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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...

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  • $\begingroup$ What you need is contained here: en.wikipedia.org/wiki/List_of_integrals_of_Gaussian_functions $\endgroup$ – kjetil b halvorsen May 21 '16 at 17:37
  • $\begingroup$ 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). $\endgroup$ – G. Ander May 21 '16 at 17:48
  • $\begingroup$ Oops, I see now it is'nt there in that list ... $\endgroup$ – kjetil b halvorsen May 21 '16 at 19:29
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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)

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    $\begingroup$ Thanks à lot for the book (and answer of course)!I have a complicated dynamic program in which i am doing value function iteration. Part of the computation require to compute this expectation as quickly as possible. That s why i would like something allowing me to go faster than numeric intégration. $\endgroup$ – G. Ander May 21 '16 at 21:35
  • $\begingroup$ It must be possible to do something better than just numerical, loking into it now. $\endgroup$ – kjetil b halvorsen May 21 '16 at 21:58
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    $\begingroup$ 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. $\endgroup$ – G. Ander May 22 '16 at 9:23

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