# Expectation of two correlated processes?

Consider the following:

$$dn_t = [\theta_n(t)-a_nn_t]dt + \sigma_ndW_{t}^n \\ dr_t = [\theta_r(t)-\rho_{r,n}\sigma_n\sigma_r-a_rr_t]dt + \sigma_rdW_{t}^r$$ Interpret $dn_t$ as the diffusion for the nominal short rate and $dr_t$ as the diffusion for the short real rate. Now write the following expectation:

$$\mathbb{E^Q}[I_{T}] = I_0\mathbb{E^Q}[e^{\int_0^Tn(s)-r(s)ds}]$$

Can it be simplified further? I considered separating the terms and calculating the expectations separately and multiply but the fact is that both processes are correlated suggests to me that this can't be done. Anyone?

I interpret the difference between $n(s)$ and $r(s)$ in the integral as the spread between the nominal and real short rates. Note that the quantity $I_t$ represents the inflation at time $t$. This makes sense intuitively: the expected inflation at a future time depends on the expected spread at the same future time.

I just want to know if I can drill further into the algebra and obtain something simpler/more elegant. Or perhaps someone knows of a discussion I could consult?

• Just a sidenote (which you might already be familiar with): add dynamics to the inflation term through: $dI_t/I_t = (n_t-r_t)dt + \sigma_I dW^I_t$ and you end up with the Jarrow-Yildirim model. – Olaf Dec 15 '16 at 11:28
• Yep. Am familiar with it. Have you ever tried implementing it? If so, I'd have a few questions for you. – baluch_stan Dec 20 '16 at 15:41
• Yes, I have implemented the model. – Olaf Dec 20 '16 at 16:01

## 1 Answer

Note that, as in this question, for $s\ge t\ge 0$, \begin{align*} n_s = e^{-a_n(s-t)}n_t + \int_t^s \theta_n(u)e^{-a_n(s-u)} du + \int_t^s \sigma_n e^{-a_n(s-u)} dW^n_u, \end{align*} and \begin{align*} r_s = e^{-a_r(s-t)}r_t + \int_t^s (\theta_r(u) -\rho_{r,n}\sigma_n\sigma_r) e^{-a_r(s-u)} du + \int_t^s \sigma_r e^{-a_r(s-u)} dW^r_u. \end{align*} Moreover, \begin{align*} \int_t^T n_s ds = \frac{1}{a_n}\Big(1-e^{-a_n(T-t)} \Big) n_t &+ \int_t^T\!\! \frac{\theta_n(u)}{a_n}\Big(1-e^{-a_n(T-u)} \Big)du \\ &+ \int_t^T \!\!\frac{\sigma_n}{a_n}\Big(1-e^{-a_n(T-u)} \Big)dW_u^n, \end{align*} and \begin{align*} \int_t^T r_s ds= \frac{1}{a_r}\Big(1-e^{-a_r(T-t)} \Big) n_t &+ \int_t^T\!\! \frac{\theta_r(u)-\rho_{r,n}\sigma_n\sigma_r}{a_r}\Big(1-e^{-a_r(T-u)} \Big)du \\ &+ \int_t^T \!\!\frac{\sigma_r}{a_r}\Big(1-e^{-a_r(T-u)} \Big)dW_u^r. \end{align*} Let $B_n(t, T) = \frac{1}{a_n}\Big(1-e^{-a_n(T-u)} \Big)$, and $B_r(t, T) = \frac{1}{a_r}\Big(1-e^{-a_r(T-u)} \Big)$. Then, \begin{align*} \int_t^T n_s ds &= B_n(t, T) n_t + \int_t^T \theta_n(u) B_n(u, T) du + \int_t^T \sigma_n B_n(u, T) dW_u^n, \end{align*} and \begin{align*} \int_t^T r_s ds &= B_r(t, T) r_t + \int_t^T (\theta_r(u)-\rho_{r,n}\sigma_n\sigma_r) B_r(u, T) du + \int_t^T \sigma_r B_r(u, T) dW_u^r. \end{align*} Moreover, \begin{align*} E\left(e^{\int_t^T (n_s-r_s) ds} \mid \mathcal{F}_t \right) &= e^{B_n(t, T) n_t-B_r(t, T) r_t+\int_t^T \theta_n(u) B_n(u, T) du -\int_t^T (\theta_r(u)-\rho_{r,n}\sigma_n\sigma_r) B_r(u, T) du}\\ &\quad \times e^{\frac{1}{2}\int_t^T \sigma_n^2 B_n(u, T)^2 du+\frac{1}{2}\int_t^T \sigma_r^2 B_r(u, T)^2 du - \int_t^T \rho_{r,n}\sigma_n\sigma_r B_n(u, T)B_r(u, T)du}. \end{align*} Moreover, \begin{align*} \int_t^T \sigma_n^2 B_n(u, T)^2 du &= -\frac{\sigma_n^2}{a_n^2}\big(B_n(t, T) -T+t\big)-\frac{\sigma_n^2}{2a_n}B_n(t, T)^2,\\ \int_t^T \sigma_r^2 B_r(u, T)^2 du &= -\frac{\sigma_r^2}{a_r^2}\big(B_r(t, T) -T+t\big)-\frac{\sigma_r^2}{2a_r}B_r(t, T)^2 \end{align*} and \begin{align*} \int_t^T B_n(u, T)B_r(u, T)du = \frac{1}{a_na_r}\left[T-t - B_n(t, T)-B_r(T, t)+\frac{1}{a_n+a_r}\left(1-e^{-(a_n+a_r)(T-t)}\right) \right]. \end{align*} Therefore, \begin{align*} E\left(e^{\int_t^T (n_s-r_s) ds} \mid \mathcal{F}_t \right) &= A_n(t, T) A_r(t, T) C(t, T) e^{B_n(t, T) n_t-B_r(t, T) r_t}, \end{align*} where \begin{align*} A_n(t, T) &= e^{\int_t^T \theta_n(u) B_n(u, T) du -\frac{\sigma_n^2}{2a_n^2}\big(B_n(t, T) -T+t\big)-\frac{\sigma_n^2}{4a_n}B_n(t, T)^2}, \\ A_r(t, T) &= e^{-\int_t^T (\theta_r(u)-\rho_{r,n}\sigma_n\sigma_r) B_r(u, T) du -\frac{\sigma_r^2}{2a_r^2}\big(B_r(t, T) -T+t\big)-\frac{\sigma_r^2}{4a_r}B_r(t, T)^2}, \end{align*} and \begin{align*} C(t, T) = e^{\frac{-\rho_{r,n}\sigma_n\sigma_r}{a_na_r}\left[T-t - B_n(t, T)-B_r(T, t)+\frac{1}{a_n+a_r}\left(1-e^{-(a_n+a_r)(T-t)}\right) \right]} \end{align*} Note that this is similar to the Hull-White zero-coupon bond pricing formula.

• Thanks. This is indeed similar to Hull and White. In fact, I started working from a framework "à la" Hull and White. I've also reworked all the steps. Learned some new and remembered a few tricks. I only have one question. Would it be possible for you to elaborate on the expression involving the covariance (i.e. \rho \sigma_n \sigma_r B_n B_r du) when doing the expectation? That's the only term I need to wrap my head around. It looks akin to what someone would do when calculating the variance of a difference (i.e. var(aX - bY) = a^2var(x) + b^2var(Y) - 2abcov(x,y)) but... – baluch_stan Dec 13 '16 at 15:49
• @baluch_stan: It is indeed the covariance term between $\int_t^T \sigma_n B_n(u, T) dW_u^n$ and $\int_t^T \sigma_r B_r(u, T) dW_u^r$ by noting that $E\left(\int_t^T \sigma_n B_n(u, T) dW_u^n \times \int_t^T \sigma_r B_r(u, T) dW_u^r \right) = \int_t^T \rho_{r, n} \sigma_n \sigma_r B_n(u, T) B_r(u, T) du$. – Gordon Dec 13 '16 at 15:57
• Yep. Actually, I worked it out like this: $$E(e^{\int_t^T\sigma_nB_ndW_u^n-\int_t^T\sigma_rB_rdW_u^r})=e^{\frac{1}{2}(\int_t^T\sigma_n^2B_n^2du + \int_t^T\sigma_r^2B_r^2du-2\int_t^T\sigma_nB_ndW_u^n\int_t^T\sigma_rB_rdW_u^r)}$$. And the last term simplifies using the fact that$$dW_u^ndW_u^r=\rho_{rn}du$$. Could be more rigorously developed I suppose but I think this works. Thanks again! – baluch_stan Dec 13 '16 at 16:08
• @baluch_stan: This is rigorous based on the Ito's cross isometry. – Gordon Dec 13 '16 at 16:10
• i knew the result but didn't know it was called the Ito isometry. you made my day ahah. – baluch_stan Dec 13 '16 at 16:13