Please solve this stats problem...

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slayer202

slayer202

Lifer
Nov 27, 2005
13,679
119
106
companys profit equation = Price x Quantity - Costs


Price = $10


Quantity's mean is 100 with a variance of 100

Cost's mean is 980 with a variance of 400

correlation of .3


find profit's variance



I'm getting some giant ass number. I spent friggin forever on this test trying to figure it out but I couldn't do it. Heres what I was doing


100^2(100) + 980^2(400) - 2(100)(980)(.3)(10)(20)

and its something like 300+ million which makes zero sense. any help?
 
A

artikk

Diamond Member
Dec 24, 2004
4,172
1
71
42, but I really can't help, we haven't done finance stats in my class just regular stats.
 
C

chuckywang

Lifer
Jan 12, 2004
20,133
1
0
Originally posted by: slayer202
companys profit equation = Price x Quantity - Costs


Price = $10


Quantity's mean is 100 with a variance of 100

Cost's mean is 980 with a variance of 400

correlation of .3


find profit's variance



I'm getting some giant ass number. I spent friggin forever on this test trying to figure it out but I couldn't do it. Heres what I was doing


100^2(100) + 980^2(400) - 2(100)(980)(.3)(10)(20)

and its something like 300+ million which makes zero sense. any help?
Click to expand...

First, let's figure out some things:
E[Q^2] = Var(Q) + E{Q]^2 = 100+100^2 = 10100
E[C^2] = Var(C) + E[C]^2 = 400 + 980^2 = 960800
E[QC] = Corr(Q,C)*sqrt(Var(C))*sqrt(Var(Q)) + E{Q]*E[C] = .3*10*20+100*980 = 98060

P=10*Q-C

Var(P) = Var(10Q - C) = E[(10Q-C)^2] - E[10Q-C]^2
=E[100Q^2 -20QC + C^2] - (10E{Q] - E[C])^2
=100E[Q^2] -20E[QC] + E[C^2] -(10*100 - 980)^2
=100*10100 - 20*98060 + 960800 - 400 = 9200

I'm sure a lot of this can be simplified using properties of variance, but I don't remember those offhand.
 
KoolAidKid

KoolAidKid

Golden Member
Apr 29, 2002
1,932
0
76
Originally posted by: slayer202
companys profit equation = Price x Quantity - Costs


Price = $10


Quantity's mean is 100 with a variance of 100

Cost's mean is 980 with a variance of 400

correlation of .3


find profit's variance



I'm getting some giant ass number. I spent friggin forever on this test trying to figure it out but I couldn't do it. Heres what I was doing


100^2(100) + 980^2(400) - 2(100)(980)(.3)(10)(20)

and its something like 300+ million which makes zero sense. any help?
Click to expand...

I got 9200 using the equation that you provided. You are using the wrong constants when you do it.
 
slayer202

slayer202

Lifer
Nov 27, 2005
13,679
119
106
Originally posted by: KoolAidKid
Originally posted by: slayer202
companys profit equation = Price x Quantity - Costs


Price = $10


Quantity's mean is 100 with a variance of 100

Cost's mean is 980 with a variance of 400

correlation of .3


find profit's variance



I'm getting some giant ass number. I spent friggin forever on this test trying to figure it out but I couldn't do it. Heres what I was doing


100^2(100) + 980^2(400) - 2(100)(980)(.3)(10)(20)

and its something like 300+ million which makes zero sense. any help?
Click to expand...

I got 9200 using the equation that you provided. You are using the wrong constants when you do it.
Click to expand...

can you look at the numbers I wrote out and show me what is different? either way 9200 doesn't seem to fit.

heres what I did

a^2(var(x)) + b^2(var(y)) - 2(a)(b)(corr(a,b))(std(a))(std(b))

a = 100
b = 980
var(x) = 100
var(y) = 400
standard devation of a = 10
std of b = 20
correlation = .3


100^2(100) + 980^2(400) - 2(100)(980)(.3)(10)(20)

 
KoolAidKid

KoolAidKid

Golden Member
Apr 29, 2002
1,932
0
76
a and b are the constants used in the profit = 10*Q - C equation.

You are trying to find var(profit) which equals var(10*Q - C) which according to your variance equation that you posted should equal...

My advice is to ignore the means of the quantity and cost distributions; they are not necessary to solve the problem.
 
B

blinky8225

Senior member
Nov 23, 2004
564
0
0
I get 9200.

Let Q = quantity and C = Cost

Corr(Q,C) = 0.3 = Cov(Q,C)/(10 * 20) ==> Cov(Q,C) = 60

Var(10Q - C) = Var(10Q) + Var(C) + 2Cov(10Q,C) = 100Var(Q) + Var(C) - 20Cov(Q,C)
=100 * 100 + 400 - 20 * 60 = 9200

All the formulas used can be found on wikipedia or derived using the definition of Variance and Covariance.

Edit: I forgot a negative sign.
 
slayer202

slayer202

Lifer
Nov 27, 2005
13,679
119
106
this problem sucks my balls. I think the teacher messed up. the standard deviation shouldnt be much higher than 10, so the variation shouldnt be much higher than 100. that is agreed at least, correct?
 
B

blinky8225

Senior member
Nov 23, 2004
564
0
0
Originally posted by: slayer202
this problem sucks my balls. I think the teacher messed up. the standard deviation shouldnt be much higher than 10, so the variation shouldnt be much higher than 100. that is agreed at least, correct?
Click to expand...
Nope. It should be pretty high actually. When you multiply a random variable by 10 (the price), you multiply the variance by 100.

So, you want to find the variance of Profit = 10Q - C. Agreed?

By definition this is E[(10Q - C - (10(uq) - uc))^2], where uq is the mean of Q and uc is the mean of C. Expectations and means always add even if the random variable are not independent.

E[(10Q - C - (10(uq) - uc))^2] = E[((10Q - 10uq) - (C - uc))^2]
=E[(10Q - 10uq)^2] + E[(C-uc)^2] - 2E[(10Q - 10uq) (C-uc)]
=100Var(Q) + Var(C) - 20Cov(Q,C) = 9200
 
B

blinky8225

Senior member
Nov 23, 2004
564
0
0
Originally posted by: slayer202
the mean of profit is 20, correct? I don't see how the variance can be so high.
Click to expand...
That's right. However, the mean and variance are independent of each other (this fact takes about an hour to prove and requires matrix algebra). Therefore, the mean being low has absolutely no bearing on the variance. The variance is defined as the expectation of the square of the difference. Thus it is a measure of how much an observation varies from the expected value of mean.

For example, you might have two random variables X and Y both with mean 0. However X might have variance 100, and Y might have variance 900.

Then the data for X might look like -5,0,5
and for Y, -20,0,20.

They will both be centered around 0, but the observations in Y will be further from 0. So, you can disregard the information about the mean when you are trying to find the variance. It's much easier to use the formulas, but you can derive if from the definition of Variance like I did above, too.

Variance of X is defined as E[(X-u)^2].
 
Born2bwire

Born2bwire

Diamond Member
Oct 28, 2005
9,840
6
71
Originally posted by: slayer202
the mean of profit is 20, correct? I don't see how the variance can be so high.
Click to expand...

Not that you should take the variance in this way at all, but just look at a possible spread in the numbers. Take the standard deviation to be 10 and 20. Then you have a "range" of quantity from 90-110 and cost from 960-1000. This gives you a profit ranging from -100 to 140, a "range" of 240. If that 240 is somewhat indicative of the standard deviation of the price then that correlates to a variation an indicative value on the order of 10,000. This is thoroughly abusing variance and standard deviation but it supposed help demonstrate how a low deviation in this case can blow up to a large variance.
 
KoolAidKid

KoolAidKid

Golden Member
Apr 29, 2002
1,932
0
76
Originally posted by: blinky8225
I get 9800.

Let Q = quantity and C = Cost

Corr(Q,C) = 0.3 = Cov(Q,C)/(10 * 20) ==> Cov(Q,C) = 60

Var(10Q - C) = Var(10Q) + Var(C) + 2Cov(10Q,C) = 100Var(Q) + Var(C) - 20Cov(Q,C)
=100 * 100 + 400 - 20 * 60 = 9800

All the formulas used can be found on wikipedia or derived using the definition of Variance and Covariance.

Edit: I forgot a negative sign.
Click to expand...

10000 + 400 - 1200 = 9200
 
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