2
$\begingroup$

I am reviewing a CIR model problem, where $r_t$ has following dynamics $$dr_t=a(b-r_t)dt+\sigma \sqrt{r_t} dW_t^* \quad \quad (1)$$ for some constants $ab>\frac{\sigma^2}{2} \quad$

Letting T be a fixed date and $f_{\lambda}$ a function defined for some constant $\lambda >0$ $$f_{\lambda}(t,r)=E^*[e^{-\lambda {r_{T}}}|r_t=r] \quad \quad (2)$$

a) derive PDE satisfied by the function $f_{\lambda}$
b) show that the function $f_{\lambda} (t,r)=e^{-A_{\lambda}(T-t)-B_{\lambda}(T-t)r_t}$ satisfies the PDE

I guess the subject function can be expressed as $$f(t,r_t)=e^{- \lambda r} \quad \quad (3)$$

My first thought was just to calculate the $df(t,r_t)$ using Ito formula and substituting for the $dr_t$ which would yield $$df(t,r_t)=f_t dt + f_r dr_t+ \frac{1}{2} f_{rr} d<r>_t=$$ $$=(f_t + a(b-r_t) f_r + \frac{1}{2} r f_{rr})dt+ \sigma \sqrt{r_t} dW_t^* \quad \quad (4)$$

Wouldn't it be already answer to a?

However the solution presents a different approach which I don't understand. It gives the Feynman-Kac equation as solution to a),

further the equation $f_{\lambda} (t,r)=e^{-A_{\lambda}(T-t)-B_{\lambda}(T-t)r_t}$ is plugged into the Feynman-Kac equation, and then using a system of 2 ODE's $A(\tau)$ $B(\tau)$ are derived.


Can anybody explain the proceedings please? I am missing "the big picture" here.
I don't understand why it starts with Feynman-Kac and why only deriving A and B proves already that they satisfy the PDE.

$\endgroup$
  • $\begingroup$ I'm assuming there is an error in your equation (2). Can you fix it? $\endgroup$ – parsiad May 6 '16 at 3:41
  • $\begingroup$ I corrected the typo in (2) $\endgroup$ – Michal May 6 '16 at 3:58
3
$\begingroup$

Here's my 2 cents:

a) Conditional expectations can always be seen as martingales (this is a direct consequence of the tower property). Thus, we here have that $$ M_t := E^*[e^{-\lambda {r_{T}}}|r_t] $$ is a martingale.

Applying Itô's lemma to $M_t = f_{\lambda}(t,r_t)$ as you did is a good starting point. But doing this, leaves you with an SDE, not a PDE.

Now, because $M_t$ is a martingale, the martingale representation theorem tells you that its drift should be strictly zero. Working from your application of Ito's lemma, equating the drift to zero gives you the following PDE which needs to be satisfied by $f_{\lambda}(t,r)$: $$ \frac{\partial f_{\lambda}}{\partial t}+ a\left(b-r\right)\frac{\partial f_{\lambda}}{\partial r}+\frac{1}{2}\sigma^{2}r\frac{\partial^{2}f_{\lambda}}{\partial r^{2}} = 0$$

This is precisely Feynman-Kac formula.

Note that your equation (3) is wrong (and useless).Instead what we could write is: $$ M_T = f_{\lambda}(T,r_T) = E^*[e^{-\lambda {r_{T}}}|r_T]=e^{-\lambda {r_{T}}} $$ which could be seen as the PDE's terminal condition.

b) To show that the proposed $$ f_{\lambda} (t,r_t)=e^{-A_{\lambda}(T-t)-B_{\lambda}(T-t)r_t} $$ is effectively a solution of the just-derived PDE, you need to plug this particular expression of $f_{\lambda} (t,r_t)$ in the PDE and verify that the resulting RHS indeed equates to zero. Here, this condition can indeed be verified (for unique values for both $A_{\lambda}$ and $B_{\lambda}$ that one can solve for), hence you have indeed found a solution of the problem.

Hope this helps.

$\endgroup$
  • $\begingroup$ to recap, the Feyneman-Kac equation is basicaly the drift of the resulting SDE equated to zero, correct? the pricing algorithm here would be to start with a function which has a stochastic variable, to calculate the df using Ito formula, to take the drift of this SDE and equate to zero (Feynman-Kac equation). solving the resulting PDE using system of two ODE's should give us the A and B. To check if the calculated functions A and B were correct we plug the $P(t,T)=e^{-A(t,T)-B(t,T)r_t}$ into the Feynman-Kac equation which should give back the drift of the SDE. is that correct? $\endgroup$ – Michal May 6 '16 at 11:01
  • 1
    $\begingroup$ No, plugging it into the PDE should give you zero on the left hand side. Also @Quantuple; you are missing a $\sigma^2$ term. See my answer below. $\endgroup$ – parsiad May 6 '16 at 12:33
  • $\begingroup$ PDE is the Feynman-Kac equation, isn't it? $\endgroup$ – Michal May 6 '16 at 12:39
  • 1
    $\begingroup$ That is correct. $\endgroup$ – parsiad May 6 '16 at 14:26
  • 1
    $\begingroup$ @Michal I have edited my answer and added a little bit more details. $\endgroup$ – Quantuple May 6 '16 at 17:48
3
$\begingroup$

Let $$ f_{\lambda}(t,r)=E^{(t,r)}\left[e^{-\lambda r_{T}}\right] $$ where $E^{(t,r)}$ denotes the expectation conditional on $r_{t}=r$. We assume $f$ is smooth for the remainder. Let $\theta=T\wedge\inf\left\{ s>t\colon\left|r_{s}-r\right|>1\right\} $. By the Markov property of $\{r_{t}\}$, $$ f_{\lambda}(t,r)=E^{(t,r)}\left[f_{\lambda}(\left(t+h\right)\wedge\theta,r_{\left(t+h\right)\wedge\theta})\right]. $$ Moving some terms around, we get $$ 0=E^{(t,r)}\left[f_{\lambda}(\left(t+h\right)\wedge\theta,r_{\left(t+h\right)\wedge\theta})-f_{\lambda}(t,r)\right]. $$ Applying Ito's lemma with $\mathcal{A}$ denoting the infinitesimal generator of the process $\{r_{t}\}$, \begin{align*} 0 & =E^{(t,r)}\left[\int_{t}^{\left(t+h\right)\wedge\theta}\left(\frac{\partial f_{\lambda}}{\partial t}+\mathcal{A}f_{\lambda}\right)(u,r_{u})du+\int_{t}^{\left(t+h\right)\wedge\theta}\frac{\partial f_{\lambda}}{\partial r}(u,r_{u})\sigma\sqrt{r_{u}}dW_{u}\right]\\ & =E^{(t,r)}\left[\int_{t}^{\left(t+h\right)\wedge\theta}\left(\frac{\partial f_{\lambda}}{\partial t}+\mathcal{A}f_{\lambda}\right)(u,r_{u})du\right]. \end{align*} The vanishing Ito integral is due to the boundedness of $r_{u}$ on $[t,(t+h)\wedge\theta]$. Multiplying by $1/h$ yields $$ 0=E^{(t,r)}\left[\frac{1}{h}\int_{t}^{\left(t+h\right)\wedge\theta}\left(\frac{\partial f_{\lambda}}{\partial t}+\mathcal{A}f_{\lambda}\right)(u,r_{u})du\right].\tag{1} $$ Since for each sample $\omega$ in some set of full measure there exists an $h_{0}(\omega)$ for which $\theta(\omega)\geq t+h$ for all $h\leq h_{0}(\omega)$, it follows from the mean value theorem that $\mathbb{P}$-almost surely, \begin{multline*} \lim_{h\searrow0}\frac{1}{h}\int_{t}^{\left(t+h\right)\wedge\theta}\left(\frac{\partial f_{\lambda}}{\partial t}+\mathcal{A}f_{\lambda}\right)(u,r_{u})du\\ =\lim_{h\searrow0}\frac{\left(t+h\right)\wedge\theta}{h}\left(\frac{\partial f_{\lambda}}{\partial t}+\mathcal{A}f_{\lambda}\right)(s(h),r_{s(h)})du=\left(\frac{\partial f_{\lambda}}{\partial t}+\mathcal{A}f_{\lambda}\right)(t,r) \end{multline*} where $s(h)$ is some number in $(t,(t+h)\wedge\theta)$. The remainder of the proof is given by taking $h\searrow0$ and applying the dominated convergence theorem in (1) to yield $$ \frac{\partial f_{\lambda}}{\partial t}+\mathcal{A}f_{\lambda}=0\text{ pointwise}. $$ where $$ \mathcal{A}f_{\lambda}\equiv a\left(b-r\right)\frac{\partial f_{\lambda}}{\partial r}+\frac{1}{2}\sigma^{2}r\frac{\partial^{2}f_{\lambda}}{\partial r^{2}}. $$

$\endgroup$
  • $\begingroup$ I assume this prof is perfectly fine but it misses the point of the problem. It doesn't state explicitly any PDE and it doesn't solve it in the analytical way in terms of A and B. I am still missing "the big picture" here $\endgroup$ – Michal May 6 '16 at 8:58
  • 1
    $\begingroup$ The PDE is on the second last line. $\endgroup$ – parsiad May 6 '16 at 12:19

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.