# Expected Value of Maximum of Two Lognormal Random Variables with One Source of Randomness

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:

Expected Value of Maximum of Two Lognormal Random Variables

• $Z = \max(X,Y)$. if $X,Y$ are independent : $$P(Z \le a) = \int_0^a F_Y(x) F_X'(x) + F_X(x) F_Y'(x) dx = F_Y(a)F_X(a)$$ so that : $$\mathbb{E}[Z] = \int_0^\infty 1-F_Y(x)F_X(x) dx$$ which shouldn't be easy to compute. Commented Jan 12, 2016 at 15:07
• @user1952009 Thanks for you suggestion. I partly understand it. Is there a x missing in the $$E(Z)$$ after the integral sign? Commented Jan 12, 2016 at 15:26
• no, if $U$ has support $[0;b]$ then $\mathbb{E}[U] = \int_0^b x f_U(x) dx = b F(b) - \int_0^b F_U(x) dx = \int_0^b (1-F_U(x))dx$ (with $F_U'(x) = f_U(x)$ and integration by part). this stays true for infinite support random variable so that $\mathbb{E}[Z] = \int_0^\infty (1-F_Z(x))dx$. finally with $F_Z(x) = P(Z \le x) = F_Y(x)F_X(x)$ you get my formula Commented Jan 12, 2016 at 17:22
• @user1952009 Could you please clarify how you get $bF(b)=\int_{0}^{b}1dx$ Commented Jan 13, 2016 at 14:28
• Given the way you have defined the problem, i.e. $$X=e^{\mu+\sigma Z};\quad Y=e^{\nu+\tau Z};\quad Z\sim N(0,1)$$ you only have 1 random variable $Z$, not two. Do you mean that you have two independent drawings of $Z$, because that is not what is described? Commented Jan 14, 2016 at 15:30

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.
• Awesome answer. Much appreciated. Please note, given my limited understanding of these concepts. Is the definition I have used for the two random variables, $X,Y$ the standard way to define two independent log normal random variables. I suppose your answer would change if this were not so. Commented Jan 15, 2016 at 6:58
• You can assume that $X=e^{\mu+\sigma Z_1}$ and $Y=e^{\nu+\tau Z_2}$, where $Z_1$ and $Z_2$ are independent. This will be another question. Commented Jan 15, 2016 at 13:51