How to take this exponentials

Question

Given an expansion of a cumulant function as follows:

$$ \kappa(t) = \frac{t^2}{2} + \frac{\rho_3 t^3}{6\sqrt{n}} + \frac{\rho_4t^4}{24n} +O\left(\frac{1}{n\sqrt{n}}\right), (*) $$

where $\rho_r=\frac{\kappa_r}{\sigma^r}$ is the standardized cumulant. To obtain the corresponding moment generating function, one only needs to take the exponential of the above formula.

$$ m(t) = e^{\kappa(t)} = \exp\left\{{\frac{t^2}{2} + \frac{\rho_3 t^3}{6\sqrt{n}} + \frac{\rho_4t^4}{24n} +O\left(\frac{1}{n\sqrt{n}}\right)}\right\} = \exp\left\{ \frac{t^2}{2}\left[ 1+\frac{\rho_3t}{3\sqrt{n}}+\frac{\rho_4t^2}{12n} + O\left(\frac{1}{n\sqrt{n}}\right) \right] \right\}. $$

However, I read the following result:

$$ m(t) = \exp\left(\frac{t^2}{2}\right) \left\{ 1+\frac{\rho_3 t^3}{6\sqrt{n}}+\frac{\rho_4t^4}{24n}+\frac{\rho_3^2t^6}{72n}+O\left(\frac{1}{n\sqrt{n}}\right) \right\}. (**) $$

I do not see how to get $(**)$ from $(*)$. What trick is used here, please? Thank you!

Answer 1

Try this with $a$ the first term and $\Sigma$ the other terms in the exponent:

$$\exp(a+\Sigma)=\exp(a)\exp(\Sigma)=\exp(a)\big(1+\Sigma+\frac{1}{2}\Sigma^2\cdots\big).$$

Answer 2

You wrote $$m(t) = \exp\left\{ \frac{t^2}{2}\left[ 1+\frac{\rho_3t}{3\sqrt{n}}+\frac{\rho_4t^2}{12n} + O\left(\frac{1}{n\sqrt{n}}\right) \right] \right\}$$ Let us expand it first; so $$m(t)=\exp \Big(\frac{t^2}{2}+\frac{\rho_3t^3}{6\sqrt n}+\frac{\rho_4t^4}{24 n}+\cdots \Big)=\exp \Big(\frac{t^2}{2}\Big)\times\exp \Big(\frac{\rho_3t^3}{6\sqrt n}+\frac{\rho_4t^4}{24 n}+\cdots \Big)$$ Now us Taylor series for the second part $$e^y=1+y+\frac{y^2}{2}+O\left(y^3\right)$$ and make $$y=\frac{\rho_3t^3}{6\sqrt n}+\frac{\rho_4t^4}{24 n}$$

I am sure that you can take from here.