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I have a joint pdf function

$$ f_{X,Y}(x,y) = \begin{cases} A, & \mbox{ } 0\le|x|+|y|<1\mbox{} \\ 0, & \mbox{otherwise } \mbox{ } \end{cases} $$

I want to find value of A by using this formula: $$\int_{x}\int_{y}f_{X,Y}(x,y)dxdy=1$$ which becomes after subsituting value: $$\int_{x}\int_{y}(A)dxdy=1$$ My question is how the values of 'x' and 'y' intervals are calculated from given information.

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up vote 1 down vote accepted

First of all, $x$ must have $|x|\le 1$, i.e. must be in $[-1,1]$ in order that $f_{X,Y}(x,y)\ne 0$, so assume $x$ goes from $-1$ to $1$, then choose the right interval for $y$ (depending on $x$).

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Thank you... so can we also do other way round, i mean pick y from -1 to 1 and choose interval for x? – Osman Khalid Feb 15 '13 at 20:29
Yes, exactly... – Berci Feb 15 '13 at 22:43
We get $|y|\le 1-|x|$, i.e. $y$ going from $|x|-1$ to $1-|x|$. – Berci Feb 15 '13 at 22:50

$0\le|x|+|y|<1$ is area (without border) of square made by points $(0,1)$, $(0,-1)$, $(1,0)$ and $(-1,0)$. With area $2$.

If I understand correctly, it means that $A=\frac{1}{2}$.

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Thank You... sorry this may be very simple question.. but can you please explain, how we get 1/2 from 2, i mean why 2 is divided by 4? – Osman Khalid Feb 15 '13 at 20:27
Interpret the integral as volume -… (from In your case it's cuboid with base area 2 (the square) and height A (we can ignore places where value of the function (or height) is $0$). And volume of this cuboid must be 1. It means $2A=1$. Or $A=1/2$. – zaarcis Feb 15 '13 at 20:44


First, plot your domain


Since your PDF is constant, you integral is simply area of that square. But if you want to truly work it out and find the limits of integration, then split your integral into four and find limits for each of them, which shouldn't be a problem since all lines then will have a form $y = kx+b$

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A picture is worth 1,000 words. – ncmathsadist Feb 16 '13 at 0:41
I share your opinion :) – Kaster Feb 16 '13 at 1:23
It is enough to split it into two: e.g. $x=-1..0$ and $x=0..1$ and use $y=x-1\,..\,1-x$ and $y=1-x\,..\,x-1$ respectively. – Berci Feb 16 '13 at 18:00
@Berci it is, I agree. – Kaster Feb 16 '13 at 20:40

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