This seems to be the underlying assumption that allows us to use the standard market model/Black's framework in order to value interest rate derivatives, but I haven't found any understandable explanation that explains why this is an assumption that can be made. Interest rates themselves don't follow a Geometric Brownian motion, which I think is implied by a log-normal distribution? So why would forward rates?


2 Answers 2


SHORT STORY: forward Libor rates need not be assumed to be log-normally distributed. For example, they can be assumed to be normally distributed (and indeed, on Bloomberg, Swaption implied vols are quoted both, in terms of normal as well as log-normal models).

The only condition required is that the forward Libor rate process needs to be a martingale under the T-forward measure (I show this below). As long as the chosen modelling process satisfies the martingality condition, the distribution of this process can (in theory) be whatever we want it to be.


Part 1: Notation:

Denote a forward Libor rate at some time $t$, that sets at time $T_i$ and matures at time $T_{i+1}$, as $L(t, T_i, T_{i+1})$ (note that it only makes sense to discuss this random variable for $t\leq T_i < T_{i+1}$). The annual fraction over which this Libor compounds is $\tau$.

To make the notation clear, some examples:

$(i) L(t_0,t_0,T_1)$ would be the spot Libor rate maturing at some future time $T_1$ (we can suppose that $T_1=t_0+\tau$)

$(ii) L(t_0,T_1,T_2)$ would be today's value of forward Libor rate setting at time $T_1$ and maturing at time $T_2$ (i.e. this would be today's FRA on that libor)

$(iii) L(t,T_1,T_2)$ would be a future value at time "$t$" of the same forward Libor as in (ii): i.e. this would be a random variable, value of which is unknown today.

Part 2: Mechanics of Libor rates:

Suppose that you can lend and borrow at these Libor rates freely: that is to say that you can agree today at time "$t_0$" to borrow or lend any amount "$x$" at time $T_i$ and you'll then have to repay (or you'll receive) an amount $x*(1+\tau L(t_0, T_i, T_{i+1}))$ at time $T_{i+1}$.

Suppose you can do this for amount $x=\frac{1}{1+\tau L(t_0, T_i, T_{i+1})}$. Then, at time $T_{i+1}$, you'd have to repay (or you'd receive) exactly 1 unit of currency: in other words, you can effectively trade (forward) zero coupon bonds that pay 1 unit of currency at some specific maturity. Suppose you can do this at any time, not just today at "$t_0$", but at any time "$t$".

Suppose that at time "$t$" you want to trade some (spot) zero coupon bonds of various maturities. Denote a zero coupon bond that matures at time $T_i$ as $P(t,T_i):=\frac{1}{1+\tau L(t, t, T_i)}$. Denote another zero coupon bond that matures at time $T_{i+1}$ as $P(t,T_{i+1}):=\frac{1}{1+\tau L(t, t, T_{i+1})}$. Note that:

$$\frac{P(t,T_i)}{P(t,T_{i+1})}=1+\tau L(t, T_i, T_{i+1})$$

I.e. the above says that we can express a forward Libor rate as a ratio of two spot zero coupon bonds (these bonds, of course, we have constructed from spot Libor rates at time $t$: so basically, we are saying that we can construct a forward Libor rate from spot Libor rates: no big deal, really).

Part 3: Martingale condition

Re-arrange the above equation as:

$$\frac{P(t,T_i)}{P(t,T_{i+1})}-1=\tau L(t, T_i, T_{i+1})$$

$$\frac{P(t,T_i)-P(t,T_{i+1})}{P(t,T_{i+1})}=\tau L(t, T_i, T_{i+1})$$

$$\left(P(t,T_i)-P(t,T_{i+1})\right)\frac{1}{\tau}=P(t,T_{i+1}) L(t, T_i, T_{i+1})$$

Now, the LHS of the above is a linear combination of traded and liquid securities (as per our assumptions): therefore, the LHS has to be a martingale under a suitable Numeraire as per the Fundamental Theorem of Asset Pricing. Choosing $P(t,T_{i+1})$ as Numeraire, we get:

$$\mathbb{E}\left[\frac{\left(P(t,T_i)-P(t,T_{i+1})\right)\frac{1}{\tau}}{P(t,T_{i+1})} | \mathcal{F}_{t_0}\right]=\frac{\left(P(t_0,T_i)-P(t_0,T_{i+1})\right)\frac{1}{\tau}}{P(t_0,T_{i+1})}$$

But $$\left(P(t,T_i)-P(t,T_{i+1})\right)\frac{1}{\tau}=P(t,T_{i+1}) L(t, T_i, T_{i+1})$$

So we get:

$$\mathbb{E}\left[\frac{P(t,T_{i+1}) L(t, T_i, T_{i+1})}{P(t,T_{i+1})} | \mathcal{F}_{t_0}\right]=\mathbb{E}\left[L(t, T_i, T_{i+1})| \mathcal{F}_{t_0}\right]=L(t_0, T_i, T_{i+1})$$

Therefore, under the $T_{i+1}$ forward measure, associated with the bond $P(t,T_{i+1})$ as numeraire, the forward Libor $L(t, T_i, T_{i+1})$ must be a martingale.

Part 4: Lognormality or Normality???

The martingale condition above does not tell us anything about the distribution of the forward Libor, except for the fact that whatever process we chose, it must be a martingale under the forward measure. Indeed, with many rates being in the negative territory, Normal models have become acceptable as well as log-normal models. On Bloomberg, Swaptions implied vols are quoted in terms of normal as well as log-normal models.


A lognormal distribution has three valuable properties (I) It ensures that the rate is only allowed to be positive; (II) the changes in the interest rate are proportional to the interest rate; and (III) the option price is analytically solvable.

BTW, just to be precise, note that in Black's model, it is an assumption that the distribution of the interest rate in the forward measure is lognormal.

In the past, (I) was considered to be essential as negative interest rates were seen as anomalous and so unlikely. How times have changed ! It was also believed that changes in higher levels of interest rates were roughly proportional to their own level in accordance with (II), although this was no longer so true at low rates. And (III) is very useful as an analytically tractable option price is faster to calculate and more numerically smooth than one computed by a tree or Monte Carlo. So for all these reasons Black's model became the industry standard for European-style interest rate options (caps/floors and European swaptions). However this does NOT mean that it is used by banks for their internal risk and pricing models.

Currently, other models such as SABR, Bachelier (normal), LMM and shifted variations of all of these models are used by banks for their internal pricing and hedging.

Black's model is mainly used for volatility quotations. In this way, Black's model is simply a translator between a volatility quote and the cap/floor price. The actual model used by the bank to hedge it must be calibrated to reprice these option prices. In doing this, we are moving beyond using Black's model as a model of interest rates and to it being used as a quotation tool which could encompass any model.

For example, we know that interest rates tend to mean-revert. Black's model does not take this into account explicitly. However Black volatilities are quoted for different expiry dates (for caps and floors) and this market term structure of volatilities embeds within it expectations of mean reversion. Also, black vols are quoted for different cap/floor strikes and this also adjusts for the shape of the true distribution and allows it to be non-lognormal.

Lastly, by definition, Black's model will exactly refit the market prices of options at the quoted strikes and maturities. It may also be used by less sophisticated users to get a quick and dirty approximate value of option greeks - as the quotation means that the price will be right, it is hoped that the delta will not be too wrong.


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