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I read a paper where the random variable $z$ is assumed to follow a log-normal distribution.

\begin{equation} \mathbb{E}\left(z^{1-\chi}\right)=\int_{0}^{\infty}z^{1-\chi}\frac{1}{z\sqrt{2\pi s^{2}}}e^{\left(-\frac{\left(\text{ln}z-m\right)^{2}}{2s^{2}}\right)}dz \end{equation}

where

\begin{equation} m=-\frac{1}{2}\text{ln}\left(1+\sigma^{2}\right) \end{equation}

(because $\mu_z=1$)

and

\begin{equation} s^2=\text{ln}\left(1+\sigma^{2}\right) \end{equation} and

The paper directly gives the result

$$\mathbb{E}\left(z^{1-\chi}\right)=\left(1+\sigma^{2}\right)^{\frac{\chi\left(1-\chi\right)}{2}}$$

But I cannot see how this result is derived. What I have tried:

I reformulate the integral by making a transformation of the variable $\text{ln}(z)=q$ as follows

$$\mathbb{E}\left(z^{1-\chi}\right)=\int_{0}^{\infty}e^{-\chi q}\frac{1}{\sqrt{2\pi s^{2}}}e^{\left(-\frac{\left(q-m\right)^{2}}{2s^{2}}\right)}dq$$

and this gives I think

$$\frac{1}{2}\left(1+\sigma^{2}\right)^{\frac{1}{2}\left(1+\chi\right)\chi}erf\left(\frac{2q+\left(1+2\chi\right)\text{ln}\left(1+\sigma^{2}\right)}{2\sqrt{2}\sqrt{\text{ln}\left(1+\sigma^{2}\right)}}\right)$$

But I am stuck at this stage.

optimal control
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  • it's expectation of a power of lognormal variable, the expression is well known, see https://en.wikipedia.org/wiki/Log-normal_distribution#Multiple,_reciprocal,_power – Aksakal Nov 19 '23 at 14:11
  • You miscalculated. Just express the integrand as a constant times $\exp(Q(q))$ where $Q$ is a quadratic function and complete the square to obtain the result. Logarithms will not be involved. Alternatively, simply recognize this integral is the moment generating function. – whuber Nov 19 '23 at 15:52

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