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Q: Probability Random Variable1 ( Answered 4 out of 5 stars,   0 Comments )
Question  
Subject: Probability Random Variable1
Category: Science > Math
Asked by: alpa101473-ga
List Price: $3.50
Posted: 27 Sep 2003 19:18 PDT
Expires: 27 Oct 2003 18:18 PST
Question ID: 260868
The random vector variable (X,Y) has the joint pdf

f(x,y) = k(x+y)     0 < x < 1, 0 < y < 1.

a.  Find k.
b.  Find the joint cdf of (X,Y).
c.  Find the marginal pdf of X and of Y.
Answer  
Subject: Re: Probability Random Variable1
Answered By: elmarto-ga on 28 Sep 2003 17:55 PDT
Rated:4 out of 5 stars
 
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
alpa101473-ga rated this answer:4 out of 5 stars

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