How do I graph and integrate a joint PDF with multiple inequalities?

[rudders93]

Homework Statement

Find λ given that the joint PDF of random variables X, Y, is given by:

$f(x,y)=\lambda x y^{2}$ where $0\leq x\leq y\leq 1$ and 0 otherwise

I have two questions:

1) How do I graph this? I'm not sure how to approach the inequality and graphing. What does this inequality actually? If you could provide a link to a site that might explain this graphing (specifically where there's multiple inequalities. I understand how to graph things with a single inequality, but I don't understand what the the multiple inequalities mean. I think it shows that both x and y are between 0 and 1 inclusive, but y is bigger than x. But how this translates graphically I'm not sure.

2) How can I integrate this? This is above question is an example, and it says that I can integrate it as follows:

$\int_0^1 (\int_x^1 \lambda x y^{2} dy) dx = 1$

where it equals 1 by definition of a pdf. But I'm not sure about why they're the integral with respect to x starts from x.

Thanks!

 

[BrownianMan]

The integral with respect to y (not x) has x as a lower integration limit. This is perfectly legitimate.

 

[rudders93]

Hi,

Thanks for reply. Yep my bad. But why does that integral work? How did they decide to do that?

 

[BrownianMan]

You are simply integrating the pdf over the region 0 <= x <= y <= 1.

 

[BrownianMan]

Doing the first part will help you understand the second part.

0 <= x <= y <= 1 is the triangular region between the y-axis and the line x=1, and y=x and y=1.

Note that lambda must be equal to 10 in order for the integral to equal 1.

 

[rudders93]

Hi, thanks for response! Could you explain how you arrived at 0<=x<=y<=1 being the triangular region between the y axis, the line x=1, y=x, and y=1?

 

[BrownianMan]

Hi, thanks for response! Could you explain how you arrived at 0<=x<=y<=1 being the triangular region between the y axis, the line x=1, y=x, and y=1?

We have 0<=x<=y<=1. First look at y. It is clearly between x and 1 (inclusive). This means that it is bounded above by y=1 and below by y=x. When y=x, 0<=x<=y<=1 becomes 0<=x<=x<=1 which becomes 0<=x<=1. When y=1, 0<=x<=y<=1 becomes 0<=x<=1. So clearly, 0<=x<=1, which means that x is bounded below by x=0 (y axis) and x=1.

 

[rudders93]

Oh, that definitely clears things up!

So using that we've restricted the region of the graph. Now if I was to graph the x,y plane, the relationship between x and y would be y = +- x^2. But as we have a restricted region, and the line y=x , y=0, x=1, x=0 bound that region, which is contained within y = +- x^2 we are left with just y = x where 0<x<1 as the required area.

Hence area would be $\int_0^1 x dx = \frac{1}{2}$? and then integrate again to get the volume of a solid. But that integral in my question ($\int_0^1 (\int_x^1 \lambda x y^{2} dy)$) is different... how do you link the two together?

Thanks again!

 

[BrownianMan]

No, just think of it as a regular double integral over a triangular region.

Once you have found the region of integration, the hard part is over. All you need to do is to integrate the density function. In general, if you want to integrate a multivariate function $$f(x,y)$$ over some region R, $$R=\left \{ a\leq x\leq b;c\leq y\leq d \right \}$$ then the integral becomes

$$\int_{a}^{b}\int_{c}^{d}f(x,y)dydx$$

In your example, the integral would then indeed be

$$\int_{0}^{1}\int_{x}^{1}\lambda xy^{2}dydx$$

 

[BrownianMan]

A good exercise to see how well you understand this, would be to change the order of integration. In other words, instead of having the inner integral be with respect to y and the outer with respect to x, make the inner integral be with respect to x, and the outer with respect to y. The limits of integration will be different, but the answer will be the same.

 

[rudders93]

Hmmm. I still don't understand this I think

So is this what is happening?

The first integral, $\int_x^1 \lambda x y^{2} dy$ finds the area by subtracting the integral of y=1 from that of y=x. in the region that 0<=y<=1. Then you somehow integrate this region over 0<x<1?

Atleast I think that's what should be happening? If so, how does the $\lambda x y^{2}$ come into it? Espcially since the restricted region does not even contain that graph?

Thanks!

 

[BrownianMan]

Have you taken multivariable calculus? I suspect that is where you are lacking. This kind of problem should be routine exercise for someone who has an understanding of calculus in multiple dimensions. The following is a good resource:

http://tutorial.math.lamar.edu/Classes/CalcIII/DoubleIntegrals.aspx
http://tutorial.math.lamar.edu/Classes/CalcIII/IteratedIntegrals.aspx
http://tutorial.math.lamar.edu/Classes/CalcIII/DIGeneralRegion.aspx

 

[rudders93]

I'll will look over that. Thanks!

 

[rudders93]

I think I've understood this now! Thanks so much BrownianMan!

Changing the order of integration, we'd get: $\int_0^1 \int_0^y \lambda x y^{2} dx dy = 1$ and the way I determined that was graphically by seeing the area over which we're cycling over. So the inner integral, we're holding y constant and cycling over the x values. When you draw the region (which you explained before), you can see that you should cycle from 0 to the line y=x to find that area, when solving with respect to y.

My main point of confusion was that $\lambda x y^{2}$ is the third axis, the 3d. I kept getting confused in thinking of it in terms of 2 dimensions.

Those articles really helped. If anyone else has this problem, http://www.khanacademy.org/math/calculus/v/double-integral-1 also helped a lot!

Thanks again!

 

[BrownianMan]

Changing the order of integration, we'd get: $\int_0^1 \int_0^y \lambda x y^{2} dx dy = 1$

Correct.

You can think of a double integral as finding the volume of the solid under the surface given by f(x,y) and above the region of integration.

 

[rudders93]

Thanks!