Deriving the Volatility Condition for One-Factor Markovian Forward Rates

# The converse to Cheyette's Ansatz


# The converse to Cheyette's Ansatz












It is well known that if the HJM volatility function $\sigma(t,T) = g(t)h(T)$ where $g(t)$ is a random process then the short rate is Markovian on two state variables. One can show that this separability condition holds if and only if

> $$\tag{RS}\frac{\sigma(u,T)}{\int_0^t \sigma(s,T) ds} = \frac{\sigma(u,t)}{\int_0^t \sigma(s,t) ds} $$ for all $u\in [0,t]$.

Ritchken and Sankarasubramanian actually state a converse in Volatility Structures of Forward Rates and the Dynamics of the Term Structure where they prove the claim above. However, they prove their result in a way that I do not understand:

> This path dependence can be captured by a single statistic common across all maturities, $T$, without imposing any additional restrictions on the initial term structure, $f(0, T)$, or on the structure for the market price of risk, $λ(t)$, providing a common “weighting scheme” exists for forward rates of all maturities. However, if a unique weighting scheme is to exists for all forward rates, it must be the case that the weighting function is independent of $T$. This in turn implies that (RS) holds.

Does anyone understand how to prove condition (RS) holds if the prices of all interest contingent claims at time $t$ are completely determined by a two-state-variable Markov process? The way I am interpreting this last claim is by saying that $P(t,T) = p(t,T,x(t),y(t))$ for some Markov process $(x(t),y(t))$ where $x(t) = r(t) - f(0,t)$, but I was not able to progress much further.

I recall reading a book or paper where condition (RS) is explained intuitively, but I cannot remember where this was done. So, a proof along of the lines of a more detailed justification of condition (RS) would also be very appreciated.

Add. I have been trying to advance of this, and this is what I have got so far.

> The condition (RS) is equivalent to: for all $t\leqslant T$ and for all $u\leqslant t$, we have that: $$ \mathrm{Corr}(\sigma(u,T), \sigma(u,t)) = 1. $$

In other words, the parametric family of processes $T\mapsto \sigma(-,T)$ is perfectly correlated across all times. This of course makes sense. Now writing

$$ x(t) = \int_0^t \sigma(s,t) \Sigma(s,t)ds + \int_0^t \sigma(s,t)dW_s $$

suggests that $\int_0^t \sigma(s,T)dW_s$ will play an important role (like the case of Carverhill when $\sigma(t,T)$ is deterministic) and indeed the function $\Lambda(t,T) = \int_0^t \sigma(u,T)du$ is what the authors say is ''independent of $T$'', I think.

So I suspect one has to use some kind of Itō isometry (maybe conditional) to say something about $\Lambda(t,T)$ that implies (RS) when $(x(t), y(t))$ is Markov where $y(t) = \int_0^t \eta(s)^2 ds$ where $\eta(t) = \sigma(t,t)$.

## Answer by Pedro (score 2, accepted)

https://quant.stackexchange.com/a/81743

Here is the answer, adapted from Modern computational finance of A. Savine. First, the most accurate statement is that the (a priori infinite dimensional Markovian process) $f(t,T)$ is one dimensional if and only if such decomposition exists. One direction is given by the Cheyette computation, and for the converse, we proceed as follows.

First, even when $\sigma(t,T)$ is random, the drift of $f(t,T)$ is deterministic (its risk neutral expectation), and thus we can consider the process $\Lambda(t,T) = \int_0^t \sigma(s,t)dW(s)$ driving the forward rate curve process. The model is one dimensional Markovian if and only if each of these is a function of the other. Since they are centered Gaussian processes, they must be linear multiples of each other: $$ \Lambda(t,T_1) = \alpha(t,T_1,T_2) \Lambda(t,T_2). $$ Computing the derivative of this equation with respect to $t$ gives us that $$ \left(\sigma(t,T_1) - \alpha(t,T_1,T_2) \sigma(t,T_2)\right)dW(t) = \alpha_t(t,T_1,T_2) \Lambda(t,T_2) dt. $$ By computing the quadratic variation, we see that $\sigma(t,T_1) = \alpha(t,T_1,T_2) \sigma(t,T_2)$, and in particular that $\alpha$ is independent of $t$. Thus, we may first change $t$ to any $u \leqslant t$, and then insert this equation into the above, set $T_2=t$, which gives $$ \frac{\sigma(u,t)}{\Lambda(t,t)}= \frac{\sigma(u,T_1) }{\Lambda(t,T_1)} $$ and the theorem is proved by the condition (RS) stated in the post.