# marginal density, conditional density, and probability

• March 24th 2008, 06:57 AM
TheHolly
marginal density, conditional density, and probability
I have some of this problem done, but I can't figure it all out, so I will post how I have tried to solve each part after I post the question. Thanks for the help!

If X is the amount of money (in dollars) that the salesperson spends on gasoline during the day and Y is the corresponding amount of money (in dollars) for which he or she is reimbursed, the joint density of these two random variables is given by
f(x,y) = { 1/25((20 - x) / x) for 10 < x < 20, x/2 <y < x
0 elsewhere.
Find a) the marginal density of X, b) the conditional density of Y given X = 12, and c) the probability that the salesperson will be reimbursed at least $8 when spending$12.

For a), I tried to find the integral from x/2 to x of f(x,y), but I ended up with [1/25(lnx - x)] evaluated from x/2 to x. There is a mistake there somewhere because I know the correct answer is supposed to be g(x) = (20 - x)/ 50 for 10<x <20.
For b), I used what I know the correct answer for a) was supposed to be and solved f(x,y) / g(x). I got 1/6 and I this is the correct answer.
For c), I tried to use the integral from 8 to 12 of f(x,y), but, again, I am not getting the right answer (which is 1/3).

Any help would be greatly appreciated!
• March 24th 2008, 06:50 PM
mr fantastic
Quote:

Originally Posted by TheHolly
I have some of this problem done, but I can't figure it all out, so I will post how I have tried to solve each part after I post the question. Thanks for the help!

If X is the amount of money (in dollars) that the salesperson spends on gasoline during the day and Y is the corresponding amount of money (in dollars) for which he or she is reimbursed, the joint density of these two random variables is given by
f(x,y) = { 1/25((20 - x) / x) for 10 < x < 20, x/2 <y < x
0 elsewhere.
Find a) the marginal density of X, b) the conditional density of Y given X = 12, and c) the probability that the salesperson will be reimbursed at least $8 when spending$12.

For a), I tried to find the integral from x/2 to x of f(x,y), but I ended up with [1/25(lnx - x)] evaluated from x/2 to x. There is a mistake there somewhere because I know the correct answer is supposed to be g(x) = (20 - x)/ 50 for 10<x <20.
For b), I used what I know the correct answer for a) was supposed to be and solved f(x,y) / g(x). I got 1/6 and I this is the correct answer.
For c), I tried to use the integral from 8 to 12 of f(x,y), but, again, I am not getting the right answer (which is 1/3).

Any help would be greatly appreciated!

a) By definition, $g(x) = \int f(x, y) \, dy = \frac{1}{25} \int_{x/2}^{x} \frac{20 - x}{x} \, dy$

$= \frac{1}{25} \, \frac{20-x}{x} \, \int_{x/2}^{x} \, dy$

$= \frac{20-x}{25 x} [y]_{x/2}^{x}$

$= \frac{20-x}{25x} \, \left( x - \frac{x}{2} \right)$

$= \frac{20-x}{50}$.

c) $\Pr(Y > 8 | X = 12) = \int_{8}^{12} \text{answer to (b)} \, dy$.

I get 2/3 .......?