U1 ~ Uniform[0,1] and U2 ~ Uniform[2, 3]. In which region of the U1-U2 plane is the joint pdf of U1,U2 nonzero?
I have not understood the meaning of U1-U2 plane here.. Can anyone please explain how do I go about solving this question?
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We actually cannot answer the question as written because you haven't specified the joint distribution of $U_1$ and $U_2.$ Are they independent? – spaceisdarkgreen Aug 08 '18 at 18:46
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Yes......they are independent – Harshada Kelkar Aug 08 '18 at 18:57
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why all the dots? If they're independent, it's important to say so. In a lot of problems, it's more or less implicit that they're independent unless otherwise noted (though lots of people will police you on this regardless since it's a very strong assumption that is often forgotten). However this problem is about 'sketching supports' which is generally a class of problem where the dependence of the variables is a main feature, so it's somewhat strange to even see a problem where they're independent. – spaceisdarkgreen Aug 08 '18 at 19:00
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Actually, nothing is mentioned in the problem.... I am not able to understand this question. Especially the U1-U2 plane. I am sorry if I am lacking clarity. Il try to get more information on this if possible and will update it. – Harshada Kelkar Aug 08 '18 at 19:11
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In my answer, your main question is answered in the first sentence and the rest is a sketch of how to do the problem. It's a plane of values for the two random variables. If you have further questions about that, let me know. – spaceisdarkgreen Aug 08 '18 at 19:29
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When you have two random variables, you can imagine each outcome as a point in the plane, since each outcome consists of a value for $U_1$ and a value for $U_2.$ For a continuous distribution, there will be a density $f_{U_1,U_2}(u_1,u_2)$ for the outcome occurring near every point $(u_1,u_2)$ in the plane, such that the probability of the outcome lying in a certain region is the integral of the density function over that region. This is what a joint PDF is.
The region of the plane where the PDF is nonzero is called the support, so this problem is asking you for the support of the PDF. Here, as you've indicated in the comments, the variables are independent, so the joint PDF is just the product of the marginal PDFs. Since they're uniform, we can write the marginal PDFs $$f_{U_1}(u_1) = 1_{[0,1]}(u_1) \\f_{U_2}(u_2) = \frac{1}{3-2}1_{[2,3]}(u_2)= 1_{[2,3]}(u_2),$$ where $1_A(u)$ is the indicator function that is $1$ when $u\in A$ and zero otherwise. By independence the joint is $$ f_{U_1,U_2}(u_1,u_2)=1_{[0,1]}(u_1)1_{[2,3]}(u_2)$$ so your task is to write down the region of the $u_1-u_2$ plane where this function is nonzero.
A couple of comments. First, I've used a lot of notation here, but really, at least in the independent case, this is pretty much common sense once you understand everything. It's basically equivalent to
$u_1$ can be any number between zero and one, and $u_2$ can be any number between $2$ and $3.$ What region of the plane can $(u_1,u_2)$ lie in?
but I'll stress this is only in the case of independence.
Second, much less important comment for completeness's sake. This is not actually all that well-posed since you can always change the PDF on a set of measure zero (including setting the PDF to zero there) and still have the same distribution. So 'places where the PDF is zero' is not quite the same thing as 'support'. This is really a minor technical point you can probably ignore for now.
spaceisdarkgreen
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Ok... so co-ordinates of this region would be (0,2) (1,2) (1,3) (0,3) assuming U1 and U2 are independent. Actually, I got confused in U1-U2 notation.. I thought its an equation of plane of the form ax+by=d. Thanks a lot. – Harshada Kelkar Aug 09 '18 at 04:17
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@Harshada Those are the corners of the region (which is a square), yes. We can write the region as either set notation $[0,1]\times [2,3]$ or as inequalities $0\le u_1 \le 1,$ $2\le u_2\le 3.$ – spaceisdarkgreen Aug 09 '18 at 08:50