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Let $$ f = \begin{cases} cx& \text{if } 1> x > y > 0\\ 0 &\text{otherwise}, \end{cases} $$ let this function be the joint density function for $X$ and $Y$, I want to determine the value of $c$ such that it is a joint denisty function, I keep getting $c=6$ rather than $3$ which is the answer. Im more concerned regarding how Im integrating to in the region $1>x>y>0$, i must be taking the integration limits wrong i presume, if someone could kindly show me how to integrate a region with constraints like this that would be great.

Thanks

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    $\begingroup$ What are your limits? They should be $0 < y < x$ and $0 < x < 1$, integrating $dy dx$. $\endgroup$ – Suugaku Dec 13 '13 at 17:50
  • $\begingroup$ The correct value is $3$. $\endgroup$ – André Nicolas Dec 13 '13 at 17:56
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Either

  • $x$ runs from $0$ to $1$ and for each fixed value of $x$, the other variable $y$ runs from $0$ to $x$; or
  • $y$ runs from $0$ to $1$ and for each fixed value of $y$, the other variable $x$ runs from $y$ to $1$.

So you have either $$ \int_0^1\left(\int_0^x\cdots\cdots\, dy\right) \, dx $$ or $$ \int_0^1 \left(\int_y^1\cdots\cdots\,dx\right) \, dy. $$ In the first case you have $$ \int_0^x cx \, dy = cx^2, $$ and then $$ \int_0^1 cx^2\,dx = \frac c 3. $$ In the second case you have $$ \int_y^1 cx\,dx = c\frac{1-y^2}{2} $$ and then $$ \int_0^1 c\frac{1-y^2}{2}\,dy = \left.c\left(\frac y 2 - \frac{y^3}6\right) \right|_0^1 = \frac c 3. $$

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The region is the triangle formed by the points $(0,0)$, $(1,0)$ and $(1,1)$. A good way to think about it is to fix an $0 < x < 1$. Then you want to go over all $y$ that is between $0$ and $x$, so it's considering all cross-sections of the triangle and then adding them all up. This gives us the integral equation: $$ \int_{0}^1 \int_0^x c x \, dydx = 1. $$ The integral on the right hand side is $$ \int_{0}^1 \int_0^x c x \, dydx = \int_0^1 c x^2 \, dx = \dfrac{c}{3} $$ so that setting this equal to 1 gives $c = 3$ as in the book.

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I am unable to solve the problem without drawing a picture. We have $0\lt y\lt x\lt 1$. In particular, $y\lt x$, so our density function lives below the line $y=x$ (which I drew). Also, $x$ is between $0$ and $1$.

So the region where our density function lives is the triangle with corners $(0,0)$, $(1,0)$, and $(1,1)$.

The integral of $cx$ over this triangle should be $1$. We express the double integral as an iterated integral in two ways: (i) integrate with respect to $y$, and then $x$, or (ii) the other way.

Method 1: We look at the triangle, and note that $y$ goes from $0$ to $x$, and then $x$ goes from $0$ to $1$. So we get $$\int_{x=0}^1 \left(\int_{y=0}^x cx \,dy\right)\,dx.$$ Integrate. The inner integral is trivial, since $cx$ does not involve $y$. So we simply get $cx^2$, Now integrate with respect to $x$. We get $\frac{c}{3}$.

Method 2: We look at the triangle. At the beginning, $x=y$, and at the end it is $1$. So our integral is $$\int_{y=0}^1 \left(\int_{x=y}^1 cx \,dx\right)\,dy.$$ The inner integral is $\frac{c}{2}(1-y^2)$. Now integrate with respect to $y$. We get $\frac{c}{3}$.

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