Hi alpa!
These are the answers to your questions.
a. We know from the exercise that the joint pdf of (X,Y) is k(X+Y). In
order to find k, we use the fact that the pdf must integrate to one
over the domain of X and Y. Therefore, we have that the following
equation must hold:
1/ 1/
| |
| | k(x+y) dy dx = 1
| |
/0 /0
(In case it's not clear - which is probably the case - those symbols
are integrals wannabees :) ). We can take the constant out of the
integral:
1/ 1/
| |
k | | (x+y) dy dx = 1
| |
/0 /0
And finally, solving the integral, which is quite easy to do with some
knowledge of calculus (please request a clarification if you have
trouble doing it), we find that:
k = 1
So k must be equal to 1 so that k(x+y) is truly a pdf; otherwise it
won't integrate to one.
b. By definition, the CDF of (X,Y) is Prob(X<x, Y<y). In order to find
the cdf of (X,Y), then, we must compute the integral:
x/ y/
| |
| | (s+t) ds dt
| |
/0 /0
which again is easy to do using calculus. The answer is:
x/ y/
| |
| | (s+t) ds dt = (1/2)*(xy^2 + yx^2)
| |
/0 /0
and that's the cdf of (X,Y).
c. The formula to find the marginal pdf given the joint pdf is given
in the following page:
Joint, Marginal and Conditional distributions
http://www.ma.utexas.edu/users/mks/384G03/jointmargdist.pdf
Thus, in order to compute the marginal distribution of X, we must
solve the integral:
1/
|
| (x+y) dy
|
/0
which simply gives x+(1/2), for x belonging to [0,1]. That's the
marginal pdf of X. Similarly, the marginal pdf of Y can be calculated
solving the integral:
1/
|
| (x+y) dx
|
/0
which is y+(1/2), for y belonging to [0,1]. You can easily verify that
both marginal pdf's integrate to one.
Google search strategy
marginal pdf
://www.google.com/search?sourceid=navclient&ie=UTF-8&oe=UTF-8&q=marginal+pdf
I hope this helps! If there's anything unclear about my answer (for
example, if you have any doubt on how the inegrals were calculated)
don't hesitate to request a clarification. Otherwise I await your
rating and final comments.
Best wishes!
elmarto |