Sam is a stockbroker with investment and revenue that are not certain. His investment, $X$, each month is a random variable with the PDF in the image. $Y$ is the amount of money he earns in a month and is distributed uniformly between 0 and twice the amount he invested that month.
(a) What is the joint PDF $f_{X,Y}(x, y)$?
(b) What is the probability that in any given month, Sam makes a profit?
(c) Sam continues his job for 10 years. What would be the approximate probability that he makes a profit in at least 63 of the months?
Attempted Solution:
a) Do I need to differentiate the marginal PDFs of both the variables and add them up. But I don't understand how the marginal PDF graph of $Y$ would look like. It would be a horizontal line with lower bound $0$, and the upper bound $2X$?
b) For this I will find $P(Y>X)$, would this be $\frac X{2X}$, since for a profit to be earned $Y$ must be greater than $X$. This would be the upper half of the marginal PDF graph of $Y$.
c) Clueless about this.
(a) Let a be the value of $f_x(x)$ at 1000 dollars.
$\int_0^{1000} axdx= 1$
$500000a = 1 $
$a = \frac{1}{500000}$
This gives us $f_x x = \frac{x}{500000}$ . From the question we know $f_{y|x} {y|x} = \frac 1{2x} $ for y ∈ [0,2x].
So we have:
$f_{x,y} {x,y} = \frac{1}{1000000}$ if 0 ≤ x ≤ 1000 and 0 < y < 2x, and $0$ otherwise.
b) You are on the right track. You need to find P(Y>X) which would be:
$\int_0^{1000}\int_x^{2x} \frac{1}{1000000} dy dx = \frac{1}{2}$
c) Approximate to normal distribution.
n = 12 (months) * 10 (years) = 120 months
np = 120 * 0.5 = 60
np(1-p) = 120 * 0.5 * 0.5 = 30
$P(x >= 63) = P( Z >= \frac{63-60}{\sqrt {30}})$
$ P( Z >= 0.548 ) = 1 - \phi(0.548) = 0.295 $