Expectation of two identical lognormal distributions

Question

I would like to compute the conditional expectation (on an interval from $c$ to $\infty$) of the minimum of two log normal distributions.

Denote $X_1$, $X_2 \sim LN(0, \sigma)$, the associated density $f()$.
$Y = min(X_1, X_2)$. From simple computation you get that $f_y(y)= 2 f(y) (1-F(y))$.

As a consequence, computing the expectation of this process comes down to computing:
$\begin{equation} \mathbb{E}[Y] = \int^\infty_{0} y 2 f(y) (1-F(y)) dy\end{equation}$

I am having trouble to compute the term:
$\int_0^\infty y f(y) F(y) dy$

Is there any math trick? Notice that I don't know how to do it either in the simply 'normal' case.
However I know the results (from numerical computation):
$\mathbb{E}[Y] = 2 \underbrace{exp(\sigma^2/2)}_{=\mathbb{E}[X]} \Phi(\sigma/\sqrt{2})$

I need to know how to derive it because in fact what I need is not to compute $\int_0^\infty y f(y) F(y) dy$ but instead to compute it from $\int_c^\infty y f(y) F(y) dy$. If I understand how the computation is done for the 'simple' case, I should be able to find the other (I hope). If you have answer directly for the final question, it would be even better.

PS: If it helps, in the gaussian case, I know that $\int^\infty_{-\infty} z f(z) F(z) dz = \frac{\sigma}{\sqrt{2}} \phi(\frac{0}{\sqrt{2}\sigma})$. But I don't know how we obtain this either...

Answer 1

First, this is a problem about (expectation of) order statistics from a iid sample from a lognormal distribution. There is a book dedicated to that topic: "Handbook of Tables for Order Statistics from Lognormal Distributions with Applications" by N. Balakrishnan and William W. S. Chen. They resort to numerical integration to find that expectation ... so maybe thats the way to go ... For what do you need the results, do you need a symbolic expression, exact or approximate, or is a numerical answer enough?

Your setup is $X_1, X_2$ independent (you do not say so, but your calculations assume it) as $\text{LN}(\mu, \sigma^2)$. Then $Y=\min(X_1, X_2)$. The density of $Y$ (as you say) $2f(y) (1-F(y))$ where $f,F$ is the lognormal density and cdf of the lognormal (with given parameters). Now the lognormal density (with given parameters) can be given as $$ f(y)=\frac{1}{\sigma y}\phi(\frac{\ln y-\mu}{\sigma}) \\ F(y) = \Phi(\frac{\ln y-\mu}{\sigma}) $$ where $\phi(), \Phi()$ are standard normal density and cdf.

Using this the density of $Y$ becomes $$ 2 f(y) (1-F(y)) = 2 \frac{1}{\sigma y}\phi(\frac{\ln y-\mu}{\sigma}) \Phi(\frac{\mu-\ln y}{\sigma}) $$ and the expectation

$$ \DeclareMathOperator{\E}{\mathbb{E}} \E Y = 2 \int_0^\infty \frac1{\sigma y}\phi(\frac{\ln y-\mu}{\sigma})\Phi(\frac{\mu-\ln y}{\sigma})\; dy $$ which by change of variable becomes $$ \E Y = 2 \int_{-\infty}^\infty e^{\sigma z+\mu} \phi(z) \Phi(-z) \; dz $$ I found a paper studying this problem. It is "Explicit Expressions for Moments of Log Normal Order Statistics" by Saralees Nadarajah (http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.501.1459&rep=rep1&type=pdf). I will try to review the basics from it, showing how to solve the problem above with the methods from the paper. (will come back to do this)