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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 $1 = \int_{-\infty}^{+\infty} f_X(x) \, dx= \int_0^2 f_X(x) \, dx= \Big[2x\Big]^2_0 = 4$.

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    $\begingroup$ $c = 2$, 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? $\endgroup$ – Brian Tung Jun 24 at 23:17
  • $\begingroup$ @BrianTung It appears that Illuminate was just adding the parts . $\endgroup$ – Graham Kemp Jun 24 at 23:20
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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:

$$\begin{align}\int_0^2f_X(x)\,\mathrm d x &= \int_0^1\int_0^\infty 2 x\mathrm e^{-2y}\,\mathrm d y\,\mathrm dx+\int_1^2\int_0^\infty 2(2-x)\mathrm e^{-2y}\,\mathrm d y\,\mathrm d x\\[1ex]&= 1\end{align}$$

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$X,Y$ are independent because the joint density is the product of a function of $x$ and a function of $y.$ A fortiori they are uncorrelated.

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