Let $(X, Y)$ be a two-dimensional random variable with density function $$f_{X,Y}(x,y)=1\;\;\;\;\;216.(0<x<1,\;-x<y<x)$$ Show that, although the correlation is null, the variables $X$ and $Y$ are not independent.
I've tried this, $$\mathbb{E}(X)=\int_0^1\int_{-x}^{x} xf_{X,Y}(x,y)\;dy\;dx = \int_0^1\int_{-x}^{x} x\;dy\; dx = \int_{0}^{1}\left[xy\right]_{-x}^{x}\;dx=\int_{0}^{1}2x^2\;dx=2\left[\frac{x^3}{3}\right]_{0}^{1}=2*\frac{1}{3}=\frac{2}{3}$$ $$\mathbb{E}(X^2)=\int_0^1\int_{-x}^{x} x^2f_{X,Y}(x,y)\;dy\;dx = \int_0^1\int_{-x}^{x} x^2\;dy\; dx = \int_{0}^{1}\left[x^2y\right]_{-x}^{x}\;dx=\int_{0}^{1}x^6\;dx=\left[\frac{x^7}{7}\right]_{0}^{1}=\frac{1}{7}$$ $$Var(X)=\mathbb{E}(X^2)-\mathbb{E}(X)^2=\frac{1}{7}-(\frac{2}{3})^{2}=-\frac{19}{63}$$ $$\mathbb{E}(Y)=\int_0^1\int_{-x}^{x}yf_{X,Y}(x,y)\;dy\;dx = \int_0^1\int_{-x}^{x} y\;dy\; dx = \int_{0}^{1}\left[\frac{y^2}{2}\right]_{-x}^{x}\;dx =\int_{0}^{1}0\;dx=0$$ $$\mathbb{E}(Y^2)=\int_0^1\int_{-x}^{x} y^2f_{X,Y}(x,y)\;dy\;dx = \int_0^1\int_{-x}^{x} y^2\;dy\; dx = \int_{0}^{1}\left[\frac{y^3}{3}\right]_{-x}^{x}\;dx=\int_{0}^{1}\frac{2x^3}{3}\;dx=\frac{2}{3}\left[\frac{x^4}{4}\right]_{0}^{1}=\frac{2}{3}*\frac{1}{4}=\frac{1}{6}$$ $$Var(Y)=\mathbb{E}(Y^2)-\mathbb{E}(Y)^2=\frac{1}{6}-0^{2}=\frac{1}{6}$$ $$\mathbb{E}(XY)=\int_0^1\int_{-x}^{x} xyf_{X,Y}(x,y)\;dy\;dx = \int_0^1\int_{-x}^{x} xy\;dy\; dx = \int_{0}^{1}x\left[\frac{y^2}{2}\right]_{-x}^{x}\;dx=\int_{0}^{1}0\;dx=0$$ $$Cov(X,Y)=\mathbb{E}(XY)-\mathbb{E}(X)\mathbb{E}(Y)=0-\frac{2}{3}*0=0-0=0$$ $$ρ_{X,Y}=\frac{Cov(X,Y)}{\sqrt{Var(X)Var(Y)}}=\frac{0}{\sqrt{Var(X)Var(Y)}}=0$$ $$f_{X}(x)=\int_{-x}^{x}1\;dy=\left[y\right]_{-x}^{x}=x-(-x)=2x$$ $$f_{Y}(y)=\int_{0}^{1}1\;dx=\left[x\right]_{0}^{1}=1$$ $$f_{X,Y}(\frac{1}{3},\frac{1}{4})=1$$ $$f_{X}(\frac{1}{3})*f_{Y}(\frac{1}{4})=2*\frac{1}{3}*1=\frac{2}{3}$$ Since $f_{X,Y}(x,y) ≠ f_{X}(x)*f_{Y}(y)$, then $X$ and $Y$ are not independent.
Am I right? Help please!