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I'm using the transformation of R.V and calculating marginal density approach and it confuses me how there's two cases.

In my working I let $U=X+Y, V=Y$

$f_{U,V} = f_{X,Y}|J|=f_{X,Y}=1$

from $0<x<1,0<y<1$ we deduct that $v<u,v>u-1,0<v< 1$ by calculating marginal density for any joint density function we calculate $\int_{Y}f_{X,Y}dy$ and here it would be $f_U=\int_{V}1dv = \int_0^u1dv+\int_{u-1}^11dv=u+2-u$.

from the graph it kinda make sense that $f_U$ is not a constant, but I can't figure out from the calculation where to introduce the two cases and why.

smaillis
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2 Answers2

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Think geometrically. Let $(X,Y)$ be a random coordinate in the unit square $[0,1] \times [0,1]$. Then if $U = X+Y \le u$, this means $(X,Y)$ is a point in the subset of the unit square that is "below" the line $X+Y \le u$. For $0 < u \le 1$, this is a triangle with vertices $(0,0), (u,0), (0,u)$. If $1 < u \le 2$, then this is a pentagon with vertices $(0,0), (1,0), (1,u-1), (u-1,1), (0,1)$. Its complement is the triangle $(1,u-1), (1,1), (u-1,1)$. What is the area of this region as a function of $u$?

Now with this insight, can you work with your method in a way that allows you to arrive at the same conclusion?

heropup
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  • I guess the main part of my confusion is still that marginal density is calculated by $f_y = \int_Y f_{X,Y}dy$ and I don't understand how to introduce the two cases by only looking at equations. Geometrically it is clear that there's two functions. but I cant see it in the equations. further, the second part of the question ask for the sum of n i.i.d U[0,1], if generalized to any distribution, how would you identify such point of discontinuity? – smaillis Feb 18 '21 at 23:13
  • @smaillis The general case is known as the Irwin-Hall distribution. See https://en.wikipedia.org/wiki/Irwin%E2%80%93Hall_distribution for more information. Your difficulty arises from applying a bivariate transformation to a distribution with finite support; therefore, you should consider using indicator functions in your calculation to help you keep track of how the support changes under such transformations. – heropup Feb 18 '21 at 23:28
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$f_{\small X,Y}(x,y)$ does not equal 1 except when the values are in the support.

Always include the support in a density function. This is usually done by using an indicator function. That is a piecewise function equal to one when the indicated condition holds, but zero when it does not: $$\begin{align}f_{\small X,Y}(x,y)&=\mathbf 1_{0<x<1,0<y<1}\\[1ex]&=\begin{cases} 1 & : & 0<x<1, 0<y<1 \\ 0 & : & \textsf{otherwise}\end{cases}\end{align}$$

So, indeed you have found the support for $\langle U,V\rangle$ is: $$\{\langle u, v\rangle: v<u<1+v, 0<v<1\}$$

Which can also be written as:

$${\{\langle u,v\rangle: 0<u<2,\max\{0,u-1\}<v<\min\{1,u\}\}\\[2ex]\text{or}\\[2ex]\{\langle u,v\rangle: 0<u<1, 0<v<u\}\cup\{\langle u,v\rangle:1\leqslant u<2, u-1<v<1\}}$$

So we have that :$$\begin{align}f_{\small U,V}(u,v) & = f_{\small X,Y}(u-v, v)\\[1ex]&=\mathbf 1_{0<v<1, v<u<1+v}\\[1ex] &= \mathbf 1_{0<u<1, 0<v<u}+\mathbf 1_{1\leqslant u<2, u-1<v<1}\end{align}$$

Thus $f_{\small V}(v)$ is found by using the former:$$\begin{align}f_{\small V}(v) &=\int_\Bbb R f_{\small X,Y}(u-v,v) \mathrm d u\\& = \int_v^{1+v} \mathbf 1_{0<v<1}\,\mathrm d u\\&= \mathbf 1_{0<v<1}\end{align}$$

Likewise $f_{\small U}(u)$ is found by using the later: $$\begin{align}f_{\small U}(u) &=\int_{0<v<u}\mathbf 1_{0<u<1} \,\mathrm d v+\int_{1-u<v<1} \mathbf 1_{1\leqslant u<2}\,\mathrm d v\\[1ex]&=u\,\mathbf 1_{0<u<1}+(2-u)\,\mathbf 1_{1\leqslant u<2}\\[1ex]&=\begin{cases} u&:& 0<u<1\\ 2-u&:& 1\leqslant u<2\end{cases}\end{align}$$

Graham Kemp
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