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Kenshinofkin

Member

Registered: 2008-07-08

Posts: 6

Variance

A dealer's profit in units of $5,000 on a new automobile is a random variable X having density function

f(x,y) = 2(1-x) for -1 <= x <= 1
a) Find the vaiance in the dealer's profit.
b) Demonstrate that chebyshev's inequality holds for k = 2 with the density function above.
c) What is the probability that the profit exceeds $500?

a) E(2(1-x)) = -4
What is f(x) in this problem? is it just x?

= -1
b) How can I show that it holds true?
c) ???

John E. Franklin

Member

Registered: 2005-08-29

Posts: 3,588

Re: Variance

First guess assumes y=1 means $5000 profit, and this first guess assumes even level probabilty between $20000 and 0:
c.) might be 39/40, since y > $500 between x=-1 and x=+0.95 (This is a total guess, I'm new to this)
b.) show the probability < 1/5 if one-sided and <1/4 if two-sided (I'm just repeating stuff I just read, a guess...)

Second guess after saw your integration:
This 2nd guess assumes y=1 means $5000 profit and y also dually means the chance the profit will happen.
This is a ridiculous 2nd guess, because y has two meanings combined:
The total area under the triangle is 4*2/2 or 4, so 4*25 = 100%
Now when x = 1/2, then y = 1, or y = $5000, is that how you do it?? (I'm guessing here, totally guessing)
I'm confused how the $5000 works into this.

Can you explain what you know about this?

If x=+0.95 is the $500 mark (guessing), then the area to the right of this is 0.05*0.10/2 or 1/400.
But total area is 4, so 100% - (1/1600) might be the answer for c.) based on area. (guessing 2nd time on c.)

So maybe c.) is 1-(1/1600) or 0.999375

Third guess:
I have a new idea, that the x-value, as it goes away from the origin, is related to the profits from 0 to $5000.
+$5000 at x=-1 and +$5000 at x=+1, and $0 profit at x=0.  This is a strange guess because the
profit goes up in both the left and right directions, and profit is never negative. (See guess #4 below if profit can go negative)
If this is true, then the triangle still has an area of 4.
So $500 is $5000/10, so find the area from x=-0.1 to x=+0.1 (again this a guess)
If so, the average height here is 2 for the trapezoid, so 2 * 0.2 is the area, and
what percentage of the total 4 area is that?  2*0.2=0.4 and 0.4/4 is 10%.
So my 3rd guess is 90% chance the profit will be above $500 for part c.)

Fourth guess: 
This 4th guess assumes the profit varies from -$5000 when x = -1, to $0 profit at x=0 and up to +$5000 profit when x = +1.   So for this scenario, find the percentage area of the triangle to the right of x=+0.1.
This should be ((1-0.1)/2)^2 or 0.2025 or 20.25% for c.)  (guess #4)

Last edited by John E. Franklin (2008-08-01 14:24:33)

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