Expected Inflation from Correlated Gaussian Short-Rate Processes
Summary
The document derives a conditional expectation of cumulative inflation when nominal and real short rates follow correlated mean-reverting Gaussian diffusions. It first writes each rate as its initial value, an integrated drift contribution, and a stochastic integral. Integrating those representations over the horizon gives the rate integrals in a form that can be combined into the inflation expression.
Because the resulting exponent is Gaussian, its expectation is an exponential of the mean plus half the variance. The formula includes a cross term from the correlation between the Brownian shocks, so the nominal and real rate expectations cannot generally be calculated separately and multiplied as if independent. The answer organizes the result into rate-specific factors and a correlation adjustment, and notes its resemblance to a Hull–White bond-pricing formula. The derivation depends on the stated diffusion model and parameterization; it is not a general formula for arbitrary rate processes, and the displayed expressions should be checked carefully for notation or transcription errors before implementation.
Key ideas
- Integrated mean-reverting Gaussian rates can be expressed using horizon-dependent loading functions.
- The exponential expectation uses the Gaussian mean and one half of its variance.
- Correlation between nominal and real rate shocks contributes a covariance adjustment.
- The result has an affine exponential form similar to Gaussian short-rate bond pricing.
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# Expectation of two correlated processes?
# 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?
## Answer by Gordon (score 6, accepted)
https://quant.stackexchange.com/a/31451
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.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.