Question

The amount of kerosene, in thousands of liters, 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 for 0 < x < y < 1. find the that the amount of kerosene sold is between 0.25 and 0.50 thousands liters given that the amount in the tank at the start of the day is 0.90 thousands liters. (Hint: f(x|y) = f(x,y) / fy(y) ).

Answer 1

Find the marginal density for $Y$ by integrating the joint PDF over all possible values of $x$:

$f_Y(y)=\displaystyle\int_{-\infty}^\infty f_{X,Y}(x,y)\,\mathrm dx=\int_0^y2\,\mathrm dx$

$\implies f_Y(y)=\begin{cases}2y&\text{for }0$

Then the density of $X$ conditioned on $Y$ is

$f_{X|Y}(x\mid y)=\dfrac{f_{X,Y}(x,y)}{f_Y(y)}=\frac1y$

for $0$ and undefined elsewhere.

Thus

$P(0.25$

= = =

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