# Expectation of two identical lognormal distributions

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...

• What you need is contained here: en.wikipedia.org/wiki/List_of_integrals_of_Gaussian_functions – kjetil b halvorsen May 21 '16 at 17:37
• My problem would be $\int_c^\infty e^y \phi(y/\sigma) \Phi(y/\sigma) dy$, I don't see anything resembling it in the link sadly. (as mentioned in my post, for $\int_{-\infty}^\infty e^y \phi(y/\sigma) \Phi(y/\sigma) dy$ I know the result (just don t know how to obtain it). But anyway I don t see it on the wikipedia link either). – G. Ander May 21 '16 at 17:48
• Oops, I see now it is'nt there in that list ... – kjetil b halvorsen May 21 '16 at 19:29

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)
• I also had an answer there: math.stackexchange.com/questions/1792792/… It basically comes down to computing $\int_{a}^{\infty} \phi(z-\sigma) \Phi(-z) dz$. Not sure it's possible to find a closed form without having to integrate this. – G. Ander May 22 '16 at 9:23