Deriving a Weighted Integral from the G2++ Ornstein–Uhlenbeck Solution

# Deriving the variance of G2++ Model


# Deriving the variance of G2++ Model












I'm studying G2++ Model in Brigo(2007)'s book.

The model constructed as follows,

$$ r(t) = x(t) + y(t) + φ(t), \quad r(0) = r_0\\ $$ with the dynamics of $dx(t)$ and $dy(t)$ described by: \begin{align} dx(t) &= -ax(t)dt + σdW_1(t), \quad x(0) = 0,\\ dy(t) &= -by(t)dt + ηdW_2(t),\quad y(0) = 0,\\ \end{align} and $dW_1(t)\cdot dW_2(t) = ρdt$.

Problem: When we calculate the variance, There is something that I cannot derive, which is described below:

$$∫^T_t(T-u)dx(u) = -a∫^T_t(T-u)x(u)du + σ∫^T_t(T-u)dW_1(u)$$ Then, $$∫^T_t(T-u)x(u)du = x(t)∫^T_t(T-u)e^{-a(u-t)}du + σ∫^T_t(T-u)∫^T_te^{-a(u-s)}dW_1(s)du.$$

I tried to derive the above integral, $\int_t^T (T-u) x(u) \: du$, but failed.

I want to know how to derive this equation. Especially, where in the world $e^{-a(u-t)}$ and $σ∫^T_t(T-u)∫^T_te^{-a(u-s)}dW_1(s)du$ came from, from the above equation?

##### Below picture is just the raw source of my text:

## Answer by Pleb (score 1, accepted)

https://quant.stackexchange.com/a/71111

### It comes from a direct application of the solution to $dx_t$

The solution for this SDE has already been derived by Gordon in this post where the only difference (in your specified SDE) is a sign-change in the drift-term. From the answer in the linked post, we observe that the solution to $dx_t$ in your case is given by:

$$ x_T = x_t e^{-a(T-t)} + \sigma\int_{t}^T e^{-a(T-s)} \: dW_s. $$

The derivations specified in Gordons answer involve the integrating factor method which is also used to derive the solution for the well-known Vasicek one-factor short-rate model.

Let $0\leq t < u < T$. Now we can compute the integral in question:

\begin{align} \int_t^T (T-u) \cdot x_u \: du &= \int_t^T (T-u) \cdot \left[x_t e^{-a(u-t)} + \sigma\int_{t}^u e^{-a(u-s)} \: dW_s \right] \: du \\ &=x_t \int_t^T (T-u) e^{-a(u-t)} \: du + \sigma\int_t^T (T-u)\int_{t}^u e^{-a(u-s)} \: dW_s \: du. \end{align}

In conclusion, $e^{-a(u-t)}$ and $\sigma\int_t^T (T-u)\int_{t}^u e^{-a(u-s)} \: dW_s \: du$ comes from inserting the solution of $dx_t$ in the above integral. I hope this provide some insight.