Ho-Lee Short-Rate Bond Pricing and the Feynman-Kac PDE

# The Ho-Lee Model (1986)


# The Ho-Lee Model (1986)












### (My question)

I solved the following questions. However, if you know the other solutions, please let me know those along with computation processes. Besides, $W_t$ is a S.B.M.

(Thank you for your help in advance.)

### (Cross-Link)

I have posted the same question on https://math.stackexchange.com/questions/3332466/the-ho-lee-model-1986?fbclid=IwAR3OzlLsRHlNIynul_JEhc2d6e9dpVHt0kDuctV3CKk6rtx4gkcz3-2DRb8

### (Original questions)

Consider a short term interest rate process $(r_t)_{ t \in \mathbb{R}_+ }$ in the Ho-Lee Model with constant coefficients: \begin{eqnarray} dr_t = \theta dt + \sigma dW_t \end{eqnarray} and let $P(t, T)$ will denote the arbitrage price of a zero-coupon bond in this model: \begin{eqnarray} P(t, T) = E^{ \mathbb{P} } \left[ \exp \left( - \int^T_t r_s ds \right) | \mathcal{F}_t \right] \end{eqnarray} where $t \in [0, T]$.

(1) State the bond pricing PDE satisfied by the function $F(t, x)$ defined via \begin{eqnarray} F(t, x) = E^{ \mathbb{P} } \left[ \exp \left( - \int^T_t r_s ds \right) | r_t=x \right] \end{eqnarray} where $t \in [0, T]$.

(2) Compute the arbitrage price $F(t, r_t) =P(t, T)$ from its expression of $P(t, T)$ as a conditional expectation.

(3) Check that the function $F(t, x)$ computed in Question (2) does satisfy the PDE derived in Question (1).

### (1) My answer

- One can derive the following PDE by Feynman-Kac Theorem. \begin{eqnarray} F(t, x) &=& E^{ \mathbb{P} } \left[ \exp \left( - \int^T_t r_s ds \right) | r_t=x \right] \\ \exp \left( - \int^t_0 r_s ds \right) F(t, x) &=& E^{ \mathbb{P} } \left[ \exp \left( - \int^T_0 r_s ds \right) | r_t=x \right] \end{eqnarray}

- One adapts It$\hat{o}$'s formula the above equation. Here, one has to pay attention that the R.H.S is zero because it is a constant value, namely an expectation value. Moreover, $F(T, x)=1$ by the R.H.S going to $1$ because of $t=T$. \begin{eqnarray} && d\left( \exp \left( - \int^t_0 r_s ds \right) F(t, x) \right) \nonumber \\ && \qquad \qquad \qquad \qquad \qquad = \exp \left( - \int^t_0 r_s ds \right) \left( -r_t F(t, x) + \partial_t F(t, x) \right) dt \nonumber \\ && \qquad \qquad \qquad \qquad \qquad \qquad + \exp \left( - \int^t_0 r_s ds \right) \partial_x F(t, x) dr_t \nonumber \\ && \qquad \qquad \qquad \qquad \qquad \qquad + \frac{1}{2} \exp \left( - \int^t_0 r_s ds \right) \partial_{xx} F(t, x) d[r_t] \end{eqnarray}

- One substitutes $r_t$ into the above equation. \begin{eqnarray} && d\left( \exp \left( - \int^t_0 r_s ds \right) F(t, x) \right) \nonumber \\ && = \exp \left( - \int^t_0 r_s ds \right) \left( -r_t F(t, x) + \partial_t F(t, x) \right) dt \nonumber \\ && \qquad \qquad \qquad \qquad \qquad \qquad \qquad + \exp \left( - \int^t_0 r_s ds \right) \partial_x F(t, x) \left( \theta dt + \sigma dW_t \right) \nonumber \\ && \qquad \qquad \qquad \qquad \qquad \qquad \qquad + \frac{1}{2} \exp \left( - \int^t_0 r_s ds \right) \partial_{xx} F(t, x) \sigma^2 dt \\ && = \exp \left( - \int^t_0 r_s ds \right) \left( -r_t F(t, x) + \partial_t F(t, x) + \theta \partial_x F(t, x) + \frac{1}{2} \sigma^2 \partial_{xx} F(t, x) \right) dt \nonumber \\ && \qquad \qquad \qquad \qquad \qquad \qquad \qquad + \exp \left( - \int^t_0 r_s ds \right) \sigma \partial_x F(t, x) dW_t \end{eqnarray}

- Since the coefficient of the drift term is zero due to the martingale property of S.D.E, the following equation is obtained. Here, one has to pay attention to $r_t=x$. \begin{eqnarray} && -r_t F(t, x) + \partial_t F(t, x) + \theta \partial_x F(t, x) + \frac{1}{2} \sigma^2 \partial_{xx} F(t, x) =0 \\ && -xF(t, x) + \partial_t F(t, x) + \theta \partial_x F(t, x) + \frac{1}{2} \sigma^2 \partial_{xx} F(t, x) =0 \end{eqnarray}

$\square$

### (2) My answer

- Compute S.I.E by the given S.D.E. \begin{eqnarray} dr_t &=& \theta dt + \sigma dW_t \\ r_t &=& r_0 + \theta t + \sigma W_t \end{eqnarray}

- Let $x=r_t$: \begin{eqnarray} && F(t, x) \\ &&= F(t, r_t) \\ &&= E^{ \mathbb{P} } \left[ \exp \left( - \int^T_t r_s ds \right) | r_t=x \right] \\ &&= E^{ \mathbb{P} } \left[ \exp \left( - \int^T_t \left( r_0 + \theta s + \sigma W_s \right) ds\right) | \mathcal{F}_t\right] \\ &&= E^{ \mathbb{P} } \left[ \exp \left( - \int^T_t \left( \left( r_0 + \theta t + \sigma W_t\right)+ \theta (s-t) + \sigma( W_s -W_t) \right) ds \right) | \mathcal{F}_t\right] \\ &&= E^{ \mathbb{P} } \left[ \exp \left( - \int^T_t \left( r_t + \theta (s-t) + \sigma( W_s -W_t) \right) ds \right) | \mathcal{F}_t\right] \\ \end{eqnarray}

- Here, one computes $exp$. \begin{eqnarray} && \exp \left( - \int^T_t \left( r_t + \theta (s-t) + \sigma( W_s -W_t) \right) ds \right) \\ && \qquad =\exp \left( - r_t \int^T_t ds \right) \cdot \exp \left( - \theta \int^T_t (s-t) ds \right) \nonumber \\ && \qquad \qquad \qquad \qquad \qquad \qquad \qquad \cdot \exp \left( - \sigma \int^T_t ( W_s -W_t) ds \right) \\ && \qquad =\exp \left( - r_t (T-t) \right) \nonumber \\ && \qquad \qquad \qquad \qquad \cdot \exp \left( - \theta \frac{1}{2} (T^2- t^2) + \theta t(T-t) \right) \nonumber \\ && \qquad \qquad \qquad \qquad \qquad \qquad \qquad \cdot \exp \left( - \sigma \int^T_t ( W_s -W_t) ds \right) \\ && \qquad =\exp \left( - r_t (T-t) \right) \nonumber \\ && \qquad \qquad \qquad \qquad \cdot \exp \left( - \theta (T-t) \left( \frac{1}{2} (T + t) - t\right) \right) \nonumber \\ && \qquad \qquad \qquad \qquad \qquad \qquad \qquad \cdot \exp \left( - \sigma \int^T_t ( W_s -W_t) ds \right) \\ && \qquad =\exp \left( - r_t (T-t) \right) \nonumber \\ && \qquad \qquad \qquad \qquad \cdot \exp \left( - \theta (T-t) \left( \frac{1}{2} (T - t) \right) \right) \nonumber \\ && \qquad \qquad \qquad \qquad \qquad \qquad \qquad \cdot \exp \left( - \sigma \int^T_t ( W_s -W_t)ds \right) \\ && \qquad =\exp \left( - r_t (T-t) \right) \cdot \exp \left( - \frac{1}{2} \theta (T-t)^2 \ \right) \nonumber \\ && \qquad \qquad \qquad \qquad \qquad \qquad \qquad \cdot \exp \left( - \sigma \int^T_t ( W_s -W_t) ds \right) \\ && \qquad =\exp \left( - r_t (T-t) - \frac{1}{2} \theta (T-t)^2 \right) \nonumber \\ && \qquad \qquad \qquad \qquad \qquad \qquad \qquad \cdot \exp \left( - \sigma \int^T_t ( W_s -W_t) ds \right) \\ \end{eqnarray}

- Substitute the above result into the Expectation. \begin{eqnarray} && F(t, r_t) \\ &&= E^{ \mathbb{P} } \left[ \exp \left( - r_t (T-t) - \frac{1}{2} \theta (T-t)^2 - \sigma \int^T_t ( W_s -W_t) ds \right) | \mathcal{F}_t \right] \\ &&= \exp \left( - r_t (T-t) - \frac{1}{2} \theta (T-t)^2 \right) \nonumber \\ && \qquad \qquad \qquad \qquad \qquad \qquad \cdot E^{ \mathbb{P} } \left[ \exp \left( - \sigma \int^T_t ( W_s -W_t) ds \right) | \mathcal{F}_t \right] \end{eqnarray}

- One computes the above expectation value as below. \begin{eqnarray} &&E^{ \mathbb{P} } \left[ \exp \left( - \sigma \int^T_t ( W_s -W_t) ds \right) | \mathcal{F}_t \right] \nonumber \\ && \qquad \qquad \qquad \qquad = E^{ \mathbb{P} } \left[ \exp \left( - \sigma \int^T_t W_{s-t} ds \right) | \mathcal{F}_t \right] \\ && \qquad \qquad \qquad \qquad = E^{ \mathbb{P} } \left[ \exp \left( \int^{T-t}_0 \left( - \sigma W_s \right) ds \right) | \mathcal{F}_t \right] \end{eqnarray}

- Here, It$\hat{o}$'s formula is applied to the exponent part, and further calculation is performed. \begin{eqnarray} d \left( - \sigma W_s s \right) &=& - \sigma W_s ds - \sigma s d W_s + \frac{1}{2} 0 d [W_s] \\ &=& - \sigma W_s ds - \sigma s d W_s \\ \int^{T-t}_0 d \left( - \sigma W_s s \right) &=& \int^{T-t}_0 \left( - \sigma W_s \right) ds - \sigma \int^{T-t}_0 s d W_s \\ - \sigma W_{T-t} (T-t) &=& \int^{T-t}_0 \left( - \sigma W_s \right) ds - \sigma \int^{T-t}_0 s d W_s \\ \int^{T-t}_0 \left( - \sigma W_s \right) ds &=& - \sigma W_{T-t} (T-t) + \int^{T-t}_0 \sigma s d W_s \\ &=& - \sigma (T-t) \int^{T-t}_0 d W_s + \int^{T-t}_0 \sigma s d W_s \\ &=& \sigma \int^{T-t}_0 \left( s- (T-t) \right) d W_s \\ &=& \int^{T-t}_0 \left( \sigma \left( s- (T-t) \right) \right) d W_s \\ \end{eqnarray}

- Substitute the above result into the expectation. \begin{eqnarray} &&E^{ \mathbb{P} } \left[ \exp \left( - \sigma \int^T_t ( W_s -W_t) ds \right) | \mathcal{F}_t \right] \nonumber \\ && \qquad \qquad \qquad \qquad = E^{ \mathbb{P} } \left[ \exp \left( \int^{T-t}_0 \left( - \sigma W_s \right) ds \right) | \mathcal{F}_t \right] \\ && \qquad \qquad \qquad \qquad = E^{ \mathbb{P} } \left[ \exp \left( \int^{T-t}_0 \left( \sigma \left( s - (T-t)\right) \right) d W_s \right) | \mathcal{F}_t \right] \\ && \qquad \qquad \qquad \qquad = \exp \left( \frac{ \sigma^2}{2} \int^{T-t}_0 \left( s - (T-t) \right)^2 ds \right) \\ && \qquad \qquad \qquad \qquad = \exp \left( \frac{ \sigma^2}{2} \left[ \frac{1}{3} \left( s - (T-t) \right)^3 \right]^{T-t}_0 \right) \\ && \qquad \qquad \qquad \qquad = \exp \left( \frac{ \sigma^2}{6} (T-t)^3 \right) \end{eqnarray}

- Substitute the above result into the Expectation of the $F(t, r_t)$ equation. \begin{eqnarray} && F(t, r_t) \\ &&= \exp \left( - r_t (T-t) - \frac{1}{2} \theta (T-t)^2 \right) \nonumber \\ && \qquad \qquad \qquad \qquad \qquad \qquad \cdot E^{ \mathbb{P} } \left[ \exp \left( - \sigma \int^T_t ( W_s -W_t) ds \right) | \mathcal{F}_t \right] \\ &&= \exp \left( - r_t (T-t) - \frac{1}{2} \theta (T-t)^2 \right) \cdot \exp \left( \frac{ \sigma^2}{6} (T-t)^3 \right) \\ &&= \exp \left( - r_t (T-t) - \frac{1}{2} \theta (T-t)^2 + \frac{ \sigma^2}{6} (T-t)^3 \right) \end{eqnarray}

$\square$

### (3) My answer

- Let $r_t=x$ into $F(t, r_t)$ of (2). \begin{eqnarray} F(t, x) = \exp \left( - x (T-t) - \frac{1}{2} \theta (T-t)^2 + \frac{ \sigma^2}{6} (T-t)^3 \right) \end{eqnarray}

- Then, one reaches the following equations. \begin{eqnarray} \partial_t F(t, x) &=& \left( x + \theta (T-t) - \frac{ \sigma^2}{2} (T-t)^2 \right) F(t, x) \\ \partial_x F(t, x) &=& - (T-t) F(t, x) \\ \partial_{xx} F(t, x) &=& (T-t)^2 F(t, x) \\ \end{eqnarray}

- Therefore, one computes the P.D.E of (1). Besides, the terminal condition is $F(T, x)=1$. \begin{eqnarray} && -xF(t, x) + \partial_t F(t, x) + \theta \partial_x F(t, x) + \frac{1}{2} \sigma^2 \partial_{xx} F(t, x) \\ && \qquad \qquad \qquad \qquad \qquad = -xF(t, x) \\ && \qquad \qquad \qquad \qquad \qquad \qquad + \left( x + \theta (T-t) - \frac{ \sigma^2}{2} (T-t)^2 \right) F(t, x) \\ && \qquad \qquad \qquad \qquad \qquad \qquad - \theta (T-t) F(t, x) + \frac{1}{2} \sigma^2 (T-t)^2 F(t, x) \\ && \qquad \qquad \qquad \qquad \qquad = 0 \end{eqnarray}

$\square$

(Thank you for your help in advance.)