Applying Itô’s Lemma to Derive Vasicek Rate Dynamics

# Obtaining the dynamics of the Vasicek model using Itô


# Obtaining the dynamics of the Vasicek model using Itô












Consider the following expression for the short-term interest rate $$r_t=r_0 e^{\beta t}+\frac{b}{\beta}\left(e^{\beta t}-1\right)+\sigma e^{\beta t}\int_0^te^{-\beta s}dW_s \tag{1},$$ which is solution of the following version of the well-known Vasicek model $$dr_t=\left(b+\beta t \right)dt+\sigma dW_t \tag{2}.$$ I am aware of how to go from (1) to (2) according to the steps followed here for example, and how to make them backward from (2) to (1) respectively.

However, I would like to get from (1) to (2) by using Ito’s lemma, or using differentiation in some other way, would it be possible? If yes, can anybody show the steps? It is written at page 328 of Klebaner (1998) that regardless of the derivation, it is easy to see that (1) satisfies (2).

## Answer by Jan Stuller (score 2, accepted)

https://quant.stackexchange.com/a/69479

Your equation (2), $dr_t = (b+\beta t)dt + \sigma dW_t$, is a short hand version of:

$$r_t=r_0+\int_{h=0}^{h=t}(b+\beta h)dh+\int_{h=0}^{h=t}\sigma dW_h$$

Ito Process is defined as:

$$X_t=X_0+\int_{h=0}^{h=t}a(X_h,h)dh+\int_{h=0}^{h=t}b(X_h,h) dW_h$$

with $a()$ and $b()$ being some square-integrable functions of $t$ and $X_t$: therefore $r_t$ is an Ito process (with $a=(b+\beta t)$ and $b=\sigma$, obviously the two $b$s are different, to make it easy, I will use $\mu$ below instead of your $b$).

Ito's lemma states that any smooth, twice-differentiable function of time and the Ito process $X_t$, i.e. $F(t, X_t)$, will be governed by the following equation:

$$F(X_t,t)=F(X_0,t_0)+\int_{h=0}^{h=t} \left( \frac{\partial F}{\partial t}+\frac{\partial F}{\partial X}*a(X_h,h) + 0.5\frac{\partial^2 F}{\partial X^2}*b(X_h,h)^2 \right)dh+\int_{h=0}^{h=t}\left(\frac{\partial F}{\partial X}b(X_h,h)\right)dW_h$$

If we want to use Ito's lemma explicitly, the trick is to set $F(X_t, t)$ to $F(r_t,t):=r_t e^{\beta t}$ and apply the lemma to this expression, as follows:

$$r_te^{\beta t}=F(r_0,t_0)_{=r_0}+\int_{h=0}^{h=t} \left( \frac{\partial F}{\partial t}_{=\beta r_h e^{\beta h}}+\frac{\partial F}{\partial r}_{=e^{\beta h}}*a(r_h,h) + 0.5\frac{\partial^2 F}{\partial r^2}_{=0}*b(r_h,h)^2 \right)dh+\int_{h=0}^{h=t}\left(\frac{\partial F}{\partial r}_{=e^{\beta h}}b(r_h,h)\right)dW_h=\\=r_0+\int_{h=0}^{h=t}\left(\beta r_h e^{\beta h}+e^{\beta h}\beta(\mu- r_h)\right)dh+\int_{h=0}^{h=t}\left(e^{\beta h} \sigma\right)dW_h=\\=r_0+\int_{h=0}^{h=t}\left(e^{\beta h}\beta\mu\right)dh+\int_{h=0}^{h=t}\left(e^{\beta h} \sigma\right)dW_h$$

Now, to get the solution for $r_t$, the final step is simply to divide both sides by $e^{\beta t}$, to isolate the $r_t$ term on the LHS, which gives:

$$r_t=r_0e^{-\beta t}+\int_{h=0}^{h=t}\left(e^{\beta(h-t)}\beta\mu\right)dh+\int_{h=0}^{h=t}\sigma e^{\beta(h-t)} dW_h$$

What they did on Quantpie is less "mechanical" and probably more elegant: probably what an interviewer would want to see in an interview :) But I sympathize with "mechanical" approaches, I was always more of a mechanical guy myself, never an elegant one :)