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April 18[edit]

Hyperbola[edit]

I'd like to know the difference between the hyperbola y=a/x and x^2/a^2-y^2/b^2=1. I can't find anything about how these equations are related if they are related. — Preceding unsigned comment added by KostasGTAV (talk • contribs) 13:01, 18 April 2013 (UTC)[reply]

y = a/x is a rectangular hyperbola, meaning it has asymptotes that are perpendicular and an eccentricity of sqrt(2). x²/a² - y²/b² = 1 is a general hyperbola with asymptotes y = ±bx/a and eccentricity given by b² = a²(e² - 1). If e = sqrt(2) and hence b² = a² you get x² - y² = a², also a rectangular hyperbola, but with asymptotes rotated 45° from the asymptotes of y = a/x. EdChem (talk) 13:12, 18 April 2013 (UTC)[reply]

Expected value problem[edit]

Suppose in a quiz I have ten photos of different people to match with a list of ten people's names. If I fill out the answer sheet randomly, using each person's name once, what is the expected number that I will get right?

I'm pretty sure the answer is 1 for all n (where n=10 in the statement of the problem above). I have proved this for n=2,3,4 by listing all the possibilities. I would like to see a proof for all n, assuming my hunch is correct, and the more simple and elegant the better. Darkhorse06 (talk) 22:15, 18 April 2013 (UTC)[reply]

There's an inductive proof at http://www.askmehelpdesk.com/advice/t-598182.html 86.128.43.105 (talk) 23:41, 18 April 2013 (UTC)[reply]

See rencontres numbers. Bo Jacoby (talk) 00:40, 19 April 2013 (UTC).[reply]

I think those are both overdoing it. Clearly the probability of getting any one photo correct is 1/n, so by linearity of expectation the total expected value is 1.--80.109.106.49 (talk) 01:04, 19 April 2013 (UTC)[reply]

No, it's not as simple as that, 80.109.106.49. You are assuming the probabilities are independent, but they are not. Your method would apply for example to the expected number of times that you would roll a six if you rolled a dice 6 times. But this situation is more complicated. Darkhorse06 (talk) 10:42, 19 April 2013 (UTC)[reply]

Linearity of expectation doesn't require independence. Expectation is an integral; integrals are linear.--80.109.106.49 (talk) 10:47, 19 April 2013 (UTC)[reply]

Okay, now I understand. You are right that linearity of expectation does not require independence, and in this case that is very handy. Here is a fuller explanation:

Let X(i)=number of correct answers to question i (0 or 1). Let X=number of correct answers to all questions=sum of X(i)

Then E(X(i))=0 × (n-1)/n + 1 × 1/n = 1/n

And by linearity of expectation E(X) = E(sum of X(i) over n) = [sum of E(X(i)) over n] = 1.

Nice.

Darkhorse06 (talk) 13:37, 19 April 2013 (UTC)[reply]