Another probability problem

boks

Joined Oct 10, 2008
218
The amount of kerosene in a tank at the beginning of any day is a random amount Y from which a random amount X is sold during that day. Suppose that the tank is not resupplied during the day so that x<=y, and assume that the joint density function of these variables is

f(x,y) = 2, 0<x<y<1
f(x,y) = 0, elsewhere

a) Determine if X and Y are independent.
b) Find P(1/4 < X < 1/2 | Y = 3/4).

a)

$g(x) = \int ^1 _{0}2 dy = 2$
$h(y) = \int ^y _{0}2 dx = 2y$

$g(x)h(y) = f(x,y)$, so they are independent.

b)

$f(x|y) = \frac{f(x,y)}{h(y)} = \frac{1}{y}$

steveb

Joined Jul 3, 2008
2,431
The amount of kerosene in a tank at the beginning of any day is a random amount Y from which a random amount X is sold during that day. Suppose that the tank is not resupplied during the day so that x<=y, and assume that the joint density function of these variables is

f(x,y) = 2, 0<x<y<1
f(x,y) = 0, elsewhere

a) Determine if X and Y are independent.
b) Find P(1/4 < X < 1/2 | Y = 3/4).

a)

$g(x) = \int ^1 _{0}2 dy = 2$
$h(y) = \int ^y _{0}2 dx = 2y$

$g(x)h(y) = f(x,y)$, so they are independent.

b)

$f(x|y) = \frac{f(x,y)}{h(y)} = \frac{1}{y}$
Your answer to part A has a logical error in it. Clearly the amounts must be dependent since you can't sell more than you have.

An example of independent amounts would be.

1 the amount in the tank at the beginning of the day
2 the amount ordered by customers

assuming that people requesting to purchase have no way of knowing what is in the tank, or if other customers have been turned away.

In your case, knowledge about either amount tells you something about the other amount.