# Joint to marginal distribution does not make sense

Let the joint distribution of $$X$$ and $$Y$$ be

$$f_{X,Y}(x,y) = \begin{cases} cxe^{-2y} &0\leq x\leq 1, y> 0 \\ c(2-x)e^{-2y} &1\leq x\leq 2, y> 0 \\ 0 &\mathrm{else} \end{cases}$$

1. Find the value of $$c$$. $$\underbrace{\int _1^2\int_0^\infty c(2-x) e^{-2y} \, dy \, dx}_{c/4} + \underbrace{\int _0^1\int_0^\infty cxe^{-2y} \, dy \, dx}_{c/4} = 1 \implies c=2$$
2. Are $$X$$ and $$Y$$ independent, are $$X$$ and $$Y$$ uncorrelated?

Hence, I calculated $$f_X(x)=c=2$$, but this does not make any sense because now $$1 = \int_{-\infty}^{+\infty} f_X(x) \, dx= \int_0^2 f_X(x) \, dx= \Big[2x\Big]^2_0 = 4$$.

• $c = 2$, yes, but why do you have $f_X(x) = c = 2$? You should have $f_X(x) = x$ for $0 \leq x \leq 1$, and $f_X(x) = 2-x$ for $1 \leq x \leq 2$, shouldn't you? – Brian Tung Jun 24 at 23:17
• @BrianTung It appears that Illuminate was just adding the parts . – Graham Kemp Jun 24 at 23:20

The integration depends on the placement of $$x$$ in the partition of the joint support.   It too is a piecewise function.
\begin{align}f_X(x) &= \int_\Bbb R f_{X,Y}(x,y)\,\mathrm d y \\[2ex] &= \mathbf 1_{x\in[0,1)}\int_0^\infty 2 x\mathrm e^{-2y}\,\mathrm d y+\mathbf 1_{x\in[1,2)}\int_0^\infty 2(2-x)\mathrm e^{-2y}\,\mathrm d y\\[3ex]&=\begin{cases}x\int_0^\infty 2\mathrm e^{-2y}\,\mathrm d y&:& 0\leqslant x< 1\\(2-x)\int_0^\infty 2\mathrm e^{-2y}\,\mathrm d y&:&1\leqslant x< 2\\0&:&\textsf{otherwise}\end{cases}\\[1ex]&~~\vdots\end{align}
And the integration of this is clearly $$1$$, as you have already found $$c=2$$ to make it be so:
\begin{align}\int_0^2f_X(x)\,\mathrm d x &= \int_0^1\int_0^\infty 2 x\mathrm e^{-2y}\,\mathrm d y\,\mathrm dx+\int_1^2\int_0^\infty 2(2-x)\mathrm e^{-2y}\,\mathrm d y\,\mathrm d x\\[1ex]&= 1\end{align}
$$X,Y$$ are independent because the joint density is the product of a function of $$x$$ and a function of $$y.$$ A fortiori they are uncorrelated.