Recently I study some interest rate models.

When I moved on to forward rate models, I see this documents


It said "HJM-type models capture the full dynamics of the entire forward rate curve, while the short-rate models only capture the dynamics of a point on the curve"

What I don't understand is that why short rate model can't not capture the full dynamics of curve? It seems that the only difference in the model is the short rate 'r(t,T)' is substituted for instantaneous forward rate 'f(t,s)'

I mean, the short rate model : $dr(t,s) = μ(t,s)dt + σ(t,s)dW_t$

and the HJM Framework : $df(t,s) = μ(t,s)dt + σ(t,s)dW_t$

So I think the short rate model can capture the dynamics of full term structure. Also any no arbitrage short rate model could make current term structure.

I tried to understand meaning of the above bold&italic sentences with the fixed income securities textbook such as Tuckman, Veronesi.. But it fails.

What I'm misunderstanding now?


1 Answer 1


Most principal component analyses (PCAs) on historical data of yield curves find that typically a yield curve

  • moves parallel

  • flips from normal to inverse (or vice versa)

  • twists (changes its curvature)

As in every PCA the drivers of these movements are uncorrelated standard normals of one dimension each.

Mathematically you could model this by a HJM model that is driven by three Brownian motions.

  • I do not believe that every HJM model captures the full dynamics of the entire forward curve. If that were true even the poorest HJM model (namely the one you get when you start with the Ho-Lee model) could capture the full dynamics.

  • What is true is that a HJM model by definition captures the shape of the current yield curve because that is just the curve $T\mapsto f(0,T)$ which is part of the model.

  • Short rate models are typically Markovian (at least those that are mostly used such as Vasicek, CIR, Black-Karasinski, etc.). This Markov property means that every conditional zero bond price $$ P(t,T)=\textstyle\mathbb E\Big[\exp\Big(-\int_t^Tr(s)\,ds\Big)\Big|{\cal F}_t\Big] $$ is a deterministic function of the single variable $r(t)$: $$ P(t,T,r(t))\,. $$ In turn, this means that the yield curve $$ Y(t,T)=-\log P(t,T)/(T-t) $$ or its continuously compounded sister $$ f(t,T)=-\frac{\partial}{\partial T}\log P(t,T) $$ must also be a deterministic function of a single variable $r(t)$.

This precludes that the dynamics of the yield curve in a Markovian short rate model can perform uncorrelated parallel shifts, flips or twists.

  • $\begingroup$ Almost understanding now.. However, what am I confused is that sentence, "because that is just the curve $T↦f(0,T)$ which is part of the model." I think not $f(0,T)$(term forward rate) but the instantaneous forward rate is the part of model. (I mean $df(t,s)$ equation) because the dynamics is defined with instantaneous forward rate. So I have now in trouble to understand your sentence. Could you please add more explanation for that sentence? $\endgroup$ Apr 20, 2022 at 0:11
  • $\begingroup$ To be precise. HJM models -as you say- the instantaneous forward rate $f(t,s)$ as a stochastic process in $t$. The starting value is $f(0,s)$ and the HJM equation in integral form is $f(t,s)=f(0,s)+\int_0^t\mu(u,s)\,du+\int_0^t\sigma(u,s)\,dW_u$. In that sense $s\mapsto f(0,s)$ is part of the model. In fact, this approach was the whole starting point of HJM. $\endgroup$
    – Kurt G.
    Apr 20, 2022 at 8:26
  • $\begingroup$ Wonderful explanation. Thank you very much. Then you mean that typical short rate model cannot capture the shape of the current yield curve. Because typical short rate model(not in HJM Framework), for example Ho-Lee, doesn't have extra independent variable. I mean the dynamics not described like $dr(t,s) = ~$ but $dr(t) = ~$. So the starting value is $r(0)$. And it couldn't consider any 'term' stuff because it is short of buffer variable to deal with it. Am I right? $\endgroup$ Apr 20, 2022 at 8:51
  • $\begingroup$ Not exactly. In a short rate model it is a bit harder to show that it can (when properly defined) capture the current yield curve. This was what made CIR famous. They found out that one just has to make the mean reversion level $\theta(t)$ time dependent. This works for Vasicek and Black-Karasinsiki also. $\endgroup$
    – Kurt G.
    Apr 20, 2022 at 9:58

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