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Why the Vasicek Short Rate Is Normally Distributed

Article Quant Q&A · Author: Michael Mark
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Summary

The document explains why the short rate in the Vasicek model has a Gaussian distribution at a fixed time. After solving the stochastic differential equation, the rate is written as a deterministic mean component plus a stochastic integral with a deterministic integrand. Such an Itô integral against Brownian motion is normally distributed, which establishes normality of the rate.

The answer identifies the mean as the initial rate’s exponentially decaying contribution plus the model’s long-run level adjustment. It gives the variance as the volatility squared times the integrated squared exponential weight, yielding a closed form. The original question’s proposed variance expression is incorrect: variance must be a deterministic quantity and includes the squared volatility. The explanation is limited to the stated Vasicek dynamics and fixed-time distribution; it does not address parameter estimation, empirical fit, or extensions with different shocks.

Key ideas

  • The Vasicek solution separates into a deterministic term and a Brownian stochastic integral.
  • An Itô integral with a deterministic integrand is Gaussian with mean zero.
  • The short rate’s mean follows from the deterministic part of the solution.
  • Its variance is the squared volatility multiplied by the integral of the squared kernel.
  • The result applies to the specified Vasicek model and does not establish empirical normality.

Tags

Fixed incomeStatistics
Full text
# How to show that this process is "normally distributed"?


# How to show that this process is "normally distributed"?












Say we have following SDE (Vasicek): $$dr(t) =(b-ar_t) dt + \sigma dW_t$$

I am able to reach an integral form of this SDE : $$r(t) = r(0) e^{-at} + \frac{b}{a}[1 - e^{-at}] + \sigma e^{-at}\int_0^t e^{as}dW_s$$

From here, I would like to conclude that $r(t)$ is Gaussian but I don't know how to proceed.

I somehow understand that

$$E[r(t)] = r(0) e^{-at} + \frac{b}{a}[1 - e^{-at}]$$

and that

$$Var[r(t)]= \sigma e^{-at}\int_0^t e^{as} dW_s$$

## Answer by user16651 (score 8, accepted)

https://quant.stackexchange.com/a/27926

First, note $$\mathbb{E^Q}\left[\int_0^t e^{-a(t-s)}dW_s\right]=0 $$ and $$\mathbb{Var^Q}\left[\int_0^t e^{-a(t-s)}dW_s\right]=\mathbb{E^Q}\left[\int_{0}^{t} e^{-2a(t-s)}ds\right]=\frac{1}{2a}(1-e^{-2at}) $$ therefore $$\mathbb{E^Q}[r_t]=r_0 e^{-at} + \frac{b}{a}(1 - e^{-at})$$ $$\mathbb{Var^Q}(r_t)=\frac{\sigma^2}{2a}(1-e^{-2at})$$ second

The Itô integral can be defined in a manner similar to the Riemann–Stieltjes integral, that is as a limit in probability of Riemann sums; such a limit does not necessarily exist pathwise. Suppose that $W_t$ is a Wiener process and that $X_t$ is a right-continuous (cadlag), adapted and locally bounded process if $I=\{t_0,t_1,\cdots,t_n\}$ is a sequence of partitions of $[0,t]$ with mesh going to zero, then the Itô integral of $X_t$ with respect to $W_t$ up to time t is a random variable $$\int_{0}^{t}X_sdW_s=\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{i=1}^{n}{X({{t}_{i-1}})(W({{t}_{i}})-W({{t}_{i-1}})})$$ Set $X_s=e^{as}$, $X_s$ is a deterministic function thus $$\int_0^t e^{-a(t-s)}dW_s\sim N\left(0\quad,\quad\frac{1}{2a}(1-e^{-2at})\right) $$

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.