# Vasicek model: joint simulation with discount factor

In Vasicek model, we have the following relation to get Discount factors given the value of short rate: $$P(t\,,T)={{e}^{A(t,T)\,-\,B(t,T){{r}_{t}}\,}}$$

So, Discount factors are known as soon as we know the short rate. But then in some references like Glasserman (pg. 115) there is a whole subsection on "Joint Simulation [of short rate] with the Discount Factor" where he talks about simulating the pair $$({r}_{t},\int_{0}^t{r(u)}du)$$.

Piterbarg's book has something similar too. So my question is - why do we need to simulate Discount factor if we have an exact analytical result.

• Read all of page 115 in Glasserman's book. Since the process X=(r,integral of r) is jointly Gaussian, it can be simulated with precision. We simulate X instead of simulating r and then estimating the integral of r with a sum.
– AXH
Dec 17, 2018 at 21:20
• @AXH thats exactly my confusion - why care about integral of r when we can calculate it's exact value analytically. Also, given we can exactly simulate r too, simulation precision cannot be the reason to simulate X. But since several references do it this way, there must be a reason. Dec 17, 2018 at 22:00

## 1 Answer

Although it's been a long time this question has been asked, I'd like to propose an answer in case someone was looking for the same thing.

First, I think there's a confusion between $$P(t,T)$$ and $$DF(t,T)$$. The former is the $$t-$$price of a contract paying $$1$$ unit of currency at date $$T$$ while the later is the (stochastic) discount factor at $$t$$ for flows occuring at $$T$$. The two are linked through the relationship $$P(t,T)=\mathbb{E}^Q[DF(t,T) | \mathcal{F}_t]$$

If $$r_t$$ is the instantaneous short rate, then $$DF(t,T)$$ is given by $$DF(t,T)=e^{-\int_t^T r_s ds}$$ and is a random variable.

Now, the argument of Glasserman is about computing $$\int_t^T r_s ds$$. In theory, since one has $$r_t$$ up to maturity on a given path, this is just a matter of doing a Riemann sum. However, this may be very "noisy" because of discretization errors. It turns ou, as AXH mentionned, that $$(r_t, \int_t^T r_s ds)$$ are jointly gaussian and can be simulated precisely.

• To complete this very good answer, as usually one needs to discount quantities, it's very useful to have the $\int_0^t r_s ds$ as well as $r_t$. For example, if you are computing the CVA of an interest rates swap, you need the values of $r_t$ to value it at $t > 0$, but you also need $\int_0^t r_s ds$ to discount the exposures to today to compute your CVA value. May 13, 2021 at 18:48