Pricing interest rate swap in Ho Lee model

In Ho Lee model, assuming risk neutral probability is not exactly 0.5, would a change in the volatility of short-term rate affect the price of an interest rate swap? My intuition tells me no as interest rate swap price should only depend on prices of zero at t=0 but my model is throwing a different answer.

Would you it be possible to have some help on both mathematical and intuitive explanations, please?

Many thanks.

• I don't understand your question. Please clarify it.
– user16651
Nov 13 '16 at 20:31

Under the Ho-lee model, \begin{align*} dr_t = \theta_t dt + \sigma dW_t. \end{align*} Then, the price at time $t$ of a zero-coupon bond with maturity $T$ and unit notional is given by \begin{align*} P(t, T) = E\left(e^{-\int_t^T r_s ds} \mid \mathcal{F}_t \right), \end{align*} where $\mathcal{F}_t$ is the information set at time $t$. Note that, for any $s\ge t \ge 0$, \begin{align*} r_s = r_t + \int_t^s \theta_u du + \sigma\int_t^s dW_u. \end{align*} Therfore, \begin{align*} \int_t^T r_s ds &=r_t(T-t) + \int_t^T\left(\int_t^s \theta_u du \right)ds+ \sigma \int_t^T\left(\int_t^s dW_u\right)ds\\ &=r_t(T-t) + \int_t^T\left(\int_u^T \theta_u ds \right)du+ \sigma \int_t^T\left(\int_u^T ds\right) dW_u\\ &=r_t(T-t) + \int_t^T (T-u)\theta_u du + \sigma \int_t^T (T-u) dW_u. \end{align*} Note that $\int_t^T (T-u) dW_u$ is independent of $\mathcal{F}_t$ and normal with zero mean and variance \begin{align*} \int_t^T (T-u)^2 du &= \frac{1}{3}(T-t)^3. \end{align*} Consequently, \begin{align*} P(t, T) &= E\left(e^{-\int_t^T r_s ds} \mid \mathcal{F}_t \right)\\ &=e^{-r_t(T-t) -\int_t^T (T-u)\theta_u du + \frac{\sigma^2}{6}(T-t)^3}. \end{align*} For $t=0$, then \begin{align*} P(0, T)=e^{-r_0 T -\int_0^T (T-u)\theta_u du + \frac{\sigma^2}{6}T^3}.\tag{1} \end{align*} From $(1)$, we see clearly that the bond price depends on the volatility $\sigma$.