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sheila
10-22-2011, 06:44 AM
Dear all,

I really need your help in solving these questions.
If X is the amount of money (in dollars) that a salesperson spends on gasoline and Y is the corresponding amount of money for which he is reimbursed, the joint density of these two random variables is given by:

_________1/25((20-x)/x) for 10<x<20, x/2<Y<X
f(X,Y) = {
_________0, Elsewhere

Find
A| The marginal density of X
B| The conditional density of Y given X=15
C| The probability that the salesperson will be reimbursed at least \$10 when spending \$15

I found the basics about marginal/conditional density and joint probability in my statitics book. However, for determining the marginal density the probabilities were already given in the book. Therefore only rows or columns were added up in order to determine the marginal density of X and Y. For me this example seems to be more complicated. Furthermore I understood there is a difference in the calculation method on discrete and continuous random variables. I think this example is with continuous variables, is this correct?
I tried to solve the integral and this is my answer:

1/25∫(20-x/x)dx = 1/25 ∫((20/x)-(x/x))dx = 1/25 ∫((20/x)-1)dx
20/25∫((1/x)-1dx=20/25(ln|x|-x)

Hope I can get a solution as I am preparing for my exam.

BGM
10-22-2011, 10:21 AM
The first thing is that you want to obtain the marginal density of X, so you should integrate with respect to the arguments except the one correspond to X. So in this case you should integrate with respect to y.

The second thing is that the integral is a definite one. You need to know the upper and lower limit from the given joint probability density function as well.

sheila
10-29-2011, 03:02 PM
Thank you for your reply. I tried to solve question A. Is this correct?
_________________x
[1/25*((20-x)/x) y]
_________________x/2
1/25*(20-x)*x - 1/25*((20-x)/x)*x/2 =

(20-x)/50

BGM
10-29-2011, 03:08 PM
Yes it is correct.

sheila
10-29-2011, 03:23 PM
Is the answer of question B, the conditional density of Y given X=15,
1/10?