Your argument for the normal is extremely elegant. I like it a lot.

I have not understood how to work out the lognormal, however. I am really confused. The $\displaystyle X$ and $\displaystyle Y$ you wrote on those lines, do I have to read that $\displaystyle X$ as $\displaystyle X|X \geq Y $ and $\displaystyle Y$ as $\displaystyle X$? Because if it so, then the argument would not work because $\displaystyle X|X \geq Y $ and $\displaystyle X$ have different variances, so you would need a different rescaling for making them independent (but I things might get complicated). I think I am missing something, sorry about that.

I attach an argument for the conditional mean under independence. It is not very nice, sorry. I think the proof is a little longer than what could be.

Attachment 9904
By the way, I could not find any reference on any book about that result, so should you find a reference, I would appreciate if you could let me know.