# Roko’s Shortform

• One weird trick for es­ti­mat­ing the ex­pec­ta­tion of Log­nor­mally dis­tributed ran­dom vari­ables:

If you have a vari­able X that you think is some­where be­tween 1 and 100 and is Log­nor­mally dis­tributed, you can model it as be­ing a ran­dom vari­able with dis­tri­bu­tion ~ Log­nor­mal(1,1) - that is, the log­a­r­ithm has a dis­tri­bu­tion ~ Nor­mal(1,1).

What is the ex­pec­ta­tion of X?

Naively, you might say that since the ex­pec­ta­tion of log(X) is 1, the ex­pec­ta­tion of X is 10^1, or 10. That makes sense, 10 is at the mid­point of 1 and 100 on a log scale.

This is wrong though. The chances of larger val­ues dom­i­nate the ex­pec­ta­tion or av­er­age of X.

But how can you es­ti­mate that cor­rec­tion? It turns out that the rule you need is 10^(1 + 1.15*1^2) ≈ 141.

In gen­eral, if X ~ Log­nor­mal(a, b) where we are work­ing to base 10 rather than base e, this is the rule you need:

E(X) = 10^(a + 1.15*b^2)

The 1.15 is ac­tu­ally ln(10)/​2.

For a product of sev­eral in­de­pen­dent log­nor­mals, you can just mul­ti­ply these to­gether, which means adding in the ex­po­nent. If you have 2 or 3 things which are all log­nor­mal, the var­i­ance-as­so­ci­ated cor­rec­tions can eas­ily add up to quite a lot.

Re­mem­ber: add 1.15 times the sum of squares of log-var­i­ances!