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Verify that the following is a bivariate density:

$F(X, Y) = \displaystyle\frac{2}{\pi}$ for $X^2+Y^2\leq1$ and $X < Y$

$F(X,Y) = 0$ otherwise

Now, I know that verifying that this is a bivariate density involves taking a double integral and showing that it is equal to 1. However, I'm not quite sure what the limits are for this integral! Here is the integral I have:

$\displaystyle\int\displaystyle\int \displaystyle\frac{2}{\pi} dx dy$

How can I find out the limits for this integral?

Thank you!

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    $\begingroup$ Integrate over all possible values of $x$ and $y$ where $F(x,y)\neq 0$. $\endgroup$ – Matthew Gunn Jan 26 '17 at 7:00
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    $\begingroup$ @MatthewGunn So, would that be $-1 \leq x < \displaystyle\frac{1}{\sqrt{2}}$ and $-\displaystyle\frac{1}{\sqrt{2}} < y \leq 1$? $\endgroup$ – Bing Jan 26 '17 at 7:06
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    $\begingroup$ Close, but is $x=-1, y=-1$ possible? the range of one variable has to be conditional on the value of the other. $\endgroup$ – Matthew Gunn Jan 26 '17 at 7:09
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    $\begingroup$ If you draw a picture you can do this one by inspection. $\endgroup$ – Glen_b Jan 26 '17 at 7:27
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    $\begingroup$ Three approaches I would consider depending on how much time you want to spend. (1) Recognize that you're after $2/\pi$ times the area of the intersection of $x^2+y^2\leq1$ and $x<y$. What is that? (2) Recognize that this particular problem is invariant to rotation so you could use the algebraicly easier constraints $x^2+y^2\leq 1$ and $y \geq 0$ or (3) integrate over the region from $x \in (-1, 1/\sqrt{2})$ and $y$ what it needs to be (may be a good algebra/geometry/careful thinking exercise, but it's kinda of an unneccesary pain). $\endgroup$ – Matthew Gunn Jan 26 '17 at 7:42
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My calculus is probably rusty.

I think it's a good idea to sketch the region over which you will integrate. Firstly, we know that:

$$X^{2}+Y^{2}\leq 1$$ which is the region bound by a unit circle. Furthermore, we have the constraint that:

$$X<Y$$ so the desired region is that bounded by a unit circle where $X$ is less than $Y$. To illustrate:

enter image description here

If you want you can convert to polar coordinates as follows:

$$\begin{align} x=r\text{cos}(\theta)\\ y=r\text{sin}(\theta) \end{align}$$ where the Jacobian is given by:

$$\begin{align} J&=\left|\begin{array}{cc} \tfrac{\partial x}{\partial r} & \tfrac{\partial x}{\partial \theta}\\ \tfrac{\partial y}{\partial r} & \tfrac{\partial y}{\partial \theta} \end{array} \right| \\ &=\left|\begin{array}{cc} \text{cos}(\theta) & -r\text{sin}(\theta)\\ \text{sin}(\theta) & r\text{cos}(\theta) \end{array} \right| \\ &=r \end{align}$$

You'll be able to determine the appropriate limits of integration from the plot. In polar coordinates, the bounds are:

$$\begin{align} 0&\leq r\leq 1\\ \tfrac{\pi}{4}&\leq \theta \leq \tfrac{5\pi}{4} \end{align}$$

The integral you want to evaluate is:

$$\frac{2}{\pi}\int_{\pi/4}^{5\pi/4}\int_{0}^{1}r\,dr\,d\theta$$

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  • $\begingroup$ I've drawn the picture myself, I just can't figure out the correct limits for the integrals. How do I determine the relationship between $x$ limits and the $y$ limits? $\endgroup$ – Bing Jan 26 '17 at 7:54
  • $\begingroup$ @Bing In polar coordinates: $r$ is bounded by $0$ and $1$. $\theta$ is bounded below by $\pi/4$ and above by $5\pi/4$. $\endgroup$ – StatsPlease Jan 26 '17 at 8:29

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