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!

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!