Wikipedia:Reference desk/Archives/Mathematics/2008 November 5

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November 5

Some Econ Questions

So I'm taking a microeconomics class, and I'm working on a homework assignment that has me totally confused. Unfortunately, it being election night, there's basically nobody else who's willing at the moment to help walk me through it. I just wanted to explain one question in particular and see if my answers are on the right track.

So the question is about a football coach who's trying to make the most money off of the game. The demand function for ticket sales is D(p) = 200,000 - 10,000p. The stadium capacity is 100,000.

The first thing they asked is what is the inverse demand function. The damn econ book tells you the idea of what an inverse demand function is, but never really specifies what it means mathematically. So I only have my best guess, which is that the inverse function here would be the price you'd need to sell to sell x number of seats. So: P(x) = (200,000 - x)/10,000, I said. Right? Wrong?

Next question is to find the equations for total revenue and marginal revenue. Total revenue is pq. I thought that meant p * D(p) here, or p(200,000 - 10,000). Right? Wrong?

Then, Marginal Revenue. That's supposed to equal change in Rev over change in p. Or q plus the change in q over the change in p. So I wrote that it equaled D(p) + (-10,000)/1 (I'm terrible at calculus -- did I just screw something up?) So I guess the question is whether my ideas here are right and, if so, whether my calculus is.

Also, more general question: Elasticity is a function of rate change. If numbers are only given to you for a single state (so there's no change), how are you supposed to calculate elasticity? Thanks, 140.247.133.136 (talk) 00:56, 5 November 2008 (UTC)

Inverse demand function says that your understanding of it is correct, and your math for it is correct. Total revenue is certainly the product of price and volume (which is equal to demand), but you left out a p: $p\,D(p)=10^{4}(20p-p^{2})$. (This quantity is also $q\,P(q)$.) Marginal revenue is defined in our article as the derivative of that with respect to q: ${\frac {d}{dq}}(q\,P(q))=P(q)+q\,P'(q)$ via the product rule. What you've written more resembles the derivative of total revenue with respect to p: ${\frac {d}{dp}}(p\,D(p))=D(p)+p\,D'(p)$; you seem to have left out a p in that formula also. Of course, given that the tickets have zero cost, what you've written (once corrected) is really the more useful quantity: solving for its zero will maximize the revenue (and thus the profit, since cost is constant). Elasticity is a description of $D'(p)$, so can't be calculated from isolated data unless you already have a parameterized model whose parameters can be determined from the data. --Tardis (talk) 02:48, 5 November 2008 (UTC)

algebra

When doing the homework of mathematics i got a problem of algebra where one condition is given. But i couldn't solve the problem. is it sufficient to show the condition with only one given condition? if yes, how? The problem is:

 —Preceding unsigned comment added by Dahalrk (talk • contribs) 04:58, 5 November 2008 (UTC) 

if 2^x=3^y=12^z then prove that

1/z=1/y+2/x —Preceding unsigned comment added by Dahalrk (talk • contribs) 04:40, 5 November 2008 (UTC)

Obviously, x, y and z can't be rational.

Topology Expert (talk) 05:38, 5 November 2008 (UTC)

Why not start by taking logarithms?

Maybe easier to start with

$2^{x}=3^{y}=12^{z}=a$

$\Rightarrow a^{1/x}=2\quad a^{1/y}=3\quad a^{1/z}=12$

Incidentally, I have made an assumption here that excludes the one rational solution; precise result should be 1/z=1/y+2/x or ... Gandalf61 (talk) 07:09, 5 November 2008 (UTC)

Multinomial logit

Is this (see section discussing error) assertion by the IP correct? I know that she is right in that we have to normalize one of the Betas, but we dont normalize so that the probabilities sum to one, we do it to identify the model. I mean, even with error in the previous denominator the probabilities would have still summed to 1. Am I missing something? Also, can anyone explain where things go wrong if we dont normalize. I understand that we have to normalize, but only in intuition, not really in a mathematical sense. Most texts I have encountered do not go into much detail, so any light would be greatly appreciated. Thanks, Brusegadi (talk) 08:24, 5 November 2008 (UTC)

Either way will work as they are only reparameterizations of each other, so long as the treatment is consistent throughout the article. In the format without using 1, the terms in the denominator are proportional to the raw group probabilities. With the 1, they are proportional to the odds ratios of each group's probability to the group corresponding to the 1 itself. The 1 is then actually just the odds ratio of a group to itself. Baccyak4H (Yak!) 16:46, 5 November 2008 (UTC)

Statistics question

I seem to have stumbled on a problem. I was wondering what the odds of rolling AT LEAST one (SIX) in seven rolls are. It seems you could add 1/6 (the odds on one roll) seven times to get 7/6, but this is non-sensical. The alternative is to do: 1-(5/6)^7, which makes more sense, but why are the results any different, I'm confused. 169.229.75.140 (talk) 21:37, 5 November 2008 (UTC)

Adding 1/6 seven times does not give the probability of at least one six being rolled, it gives the expected number of sixes rolled. Since it is possible to roll more than 1 six in seven rolls, this is larger than the probability of at least one six being rolled. Algebraist 21:43, 5 November 2008 (UTC)

This is a binomial distribution with 7 trials and probability of a success equal to 1/6. So the probability of rolling at least one six equals, 1-P(X=0) (or 1 minus the probability of rolling 0 sixes) which is just what you wrote.

The reason why the two are different is because even though the probability of getting one six is 1/6, you are not considering the fact that you can't get sixes on the other rolls (because you are aiming for exactly one six out of seven rolls). Now find the probability of precisely two sixes in seven rolls which works the same way (you have to consider that you are getting two sixes, and five non-sixes. By writing just (1/6)^2, you are not considering that you can't get sixes on the other rolls) etc... until you get to seven sixes out of seven. Now sum them up.

Topology Expert (talk) 02:24, 6 November 2008 (UTC)

Adding in that way makes sense only when the events involved are mutually exclusive. The event of getting a 6 the first time, and the event of getting a 6 the second time, and so on, are not mutually exclusive. Michael Hardy (talk) 02:25, 6 November 2008 (UTC)