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For any $s \geq t$, note that \begin{align*} r_s = r_t + \sigma\int_t^s dW_u + \int_t^s \theta_u du. \end{align*} Then, \begin{align*} \int_t^T r_s ds &= (T-t)r_t + \sigma\int_t^T\int_t^s dW_u ds + \int_t^T \int_t^s\theta_u du ds\\ &=(T-t)r_t + \sigma\int_t^T\int_u^T ds\, dW_u +\int_t^T\int_u^T\theta_u ds du\\ &=(T-t)r_t + \sigma\int_t^T (T-u)...
Let $$Z_t = \exp(-X_t)$$ with $$X_t = \sigma(T-t)W_t+\sigma\int_0^tW_sds+\int_0^Tf(0,u)du+\int_0^t\int_s^T\alpha(s,u)du ds$$ and $W_t$ a standard Brownian motion, along with the usual assumptions. We can write $X_t=f(t,W_t)$ and apply Itô's lemma to get: $$dX_t = \frac{\partial f}{\partial t}(t,W_t) dt + \frac{\partial f}{\partial W_t} (t,W_t) dW_t + \... 5 This is known as the classical Leibniz rule. The link sends to Wikipedia, where you can find a proof. It allows to differentiate under the integral sign. A general statement of the formula is:$$\text{d}\left(\int_{g(x)}^{h(x)}f(x,s)\text{d}s\right)=h'(x)f(x,h(x))\text{d}x-g'(x)f(x,g(x))\text{d}x+\int_{g(x)}^{h(x)}\text{d}f(x,s)\text{d}s4 Here we provide another answer using Ito's calculus. It appears involved, but it also has its own interest. Given the short rate dynamics \begin{align*} dr_t = \nu(r_t, t) dt + \rho(r_t, t) dW_t, \end{align*} we define the function \begin{align*} g(x, t, T) = -\ln E\left(e^{-\int_t^T r_s ds} \,\big|\, r_t = x\right). \end{align*} The forward rate f(t, T) ... 4 Based on this question, for the Hull-White model of the form \begin{align*} dr_t = (\theta(t)-a r_t) dt + \sigma dW_t, \end{align*} where a and \sigma are constants, a(t) is a deterministic function, and W_t is a standard Brownian motion, the price at time t of a zero-coupon bond with maturity T and unit face value is given by \begin{align*} P(t, ... 4 Keeping it simple, you know HJM SDE gives the dynamics of an instantaneous forward referencing a fixed maturity T, f\left(t, T\right), but there is a continuum of such maturities - the whole forward curve as a function of T. You can have each forward driven by a different brownian for example, so in general the HJM approach will be infinite dimensional. ... 3 OK, your framework on this is right. Long-term yields embed "expectations" about future short rates. That is: what do I think I'll get if I sat for T years with cash in the bank for swaps, or the same rolling Govvie Bills with respect to Bonds? Plus a cherry-on-top called "term premium". One can rationalise the existence of this in lots of intuitive ways, ... 3 You almost get there. However, you ca not conclude that \rho^2 is a constant based on (10). Note that, from your (7) and (8), \begin{align*} \frac{\rho(z_t)^2}{\beta} e^{\beta \tau} (e^{\beta \tau} - 1) = -h'(\tau)+e^{\beta \tau}h'(0). \end{align*} Taking derivative with respect to \tau on both sides, we obtain that \begin{align*} \frac{\rho(... 2 If you have a simple instrument, short rate models capture all the key variance, but they impose structure on the shape of forward volatility curves (and, usually, forward tilt) that is often far from realistic. If you have instruments whose value is sensitive to what might happen with tilt or forward volatility, you need a multifactor model like HJM. ... 2 I am not sure about this specific algorithmic implementation, but I am a bit confused by your indexes and suspect you might be as well (e.g. M not defined, you're showing cases of i looping when it seems you mean j). I think it would be useful to revisit the basics: Let D_t \in (0,1] be the present value factor for a cash flow at time t. By ... 2 This is my first ever answer so please bear with me. Apologies in advance for terrible formatting. Also fyi, you have some typos in your post that may be making things more confusing. The issue here is that \int_{0}^{t}\int_{s}^{T}\sigma(s,u)dudW_s is itself a stochastic process and trying to take its partial derivative with respect to time (which you ... 2 We assume that \begin{align*} d \ln P(t,T) = \mu(t, T) dt - \sigma (t,T) dW(t). \end{align*} Then, \begin{align*} \ln P(t,T) = \ln P(0,T) + \int_0^t \mu(s, T) ds - \int_0^t \sigma (s,T) dW(s). \end{align*} Moreover, \begin{align*} f(t, T) &= -\frac{\partial\ln P(t,T)}{\partial T} \\ &= -\frac{\partial\ln P(0,T)}{\partial T} - \int_0^t \frac{\partial\... 2 In the real world, many people believe that the yield of a government bond consists of an expectations part plus a term premium. However financial models such as HJM are built in a “risk -neutral” probability measure, with respect to which every asset is the expectation of future cash flows without regard to any term premium. This is done to ensure the ... 2 It is just an application of the Leibniz integral rule, written in differential form. Please see here: https://en.m.wikipedia.org/wiki/Leibniz_integral_rule Capital T is constant, t is changing, so the second term on the right hand side is the exchange of integral and differential, the first term on the right hand side is the function value at the lower ... 1 Note that \begin{align*} f(t, T) = f(0, T) + \int_0^t\alpha(u,T)du+\int_0^t\sigma e^{-a(T-u)}dW_u, \end{align*} where, based on this question, \begin{align*} f(0, T) = \int_0^T \theta(u) e^{-a(T-u)} du - \frac{\sigma^2}{2a^2}\big(e^{-a T} -1\big)^2 + e^{-a T} r_0. \end{align*} Note also that \begin{align*} \int_0^t\alpha(u,T)du &= \int_0^t\sigma(u,T)\... 1 In the typical HJM model, during a single time step forward rates evolve according to their individual volatilities and according to pairwise correlations which can be specified. That arrangement is arbitrage free within the time step. In addition, different time steps are independently generated. This latter feature ensures that the model does not ... 1 OK, so I think I have figured it out. I assumed that we need to use Ito's lemma here, however, it seems the authors mean to use Ito's isometry, which must be used to prove below equality\mathbb{E}_Q\Big(e^{-\sigma\int_t^T (T-u)dW_u} \mid r_t\Big)=e^{\frac{\sigma^2}{2}\int_t^T(T-u)^2 du}$$We know that for normal-distibuted random variable X (with mean ... 1 Your issue is that you misinterpreted the NA criterion, it reads:$$ a_t(x) \triangleq \sum_{i=1}^{\infty} \left(b^i_t(x) \int_0^x (b_t^i(u))^T du\right)e_i,  where $b^i_t$ denotes the $i^{th}$ column of the volatility matrix $b_t$, $^T$ the transpose and $e_i$ the $i^{th}$ standard basis vector in $\ell^1$. In other words the $i^th$ coordinate of the ...