Variance

**kenshinofkin** · August 1st 2008, 12:43 AM

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
$\int-1,1 (-2x-1)f(x)dx$ What is f(x) in this problem?

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

**CaptainBlack** · August 1st 2008, 09:40 AM

Originally Posted by kenshinofkin

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
$\int-1,1 (-2x-1)f(x)dx$ What is f(x) in this problem?

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

$f(x)=2(1-x)\ x \in [-1,1]$

is not a density as it integrates up to 4 not 1, are you sure that the range is not $[0,1]$ . Also it seems unlikely that a negative profit would be acceptable on a new car (even less a mean profit that is negative).

RonL

**kenshinofkin** · August 1st 2008, 10:03 AM

It could be a typo. So if I go with (0,1) am I doing it right so far. f(x) = x?
So what about the other parts of the question.
b) How can I show that it holds true?
Do I do (1,2) which would show a negitive profit?
c) ???
Not sure about this one

**mr fantastic** · August 1st 2008, 04:57 PM

Originally Posted by kenshinofkin

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 0 <= x <= 1 Mr F says: I don't know why you have f(x , y) ....?
a) Find the vaiance in the dealer's profit.
b) Demonstrate that chebyshev's inequality holds for k = 2 with the density function above.
[snip]

a) E(2(1-x)) = -4
$\int-1,1 (-2x-1)f(x)dx$ What is f(x) in this problem?

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

b) Chebyshev's Inequality: $\Pr(|X - \mu| \geq k \sigma) \leq \frac{1}{k^2}$ .

So you first need to calculate:

1. $\mu = \int x f(x) \, dx = \int_{0}^{1} x [2(1 - x)] \, dx = 2 \int_{0}^{1} x - x^2 \, dx = \frac{1}{3}$ .

2. $\sigma^2 = E(X^2) - \mu^2$ ,

where $E(X^2) = \int x^2 f(x) \, dx = \int_{0}^{1} x^2 [2(1 - x)] \, dx = 2 \int_{0}^{1} x^2 - x^3 \, dx = \, .......$ (which will answer part a) by the way).

Therefore $\sigma = \frac{1}{3 \sqrt{2}}$ .

Substitute $\mu = \frac{1}{3}$ , $\sigma = \frac{1}{3 \sqrt{2}}$ and k = 2 into the left hand side of Chebyshev's Inequality:

$\Pr \left ( \left |X - \frac{1}{3} \right| \geq \frac{\sqrt{2}}{3} \right)$

$= \Pr \left( X \geq \frac{\sqrt{2} + 1}{3} \right)$

$= \int_{(\sqrt{2} + 1)/3}^{1} 2 (1 - x) \, dx = \, .... \leq \frac{1}{4}$

which demonstrates the inequality.

Most of this is basic application of routine definitions and formulae.

NB: I reserve the right for the arithmetic details to contain errors so make sure you carefully check all numerical results given above.

**mr fantastic** · August 1st 2008, 05:05 PM

Originally Posted by kenshinofkin

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 0 <= x <= 1
[snip]
c) What is the probability that the profit exceeds $500?

[snip]

Note that a profit of $500 is equivalent to $X = \frac{1}{10}$ since X is in units of five thousand dollars.

After this it's all down hill:

$\Pr \left( X > \frac{1}{10} \right) = \int_{1/10}^{1} 2 (1 - x) \, dx = \, ....$

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