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https://i.stack.imgur.com/8j96g.png

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.

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  • $\begingroup$ First you find the marginal pdf of $X$ from the given graph - it is linear, passing through the origin, the support is $(0, 1000)$, and the area under the curve (triangle) is equal to 1. Then you also know the conditional pdf of $f_{Y|X=x}(y|x)$ which is given. Multiply them you obtain the joint for part a). $\endgroup$
    – BGM
    Oct 25, 2017 at 17:03

1 Answer 1

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(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 $

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