Question

If two random variables have the joint density f(x1,x2)=x1x2 for 0 12 years ago

Answer 1

Have you considered drawing the region and cutting it in half both horizontally and vertically - and then pondering the geometry?

Answer 2

I guess that I don't know how I would draw the region?

Answer 3

We get to use integration, right??

Check to see that you have it: \(\int\limits_{0}^{1}\;\int\limits_{0}^{2}x_{1}\cdot x_{2}\;dx_{1}\;dx_{2}\;=\;1\) -- Always a good thing to check.

Solve the first one. \(\int\limits_{0}^{1}\;\int\limits_{0}^{1}x_{1}\cdot x_{2}\;dx_{1}\;dx_{2}\)


Solve the second one. \(\int\limits_{0}^{1}\;\int\limits_{0}^{x_{2}}x_{1}\cdot x_{2}\;dx_{1}\;dx_{2}\)

Just look REALLY carefully at the limits. Tell if if the second one is just wrong!

Answer 4

I get 1/8 for the second one

Answer 5

and 1/ 4 for the first

Answer 6

Okay, then it is time to decide who it right. How should it work? What does the setup all mean? Have we wandered off, somewhere?

Answer 7

I am sorry. I may have misunderstood your previous post. What are you referring to in the second integral?

Solve the second one. ∫01∫0x2x1⋅x2dx1dx2 Just look REALLY carefully at the limits. Tell if if the second one is just wrong!

Answer 8

There are 3 integrals.
What changed from the first (just check) to the second (answering question #1)?

Answer 9

The bounds?

Answer 10

Right. Does this solve the problem? If so, how? If not, why not?

Answer 11

Are you referring to the problem of me not knowing how to draw a picture of the region? If so, would the first part be the entire area = the entire probability from 0 to 1? Then the second part would take up approximately .25 of the area. The second part would take up 1/24 and from examining this geometrically, I should be able to come up with the marginal density?

Answer 12

I am sorry if I seem really dim. But, I really appreciate the help

Answer 13

If the joint density were uniform (1/2 throughout). it would be as simply as drawing the picture and using the very square geometry. As it is NOT uniform, the geometry gives only a clue. This is why we need the integral.

Solving the first: It is the VOLUME under the surface that is important. If we choose 0 < x1 < 1 and 0 < x2 < 1,clearly this is 1/2 the total area for a UNIFORM distribution. Upon integration with the actual joint density, we see that the volume in the first square (adjacent to the Origin) is only 1/4 - Leaving a whopping 3/4 in the other square.

Solving the second: x1 * x2 is symmetrical about x1 = x2 in the first square., so this is easily answered by 1/2 of whatever we answered on the first question. x2 never exceeds Unity, so there can't be any in the 2nd square.

I have to find exception with the given answer.

Answer 14

you find exception with the given answer 1/24?