I recently stumbled upon this paper where the author gives a closed form for the Arrow--Debreu function (also known as Green's function) of the Hull--White model. The mathematics in it seem too complicated for such a simple Gaussian model, but it seems that the author has more ambitious goals in mind (like applying the methods there to more complex models) which justify the approach. Having said this, let me show you a very simple way to compute this function that only depends on knowing how to compute the conditional law of a two-dimensional Gaussian.
Following the previous post, if we consider a short rate process of the form $r(t) = x(t) + f(0,t) + \frac 12 v'(t)$ with $ dx = -k x dt + \sigma dw$ where the deterministic coefficients depend only on time, then we want to compute the function $$ U(t,x) = \mathbb E\left( e^{-\int_0^t r(s)ds} \mid x(t) = x\right) p_t(x)$$ where $p_t(x)$ is the density of $x(t)$. The reason is quite simple: this function acts as a kernel to compute the payoff of any European claim contigent on the state of $r$ at some future time, since conditioning on the value of $x(t)$, we have that $$ \mathbb E\left( e^{-\int_0^t r(s)ds} f(x(t))\right) = \int_{\mathbb R}U(t,x) f(x)dx. $$ Let us prove that (using the notation in the previous post), we have that $$U(t,x) = \frac{P(0,t)}{\sqrt{2 \pi V(0,t)}} \exp\left(-\frac12\frac{ (x +\frac 12 v'(t))^2}{V(0,t)} \right).$$
We use the following elementary fact: if $X$ and $Y$ are Gaussian with mean zero and covariance $c$ and variances $\sigma_X$ and $\sigma_Y$. Then $$ \mathbb E(e^{-Y} \mid X = x) = \exp\left(-\frac12\frac{c^2 + 2c x}{\sigma_X^2} + \frac 12 \sigma_Y^2\right).$$ This follows by using the joint law of $(X,Y)$ to derive the conditional law of $Y\mid X=x$, and then apply it to the computation above: this variable is Gaussian with mean $cx/\sigma_X^2$ and variance $\sigma_Y^2 - c^2/\sigma_X^2$. Evaluating the moment generating function of this Gaussian at $t=-1$ gives us what we want.
To apply this result, we just need to go back to the previous post and recall the that $y(t) = \int_0^t x(s)ds$, and that $x(t)$ and $r(t)$ differ by the forward rate term and the convexity term $v'(t)$. Thus, we can incorporate the factor $P(0,t) e^{-\frac 12 v(t)}$ that appears after integrating $r(t)$ later, and focus on $$ \mathbb E(e^{-y(t)} \mid x(t) = x).$$ From the previous post it follows that $x(t)$ and $y(t)$ are both Gaussian of mean zero, with variance $V(0,t)$ and $v(t)$, respectively. Moreover, $\mathbb E(x(t) y(t))$ is just the derivative of half the variance of $y(t)$, which we computed to be $\frac 12 v'(t)$, the convexity term we just mentioned.
Applying the lemma, we get that $$ \mathbb E\left( e^{-\int_0^t x(s)ds} \mid x(t) = x \right) = \exp\left(-\frac18\frac{ v'(t)^2 + 4 v'(t) x}{V(0,t)} + \frac 12 v(t)\right)$$ The last term in the exponential obviously cancels with the term in $r(t)$ coming from integrating $\frac 12 v'(t)$ (recall that $y(t)$ has variance $v(t)$!). On the other hand, the density of $x(t)$ is $$p_t(x) = \frac{1}{\sqrt{2 \pi V(0,t)}} \exp\left( -\frac 12 \frac{x^2 }{ V(0,t)} \right) $$ so multiplying this with the above expression, we get $$ e^{-\frac 12 v'(t)} \mathbb E\left( e^{-\int_0^t x(s)ds} \mid x(t) = x \right)p_t(x) = \\ \frac{1}{\sqrt{2 \pi V(0,t)}} \exp\left( -\frac 18 \frac{ v'(t)^2 + 4 v'(t) x + 4x^2 }{ V(0,t)} \right). $$ The only thing left to do is to notice the denominator is $(v'(t)+2x)^2$ and that the term $f(0,t)$ contributes the term $P(0,t)$ to the integral. The takeaway is you can now use Gaussian quadrature (or your favourite quadrature method) to quickly compute European payoffs using $U(t,x)$.
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