Recent content by jjhyun90

I am trying to reproduce results of a paper. The model is: dS = (v-y-\lambda_1)Sdt + \sigma_1Sdz_1 \\ dy = (-\kappa y - \lambda_2)dt + \sigma_2 dz_2 \\ dv = a((\bar{v}-v)-\lambda_3)dt + \sigma_3 dz_3 \\ dz_1dz_2 = \rho_{12}dt \\ dz_1dz_3 = \rho_{13}dt \\ dz_2dz_3 = \rho_{23}dt \\...

The condition I posted on #5 is what I desire to have, rather than what will follow from the assumption. Sorry for the confusion. By the assumption that X is symmetric around 0 and thus its expected value is 0, it follows g'(0)=0, so it is necessary to use l'Hopital's rule again. Thus it seems...

I must be confused. If I'm not mistaken, the first evaluates to -\lambda g'(\lambda t) and the second -g'(t).

Using l'Hopital's rule and chain rule, the equivalent condition I need will be \lim_{t\to0}\frac{g'(\lambda t)}{g'(t)} = \lambda .

g(0)=1 since it is a characteristic function.

Homework Statement The problem is to find sufficient and preferably also necessary conditions on random variable X such that its characteristic function g(x) satisfies the limit property: \lim_{t\to0}\frac{1-g(\lambda t)}{1-g(t)}=\lambda^2 I may assume X is symmetric around 0, so the...

Thank you for the reply, but the e^t on the denominator is multiplied to the first term only, so the simplification doesn't quite work.

Homework Statement A function g is \alpha-regularly varying around zero if for all \lambda > 0, \lim_{x\to 0} \frac{g(\lambda x)}{g(x)}=\lambda^{\alpha} For real s and \alpha \in (0,1), define f: f(s)=1-\alpha \int_{0}^{\infty} e^{\alpha t}...