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I hope everyone is having a great morning. My question here is about how the integrals were set up.

Question: Find $c$ if $f_{X,Y}(x,y) = cxy$ for $X,Y$ defined over the triangle whose vertices are the points $(0,0),(0,1)$ and$ (1,1)$.

Solution: Solve for $\int_0^1 \int_0^y (cxy)dxdy= 1 \Rightarrow c = 8$

My question: We are given points where $X$ and $Y$ are defined. $(0,0),(0,1),(1,1)$. Why was the integrals set up as $\int_0^1 \int_0^y$ and not $\int_0^1 \int_0^1$? we know both $x$ and $y$ are defined under $0-1$. Thank you in advance.

Emad Zamout
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2 Answers2

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You were given that $(X,Y)$ is defined in the triangle with vertices $(0,0)$, $(0,1)$, and $(1,1)$. If you draw that triangle, you will see that $(X,Y)$ is not defined for every possible value of $X$ and every possible value of $Y$ both between $0$ and $1$. For instance, the point $(0.75,0.25)$ lies outside this triangle, so the bounds of the integral should not contain that point.

Instead, we see that the triangle is bounded by the equations $x=0$, $y=1$, and $y=x$. The outer variable in the integral, $y$, is allowed to range from $0$ to $1$, but given a value of $y$, $x$ can only range from $0$ to $y$. If $x>y$ then you are outside the triangle.

kccu
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HINT If you do the integrals $\int_0^1 \int_0^1dxdy$ you would be integrating over the square that has vertices in $(0,0), (1,0), (0,1), (1,1)$. But now in your situation you are dealing with a triangle which is of course a bit different. Now draw the triangle and let $y$ attain all the values between $0$ and $1$. Then you have to restrict $x$ somehow so that you don't integrate over the whole square. $\int_0^1 \int_0^y dx dy$ is exactly what you should get.

Harto Saarinen
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