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Let $(X,Y)$ be a point chosen at random from the triangle $\{x,y:0\leq x\leq y\leq 1\}$. $f_{X,Y}(x,y)=2$ if $(x,y)$ is in the triangle, and it is 0 otherwise. Find the probability density function for $X$.

What confuses me about this problem is understanding how $x$ and $y$ make up the triangle. If I'm understanding this correctly, then the biggest triangle we can make has vertices $(0,1),(1,1),$ and $(0,0)$. If this is the case then the probability should be 1. As $x$ increases, $y$ can at least be $x$ which means that $y$ is dependent on $x$. From this though I'm not sure where to go from here.

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  • $\begingroup$ Note the marginal density property $f_X(x)=\frac{f_{X,Y}(x,y)}{f_Y(y\mid X=x)}$. You may find this helpful. $\endgroup$ – TheSimpliFire Apr 16 at 19:35
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If I'm understanding this correctly, then the biggest triangle we can make has vertices $(0,1),(1,1),$ and $(0,0)$.

You're right, that's the support of your random (multivariate) variable.

If this is the case then the probability should be 1.

Huh... what? What you know is that "total" probability is $1$, i.e. $\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f_{X,Y}(x,y) dx dy=1$. Now, because the density here is constant, denoting by $S$ the support region and by $A_S$ its area we get $\int_{S} 2 dx dy= 2 \, A_S= 1$ and this is indeed true, because the area of the triange is $\frac12$. Then, it's all right.

As $x$ increases, $y$ can at least be $x$ which means that $y$ is dependent on $x$.

Yes. In fact, if you have a bounded support that it's not a rectangle, (or a cartesian product of rectangles) then the variables are dependent.

You have the joint density $f_{X,Y}$. To get the single variable ("marginal") density, you sum (integrate) over the other variable ("marginalize") :

$$ f_X(x)=\int f_{X,Y}(x,y) dy $$

Because the density is constant, and the support is known, what remains is just to get the integrations limits right, i.e. which is the range for the integrating variable ($y$) for each fixed $x$. Can you go on from here?

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  • $\begingroup$ I gave this a bit more thought. Would we be doing a double integral over the triangle? The first from 0 to $x$ then the second from $x$ to $y$? $\endgroup$ – Peetrius Apr 17 at 19:15
  • $\begingroup$ No, the double integral is what gives 1. To marginalize you need only to integrate with respect to $y$, as I explained. $\endgroup$ – leonbloy Apr 17 at 19:17
  • $\begingroup$ So is $f_{X,Y}(x,y)$ simply $(1-x)$ which means I take $\int_0^1(-x+1)dy$ to find $f_X(x)$? Sorry, this feels far beyond my skills in probability. I'm going to offer a bounty on this problem. $\endgroup$ – Peetrius Apr 19 at 0:14
  • $\begingroup$ Using the last formula $$f_X(x)=\int_x^1 2dy = 2(1-x).$$ $\endgroup$ – d.k.o. Apr 19 at 0:28
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    $\begingroup$ NB: You should always include the support.$$f_X(x)= 2(1-x)~\mathbf 1_{0\leqslant x\leqslant 1}$$ $\endgroup$ – Graham Kemp Apr 19 at 2:05

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