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As we all know, all affine term-structure models are members of HJM model. Under HJM model, there is a unique risk-neutral measure in both forward-rate process and bond evolving process. Hence, the model is complete. However, there is no unique risk-neutral measure in short rate models like Vasicek, CIR model (the measure is adjusted by the parameter lambda). Thus, the model is incomplete.

The question is: How to justify the existence and absence of a unique risk-neutral measure in forward rate models (HJM) and short-rate models (Vasicek) respectively? Are there any contradictions?

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The Vasicek and other short rate models are only "incomplete" until they are calibrated to market data. If rates actually followed Vasicek processes, it would be trivial to estimate the "Real world" parameters from historical data and compute the "Risk neutral" parameters from the yield curve. In such a case the HJM and Vasicek models are simply two way of looking at the same thing: there is no contradiction. HJM is a "family" of models in which Vasicek fits nicely.

Of course, it is pretty easy to see that the Vasicek model does not, in fact, fit the empirical yield curve. In this case the HJM model will disagree with the Vasicek model since HJM perfectly fits the initial term structure.

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For any given process for the short rate $\{r_t,, t >0\}$, the price at time $t$ of a zero-coupon bond with maturity $T$, where $t\le T$, is given by \begin{align*} P(t, T) = E\left(e^{-\int_t^T r_s ds}\,\big|\, \mathcal{F}_t\right). \end{align*} Since, for $t\le T$, \begin{align*} \frac{P(t, T)}{e^{\int_0^tr_s ds}} = E\left(e^{-\int_0^T r_s ds}\,\big|\, \mathcal{F}_t\right) \end{align*} is a martingale under the risk-neutral measure, we can assume that the dynamics for $r_t$ is already defined in the risk-neutral measure.

For the forward rate $f(t, T)$, we note that $r_t = f(t, t)$ and \begin{align*} P(t, T) = e^{-\int_t^T f(t, u)du}. \tag{1} \end{align*} We assume that $f(t, T)$ follows, under the risk-neutral measure, the HJM model, that is, \begin{align*} df(t, T) = \alpha(t, T) dt + \sigma(t, T) dW_t, \end{align*} where $\{W_t, \, t \ge 0\}$ is a standard Brownian motion. From $(1)$, \begin{align*} d\ln P(t, T) &= f(t, t) dt -\int_t^T df(t, u) du\\ &=r_t dt - \left(\int_t^T \alpha(t, u) du\right)dt - \left(\int_t^T \sigma(t, u) du\right)dW_t. \end{align*} Then \begin{align*} \frac{dP(t, T)}{P(t, T)} &= \frac{1}{P(t, T)}d\left(e^{\ln P(t, T)} \right)\\ &=\frac{1}{P(t, T)}\left(e^{\ln P(t, T)} d\ln P(t, T) + \frac{1}{2}e^{\ln P(t, T)} d\langle \ln P, \ln P\rangle_t\right)\\ &=\left(r_t - \int_t^T \alpha(t, u) du +\frac{1}{2}\left(\int_t^T \sigma(t, u) du\right)^2 \right)dt - \left(\int_t^T \sigma(t, u) du\right)dW_t. \end{align*} Note that, under the risk-neutral measure, the drift term of $dP(t, T)$ is $r_t$. That is, \begin{align*} \int_t^T \alpha(t, u) du = \frac{1}{2}\left(\int_t^T \sigma(t, u) du\right)^2. \end{align*} Consequently, \begin{align*} \alpha(t, T) = \sigma(t, T)\int_t^T \sigma(t, u) du. \end{align*}

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