# joint/marginal/conditional density

• Oct 15th 2011, 01:29 AM
Naomi2011
joint/marginal/conditional density
Dear all,

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

Can anybody help me on my way to solve the questions? I tried to look for some basic information regarding marginal/conditional/joint density functions, but I couldn't find some basic information that could help me solving this question.
Thank you very much in advance.
• Oct 15th 2011, 03:34 PM
harish21
Re: joint/marginal/conditional density
information regarding marginal/conditional distributions can be found in your book/class notes

to find the marginal distribution of A, integrate the joint pdf with respect to y

$f_X(x)\;=\; \int_Y f(x,y)\;dy \;=....$
• Oct 18th 2011, 11:48 AM
Naomi2011
Re: joint/marginal/conditional density
Thank you for your reply. I found the basics about marginal/conditional and joint probability in my statitics book. However, for determining the marginal probability the probabilities were already given in the book. Therefore only rows or columns were added up in order to determine the marginal probabiltiy of X and Y. As I understood correctly this example is more advanced. Furthermore I understood there is a difference in the calculation method on discrete and continuous random variables. As I understood correctly, this example is with continuous variables?
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)
Is this answer correct so far?
• Oct 22nd 2011, 06:45 PM
harish21
Re: joint/marginal/conditional density
Quote:

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)
Is this answer correct so far?

NO.

if you are finding the marginal distribution of X, you should integrate that function with respect to y as stated in post no 2.

$f_X(x) \;=\; \int_{x/2}^x f(x,y)\;dy\;=\;\int_{x/2}^x \dfrac{1}{25}\cdot \dfrac{20-x}{x}\;dy \\ f_X(x) \;=\; \dfrac{1}{25}\dfrac{20-x}{x} \int_{x/2}^x \;dy= ...$