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Let $X$ be a lognormal random variable with pdf $f(X)$ and cdf $F(X)$. The mean and variance of $X$ are assumed to be $\mu$ and $\sigma^2$ respectively. If we assume any constant $k \ge 0$ and $0 < \alpha < 1$, how can we calculate following partial expectation?

$$\int_k^\infty x^{1-\alpha} f(x) dx$$

kang taesu
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  • Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. – José Carlos Santos Feb 19 '18 at 18:14
  • I know that $X^{1-\alpha}$ is also lognormal and its mean and variance are $\mu (1-\alpha)$ and $(\sigma (1-\alpha))^2$. – kang taesu Feb 19 '18 at 18:54
  • And partial expecation of lognormal can be written as a function of standard normal cdf $\Phi(\cdot)$ as follows: $$\int_k^\infty x f(x) dx = e^{\mu + \frac{1}{2} \sigma^2} \Phi \left( \frac{\mu + \sigma^2 - \ln k}{\sigma} \right)$$ – kang taesu Feb 19 '18 at 18:57
  • Do you mean that the mean and variance of $\ln X$ are assumed to be $\mu$ and $\sigma^2$ respectively? – Gordon Feb 22 '18 at 15:34

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We assume that the the mean and variance of $\ln X$ are $\mu$ and $\sigma^2$ respectively. That is, $X=e^{\mu+\sigma\xi}$, where $\xi \sim N(0, 1)$ is a standard normal random variable. Then \begin{align*} \int_k^\infty x^{1-\alpha} f(x) dx &= \mathbb{E}\left(\mathbb{I}_{X\ge k} X^{1-\alpha} \right)\\ &=e^{(1-\alpha)\mu}\mathbb{E}\left(\mathbb{I}_{\xi\ge \frac{\ln k-\mu}{\sigma}} e^{(1-\alpha)\sigma \xi} \right)\\ &=e^{(1-\alpha)\mu}\int_{\frac{\ln k-\mu}{\sigma}}^{\infty}\frac{1}{\sqrt{2\pi}}e^{(1-\alpha)\sigma x} e^{-\frac{x^2}{2}}dx\\ &=e^{(1-\alpha)\mu}\int_{\frac{\ln k-\mu}{\sigma}}^{\infty}\frac{1}{\sqrt{2\pi}}e^{-\frac{(x-(1-\alpha)\sigma)^2}{2}+ \frac{(1-\alpha)^2\sigma^2}{2}}dx\\ &=e^{(1-\alpha)\mu + \frac{(1-\alpha)^2\sigma^2}{2}}\Phi\left(\frac{-\ln k +\mu + (1-\alpha)\sigma^2}{\sigma} \right), \end{align*} where $\Phi$ is the cumulative distribution function of a standard normal random variable.

Gordon
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