[I’m writing this note while sitting at my son’s archery class. This is more of an attempt to use some “new to me” results to reconcile a puzzle I’ve been thinking about for 20 years than it is to write up something perfectly.]

In the mid 1990’s Zvi Bodie published a fascinating study on the price of long dated put options. The specific question he asked was about the cost of put that would be struck at a price that would guarantee an investment would at least achieve a return equal to the risk free rate (assuming 20% volatility). The paper is here:

The answer to the question was surprising -> the cost of the insurance increased with the length of time:

An additional surprise is that the cost of the put increases to the value of the investment – i.e. it costs roughly $100 to insure that a $100 investment will at least grow at the risk free rate for 1,000 years.

The paper caught my attention in the 1990s because the math was so simple and the result seemed completely non-intuitive.

An old (but new to me!) talk by Ole Peters about the difference between time averages and ensemble averages helps explain the strange result:

The solution is the ensemble average growth rate will be equal to the risk free rate, r, but the time average growth rate will be the geometric growth rate . That difference means that as time grows larger the time average growth rate will fall farther and farther below the ensemble average – just as in the example in Peters’s video. Eventually (as time goes to infinity) the difference will be so large that the time average is essentially 0.

Saying that a bit more mathematically, as the expression

if

Thinking of the cost of the put more an insurer would think of an insurance premium also reconciles with the ideas from Peters’s lecture. Assuming a $100 initial investment, the present value of the maximum payout is $100 (since you are insuring risk free growth) . The only way the cost of the put can grow to $100 is that if as time grows you have a near 100% chance of a near 100% loss. Peters’s lecture shows how that is exactly what is happening with the time average.

Anyway, happy to have run across Peters’s work. It has given me a way to understand Bodie’s paper after pondering the odd result for 20 years.