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Let ($X$, $Y$) be a uniformly random point in the triangle in the plane with vertices ${(0, 0),(0, 1),(1,0)}$. Find the the marginal PDF of $X$.

I disagree with the solution (see answer to Problem 7, Homework 7) to this problem, which claims that the upper limit of integration is $1-x$. I think that the upper limit of integration should be $x$.

My reasoning:

  1. First, we can see that the joint PDF is $f_{X,Y}(x,y) = 2$ for $x \in [0,1]$ and $0 \leq y \leq x$.
  2. Marginalizing out $Y$ means we're squishing the joint density down onto the $x$-axis. We should be integrating the joint PDF from $0$ to $y$ for each "slice" $X=x$ of the joint distribution, but $y = x$, so the upper limit of integration is $x$:

$$ \int_{0}^{x} 2 dy = 2x, \text{for $x \in [0, 1]$} $$

Sanity check: Plugging in values should show that as $x$ approaches $1$, the PDF of $X$ gets larger for my solution, which is what we would expect since the triangle's height above the $x$-axis is greater as $x$ approaches $1$.

Which is correct and why?

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The textbook solution is right

Here is your bivariate domain

enter image description here

If you want to derive the marginal density $f_X(x)$ you have to integrate Y, thus

$$f_X(x)=\int_0^{1-x}2dy=2(1-x)\mathbb{1}_{(0;1)}(x)$$

Same reasoning for the other marginal density

$$f_Y(y)=\int_0^{1-y}2dx=2(1-y)\mathbb{1}_{(0;1)}(y)$$


Your reasoning would be right if the triangle's vertices were

$(0;0)$, $(1;0)$,$(1;1)$

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    $\begingroup$ Oh, wow, such a silly mistake -- thanks for taking the time @tommik, that makes complete sense. $\endgroup$
    – Per48edjes
    Oct 17 '20 at 21:03

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