Deriving the CIR Conditional Expectation PDE with Feynman–Kac

# CIR model problem - deriving PDE, Feynman-Kac


# CIR model problem - deriving PDE, Feynman-Kac












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.

## Answer by Quantuple (score 3, accepted)

https://quant.stackexchange.com/a/25867

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.

## Answer by parsiad (score 3)

https://quant.stackexchange.com/a/25864

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 $$ \mathcal{A}f_{\lambda}\equiv a\left(b-r\right)\frac{\partial f_{\lambda}}{\partial r}+\frac{1}{2}\sigma^{2}r\frac{\partial^{2}f_{\lambda}}{\partial r^{2}}. $$