# In a Black-Scholes world, why must volatility be strictly increasing in time-to-expiration?

This question is from Rebonato's Volatility and Correlation 2nd Edition. Rebonato states that if $\sigma_T^2T$ is not strictly increasing, it would be simple to set up an arbitrage. Unfortunately (for me) I can't figure out how to do this. Could someone please point me in the right direction.

That page in the book can be found in section 3.6.1 on Google Books here: http://goo.gl/RC4Wur

-
It's not the instantaneous vol, it's just the accumulated vol. –  athos Jun 25 at 1:30

The author argues, that the Black Option price increases with $\sigma^2_TT$, which means that if this input would decrease with maturity, and one would have two options, the option with longer maturity would be worth less than the option with shorter maturity. But the option with the longer maturity contains the maturity of the shorter option, so it must be worth at least as much. This is a contradiction: You cant have an option worth less (by decreasing $\sigma^2_TT$) in the model if by nature you have that it must be worth more.

More maturity always increases the value of the option, because of its asymmetric payoff you get additional potential gain while your loss over that time is bounded. Any model that does not include this would imply arbitrage.

You may also argue on the case of constant $\sigma^2_TT$: You cannot have an option with longer maturity, that is worth the same. You would then buy the option with longer maturity and sell the one with shorter maturity, to get more "hedge" for the same price. Nobody would buy the option with the shorter maturity, if you can have a longer maturity for exactly same price, so any such option would vanish/cease to exist.

So $\sigma^2_TT$ must be strictly increasing to avoid arbitrage.

--

For a specific arbitrage portfolio, we would need additional assumptions on the dynamics and type of the option. For a European call, if the options with two maturities have same price, we sell the option with shorter maturity and buy the option with longer maturity for it (same price, zero initial value):

At expiration of the first option, if the options are out/at the money, the first option expires worthless, but we can sell the option with the higher remaining maturity immediately to make a profit. If the options are in the money, the first option we sold is exercised, but we are hedged with an option that is worth more than the exercise value (intrinsiv value+time value), so we can again sell the option with the higher maturity to finance the exercise of the sold call and receive in addition the time value premium.

This is therefore a riskfree arbitrage profit, if $\sigma^2_TT$ was constant over time. If $\sigma^2_TT$ was even decreasing, the call with higher maturity would be even cheaper than the one with shorter maturity, so we would make immediate profit at the beginning, and at the end.

Hence the Call value must have $\sigma^2_TT$ strictly increasing.

For puts and American options the proof would go analogous.

-
The points you raise make a lot of intuitive sense. However all of them allude to some sort of "soft" arbitrage. I was hoping someone could create some sort of arbitrage portfolio of the sort typically quoted when talking about option pricing (eg. zero initial value with non-zero future value). Also regarding point: "But the option with the longer maturity contains the maturity of the shorter option, so it must be worth at least as much." I can see how that could apply to American options but Black implies European and the later expiring option cannot subsume the earlier expiring one? –  JohnC Jun 25 at 16:26
Ok I just added an example on European Call arbitrage. –  emcor Jun 25 at 16:42