Variance of the log returns in jump diffusion with time-varying jump sizes

Question

I'm trying to calculate the variance $\mathrm{var}\left(\log\frac{S\left(t\right)}{S\left(0\right)}\right)$, where the dynamics of the stock $S$ follows a jump-diffusion process given by $$\frac{dS\left(t\right)}{S\left(t-\right)} = \left(\alpha-\lambda \kappa\right)dt+\sigma dZ\left(t\right)+\left(Y\left(t\right)-1\right)dN\left(t\right),$$ where the jumps are driven by an independent Poisson process $N\left(t\right)$ with constant intensity $\lambda$ and random jump size $Y$, and the jump sizes are lognormally distributed with parameters $\mu\left(t\right)$ and $\delta\left(t\right)$. $\kappa = E\left(Y\left(t\right)-1\right)$ is the expected relative jump of $S\left(t\right)$.

If $\mu\left(t\right)=\mu$ and $\delta\left(t\right)=\delta$, this is just a regular Merton jump-diffusion model, and e.g. Navas (2003) shows that the variance is $$\begin{align}\mathrm{var}\left(\log\frac{S\left(t\right)}{S\left(0\right)}\right) &=\mathrm{var}\left(\sigma Z\left(t\right)\right) + \mathrm{var}\left(\log Y\left(n\left(t\right)\right)\right) \\ &= t \sigma ^2 + t \lambda \left(\mu^2 + \delta^2 \right),\end{align}$$ because the Poisson process is independent of the diffusion.

I'm trying to calculate this quantity if the jump sizes and volatilities are not constant and, instead, are both functions of time $t$, as discussed above. I've been following Navas's derivation: $$\begin{align}\mathrm{var}\left(\log Y\left(n\left(t\right)\right)\right) &= -E \left(\log Y\left(n\left(t\right)\right)\right)^2 + E \left(\left(\log Y\left(n\left(t\right)\right)\right)^2\right) \\ &= -\left(\lambda\int \mu\left(t\right) dt \right)^2+ E \left(\left(\log Y\left(n\left(t\right)\right)\right)^2\right), \end{align}$$ but I'm stuck on the last expectation. Would anyone be able to help, please?

EDIT (after bountying this): the functions $\mu\left(t\right)$ and $\sigma\left(t\right)$ are, in my example, piecewise functions in the style of $$\mu\left(t\right) = \begin{cases} \mu_1, & 0 < t \leq 1 \\ \mu_2, & 1 < t \leq 2 \\ \vdots & \vdots \\ \mu_T, & T-1 < t \leq T. \end{cases}$$

Reference:

Navas, Javier F., Calculation of Volatility in a Jump-Diffusion Model. Journal of Derivatives, Vol. 11, No. 2, 2003, Available at SSRN: https://ssrn.com/abstract=1031196

Answer 1

Note: Time-dependent parameters can be introduced quite easily into affine jump diffusion models. Even if the corresponding (time) integrals cannot be solved in closed form, option pricing and moment estimation can always be performed, up to the solution of two ODEs.

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In my answer, I follow an ansatz that involves the moment generating function (mgf) of the return process in your (generalised) Merton model. In order to find the mgf, we make use of the machinery in Duffie, Pan, Singleton (2000). Once the mgf is found, we compute the first and second moment of the return distribution in order to find the log-return variance.

Step 1: Obtaining the log-return process

Without loss of generality, let's set $S_0=1$ and find the process for $y\equiv \log(S)$:

$$ dy=d\log(S)=\left(\alpha-\lambda\kappa(t)-\frac{1}{2}\sigma^2\right)dt+\sigma dZ(t)+Y(t)dN(t) $$

In the (generalized) Merton model, gross return jumps are distributed lognormally and log-return jumps are hence distributed normally, $Y(t)\sim \mathrm{N}\left(\mu(t),\delta(t)\right)$. The time-dependent parameter $\kappa(t)$ is, of course, not to be transformed and equals $\kappa(t)=e^{\mu(t)+\frac{1}{2}\delta(t)^2}-1$.

Step 2: The moment generating function of the return process

Duffie, Pan, Singleton (2000) lay out a straightforward path for the calculation of conditional expectations of affine jump diffusion models (with time-varying parameters). We shall use these methods and find the moment generating function by solving some differential equations. In all brevity, DPS2000 offer a framework to solve expectations of the (simplified) form

$$ \mathrm{E}\left(e^{uy_T}|\mathcal{F}_t\right)=e^{a(t)+b(t)y_t} $$ At closer inspection, we see that this is the moment generating function of $y_T$. This expectation is solved by cleverly solving ODEs for the parameters $a(t),b(t)$, subject to the boundary condition $a(T)=0, b(T)=u$

The solution for $b$ is, simply, $b(t)=u$. The solution for $a$ is found by solving the following ODE:

$$ \frac{\mathrm{d}a}{\mathrm{d} t}=-u\left(\alpha-\lambda\kappa(t)-\frac{1}{2}\sigma^2\right)-\frac{1}{2}u^2\sigma^2-\lambda\left(e^{u\mu(t)+\frac{1}{2}u^2\delta(t)^2}-1\right) $$

Letting $\tau = T-t$, the solution is

$$ a=u\tau\left(\alpha-\frac{1}{2}\sigma^2\right)+\frac{1}{2}u^2\sigma^2\tau-\lambda u\sum_i\Delta_i\kappa_i+\lambda\sum_i\Delta_i\left(e^{u\mu_i+\frac{1}{2}u^2\delta_i^2}-1\right) $$ where we let $\kappa_i=e^{\mu_i+\frac{1}{2}\delta_i^2}-1$ and $\Delta_i=t_i-{t_{i-1}}$ the duration during which the parameter is 'valid'. In your example, $\Delta_i=\Delta=1$.

We have thus found the moment generating function of $y_T$ $$ \mathrm{M}_{y_T}(u)=e^{a(t,T,u)+uy_t} $$

The moments of $y_T\equiv\log S_T$

Given the moment-generating function, we can find the $k$th moment of the return distribution as:

$$ \mathrm{E}\left(y_T^k\right)=\left.\frac{\partial^k \mathrm{M}_{y_T}\left(u\right)}{\partial u^k}\right|_{u=0} $$

The first derivative of $M$, evaluated at $u=0$ is the first moment:

$$ \mathrm{E}(y_T)=\left(\alpha-\frac{1}{2}\sigma^2\right)\tau-\lambda\sum_i\Delta_i\kappa_i+\lambda\sum_i\Delta_i\mu_i $$

The second derivative of $M$, evaluated at $u=0$ is the second moment:

$$ \mathrm{E}(y_T^2)=\left[\left(\alpha-\frac{1}{2}\sigma^2\right)\tau-\lambda\sum_i\Delta_i\kappa_i+\lambda\sum_i\Delta_i\mu_i\right]^2+\sigma^2\tau+\lambda\sum_i\Delta_i\left(\mu_i^2+\delta_i^2\right) $$

The variance is found as

$$ \mathrm{Var}(y_T^2)=\mathrm{E}(y_T^2)-\mathrm{E}(y_T)^2=\sigma^2\tau+\lambda\sum_i\Delta_i\left(\mu_i^2+\delta_i^2\right) $$

Comparing this to the result of the standard Merton model, we find that the variance contributed by the jump component is the weighted sum of the piecewise components.