx=\frac{12}{11},0 \le y \le 1$

$E(Y^2)=\int_{0}^{1}y^2f_Y(y)dy=\int_{0}^{1}\frac{12y^2}{11}dy=\frac{12}{33}$

The correct answer for part (b) is $\frac{17}{55}$. But I'd like to know why? I may have the wrong integral for the marginal pdf of $Y$.

Solution 1:

The domain is trapezoid given by its vertices which translates to,

$ \displaystyle f(x,y)=\frac{6x}{11}, ~~y \lt x \lt 2, 0 \lt y \lt 1$

Your work on $(a)$ is correct. For $(b)$, please note that support of $x$ is not $[0, 2]$. Rather, the lower support changes with $y$.

$ \displaystyle f_Y(y) = \int_y^2 \frac{6x}{11} ~ dx = \frac{3}{11} (4-y^2)$

$ \displaystyle E[Y^2] = \int_0^1 y^2 f_Y(y) ~dy = \frac{17}{55}$

Solution 2:

The marginal pdf of $Y$ is given by

$$f_{Y}(y)=\int_{y}^{1}f\,dx+\int_{1}^{2}f\,dx=\int_{y}^{2}\frac{6x}{11}\,dx=\frac{3}{11}(4-y^{2})\,,0<y<1 $$.

That is why your answer was not matching.