We have two random variables $X$ and $Y$ which are log normally distributed, with suitable parameters, what is the expected value for $\max(X,Y)$?
Given,
$$ X=e^{\mu+\sigma Z};\quad Y=e^{\nu+\tau Z};\quad Z\sim N(0,1) $$
We need to find an expression for $$E[\text{max}(X,Y)]$$
$X,Y$ are independent drawings.
Please note I have reached the step below, but am unsure how to proceed further.
Steps Tried
\begin{eqnarray*} E\left[\max\left(X,Y\right)\right]=\int_{0}^{\infty}xf_{Y}\left(x\right)F_{X}\left(x\right)dx+\int_{0}^{\infty}yf_{X}\left(y\right)F_{Y}\left(y\right)dy \end{eqnarray*} \begin{eqnarray*} \int_{0}^{\infty}xf_{Y}\left(x\right)F_{X}\left(x\right)dx{\displaystyle =\int_{0}^{\infty}\frac{1}{\tau}\phi\left(\frac{\ln x-\nu}{\tau}\right)}\Phi\left(\frac{\ln x-\mu}{\sigma}\right)dx \end{eqnarray*} \begin{eqnarray*} {\displaystyle =\int_{0}^{\infty}\frac{1}{\tau}\phi\left(\frac{\ln x-\nu}{\tau}\right)}\Phi\left(\frac{\ln x-\mu}{\sigma}\right)dx\quad\text{, Substitution }u=\left(\frac{\ln x-\nu}{\tau}\right)\Rightarrow du=\frac{1}{x\tau}dx \end{eqnarray*} \begin{eqnarray*} {\displaystyle =\int_{-\infty}^{\infty}e^{u\tau+\nu}\phi\left(u\right)}\Phi\left(\frac{u\tau+\nu-\mu}{\sigma}\right)du \end{eqnarray*} \begin{eqnarray*} {\displaystyle =e^{\nu}\int_{-\infty}^{\infty}e^{u\tau}\phi\left(u\right)}\Phi\left(\frac{u\tau+\nu-\mu}{\sigma}\right)du \end{eqnarray*}
Related Question
Please note, this present question was mis-phrased due to my limited knowledge; but still provides an interesting and instructive solution. The more general case has been made into a new question:
If $\sigma = \tau$, then \begin{align*} E(\max(X, Y)) &= E\left(e^{\sigma Z} \max\big(e^{\mu}, e^{\nu}\big) \right)\\ &=e^{\frac{1}{2}\sigma^2}\max\big(e^{\mu}, e^{\nu}\big). \end{align*}
WLOG, we assume that $\sigma < \tau$ below. Let $\lambda = \frac{\mu-\nu}{\tau-\sigma}$. Then, \begin{align*} E(\max(X, Y)) &= E\left(\mathbb{I}_{Z \le \lambda} X + \mathbb{I}_{Z > \lambda} Y\right)\\ &=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\lambda} e^{\mu+\sigma x - \frac{1}{2}x^2} dx +\frac{1}{\sqrt{2\pi}}\int_{\lambda}^{\infty} e^{\nu+\tau x - \frac{1}{2}x^2} dx. \end{align*} The remaining is now routine calculus.