2
$\begingroup$

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).

$\endgroup$
1
  • 1
    $\begingroup$ 4. line of your link is essentially using Ito's lemma. But in the differential for $e^{\kappa t} r_t$, the second order ito term is 0. $\endgroup$
    – fes
    Jan 14 at 7:45

1 Answer 1

2
$\begingroup$

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 :)

$\endgroup$
3
  • $\begingroup$ Jan: This was great. But how did you know to set $F(r_{t}, t)$ to $r_r e^{\beta t}$ initially. Since the LHS of the OP's expression just had $r_t$, I wouldn't have known to do that. Did you know that because the $F( )$ needs a $t$ and you knew that you could divide through at the end to get rid of it. Thanks. $\endgroup$
    – mark leeds
    Jan 14 at 22:24
  • $\begingroup$ @markleeds: I'll be very honest with you: imo one needs to learn it through experience and memorize it. The first person who solved the ornstein-uhlenbeck equation analytically (i.e. the Vasicek model) no doubt thought it out, but unless you're as smart as the guy who solved it first, you'll have to memorize these tricks. I do notice that the more you memorize, the easier it becomes to come up with your own, but it's deff not an easy thing to do; at least it never was for me. $\endgroup$ Jan 15 at 6:15
  • 1
    $\begingroup$ Thanks. That's quite an interesting answer. I was never good at tricks unfortunately. Also, you sell yourself very short. I can tell from this and your other answers that you know what the heck you're doing FOR SURE and also know the material quite well. All the best. $\endgroup$
    – mark leeds
    Jan 16 at 14:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.