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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.

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    $\begingroup$ 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. $\endgroup$
    – AXH
    Dec 17, 2018 at 21:20
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    $\begingroup$ @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. $\endgroup$
    – InnocentR
    Dec 17, 2018 at 22:00

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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.

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    $\begingroup$ 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. $\endgroup$
    – byouness
    May 13, 2021 at 18:48

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