Jump to content

Wikipedia:Reference desk/Archives/Mathematics/2015 January 14

From Wikipedia, the free encyclopedia
Mathematics desk
< January 13 << Dec | January | Feb >> January 15 >
Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.


January 14

[edit]

why does VaR (U2)=VaR2(U) when U is a standard uniformly distributed variable? please explain. Thank you! — Preceding unsigned comment added by 134.184.120.210 (talk) 03:42, 14 January 2015 (UTC)[reply]

I'm not quite sure I understand your question. As you will see from our article on Value at Risk, VaR is calculated for an investment or portfolio of investments, and not for a distribution. Is it possible that VaR here means something other than "Value at Risk"? I suspect this is a homework question; if so, please would you supply some additional context. Thanks. RomanSpa (talk) 18:28, 14 January 2015 (UTC)[reply]

Comment requested on a power series solution

[edit]

Talk:Trigonometric functions#Check my work. To me it isn't the most elegant way to get that series, particularly since the time computing each term grows with k2.--Jasper Deng (talk) 06:13, 14 January 2015 (UTC)[reply]

Normal distribution question

[edit]

I'm trying to work out where . I've obtained , but am convinced it's wrong. The reason I think that it's wrong is that the expectation value I'm trying to find does not depend on the sign of , which has to be nonsense. Where am I going wrong?--Leon (talk) 11:49, 14 January 2015 (UTC)[reply]

You are considering a variable, X, with probability distribution P(X)=0 for X<x and P(X)=k⋅exp(-X^2/2) for X>x. Choose k such that the total probability is one. Bo Jacoby (talk) 14:01, 14 January 2015 (UTC).[reply]
Also, should be rather than (doesn't matter much since it cancels out in normalization). So what you're looking for is
-- Meni Rosenfeld (talk) 17:03, 14 January 2015 (UTC)[reply]
Also the value does depend on the sign of x. If x is less than 0 you're including the center the distribution whereas if it is greater you're just including a tail.
That was the OP's point - the final result should have depended on the sign of x. But because he didn't normalize, he got a result which does not - the integrand is odd, so the integral over a symmetric area around the center is 0. -- Meni Rosenfeld (talk) 21:31, 14 January 2015 (UTC)[reply]
Why do you have that extra in the integrand?--80.109.80.31 (talk) 19:44, 14 January 2015 (UTC)[reply]
Because we're calculating the mean. -- Meni Rosenfeld (talk) 21:31, 14 January 2015 (UTC)[reply]