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

= January 14 =

Value at risk
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)


 * 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)

Comment requested on a power series solution
Talk:Trigonometric functions. 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)

Normal distribution question
I'm trying to work out $$E(X|X>x)$$ where $$X\sim N(0,1)$$. I've obtained $$E(X|X>x) = \int^\infty_x dy\, \frac{y \exp(-y^2/2)}{2 \pi}=\frac{\exp(-x^2/2)}{2 \pi}$$, 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 $$x$$, which has to be nonsense. Where am I going wrong?--Leon (talk) 11:49, 14 January 2015 (UTC)
 * You are considering a variable, X, with probability distribution P(X)=0 for Xx. Choose k such that the total probability is one. Bo Jacoby (talk) 14:01, 14 January 2015 (UTC).
 * Also, should be $$\sqrt{2\pi}$$ rather than $$2\pi$$ (doesn't matter much since it cancels out in normalization). So what you're looking for is
 * $$\frac{\int^\infty_x \frac{y \exp(-y^2/2)}{\sqrt{2\pi}}dy}{\int^\infty_x \frac{\exp(-y^2/2)}{\sqrt{2\pi}}dy}=

\frac{\int^\infty_x y \exp(-y^2/2)dy}{\int^\infty_x \exp(-y^2/2)dy}$$
 * -- Meni Rosenfeld (talk) 17:03, 14 January 2015 (UTC)
 * Also the value does depend on the sign of x. If x is less than 0 you're including the center the distribution whereas if it is greater you're just including a tail.
 * That was the OP's point - the final result should have depended on the sign of x. But because he didn't normalize, he got a result which does not - the integrand is odd, so the integral over a symmetric area around the center is 0. -- Meni Rosenfeld (talk) 21:31, 14 January 2015 (UTC)
 * Why do you have that extra $$y$$ in the integrand?--80.109.80.31 (talk) 19:44, 14 January 2015 (UTC)
 * Because we're calculating the mean. -- Meni Rosenfeld (talk) 21:31, 14 January 2015 (UTC)