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Find the density of random variable $X+Y$ for $f(x,y) = 6(x-y)$ if $0 \leq y \leq x \leq 1$.

I am able to use method of transformation to convert $f(x,y) = 6(x-y)$ if $0 \leq y \leq x \leq 1$, and the result is $g(v,w) = 6(v-2w)$, where $v = x + y$ and $w = y$. However, I am really stuck at the integration to find the marginal distribution of $g(v)$, particularly the limit of integration.

I think I need to divide the interval for $v$ into $[0,1]$ and $[1, 2]$ but I am not sure how to do it.

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    $\begingroup$ It may be helpful to sketch the region $0\leq y\leq x\leq 1$, along with level sets of $x+y$. $\endgroup$
    – Semiclassical
    Commented Jan 6 at 0:16
  • $\begingroup$ @Semiclassical Thanks for the comment! I did sketch it and I think it is the upper half of an parallelogram (0,0),(1,1),(1,2)(0,1). But I am not sure how to use it as I haven't done it before. Would you mind sharing with me any resources/examples you know? $\endgroup$
    – TJT
    Commented Jan 6 at 0:27
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    $\begingroup$ Maybe it would be easier to change the variables to $(x, x + y),$ as in this case it is rather clear that $0 \leq x \leq 1$ and for each fixed $x,$ the sum $x + y$ will go from $x$ to $2 x.$ $\endgroup$
    – Actually Fritz
    Commented Jan 6 at 1:08

1 Answer 1

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I am able to use method of transformation to convert $f(x,y)=6(x−y)$ if $0≤y≤x≤1$, and the result is $g(v,w)=6(v−2w)$, where $v=x+y$ and $w=y$.

More correctly, $g(v,w) = 6\,(v-2w)$ if $0\leq w\leq v\leq 1+w\leq 2$, which is $0\leq v\leq 2$ and $\max\{0,v-1\}\leq w\leq\min\{1,v\}$ so $$\begin{align} g(v) &= \int_{\max\{0,v-1\}}^{\min\{1,v\}} 6\,(v-2w)\,\mathbf 1_{0\leq v\leq 2}\,\mathrm d w \\[1ex] &= 6\;\mathbf 1_{0\leq v\leq 1}\int_0^v(v-2w)\,\mathrm dw+6\;\mathbf 1_{1\lt v\leq 2}\int_{v-1}^1(v-2w)\,\mathrm d w\\[1ex] &~~\vdots \end{align}$$

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  • $\begingroup$ Thanks you so much! I saw a similar question with similar answer like yours (putting min and max in limit of integration). I didn't understand that but your answer helped me indeed! $\endgroup$
    – TJT
    Commented Jan 11 at 17:05

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