Method of Moments for Merton’s Jump Diffusion Model

Under Merton’s Jump diffusion model, the stock price is assumed to follow \begin{align*} S_t = S_0 e^{\left(\mu - \frac{\sigma^2}{2}\right)t + \sigma W_t + \sum_{i=1}^{N_t}X_i}, \end{align*} where \(W_t\) is the standard Brownian motion, \(N_t\) is the Poisson process with parameter \(\lambda\), and \(X_i\)’s are iid normal random variables with mean \(\alpha\) and variance \(\beta^2\). Note that this model has 5 parameters to estimate: \(\mu, \sigma, \lambda, \alpha\) and \(\beta\). Given any integer \(n\) and a fixed time horizon \(T\), define \(\Delta t = T/n\) and \(t_i = i\Delta t\). Under Merton’s assumption, the log return \(R_i = \log(S_{t_{i+1}}/S_{t_i})\) can be shown to be iid with MGF \begin{align*} \phi_R(u) = E\left[e^{uR_i}\right] = \exp\left(\Delta t\left[u\left(\mu - \frac{\sigma^2}{2}\right) + \frac{\sigma^2 u^2}{2} + \lambda\left(e^{\alpha u + \frac{\beta^2 u^2}{2}} - 1\right)\right]\right), \end{align*} the \(j\)-th derivative of which at 0 gives the \(j\)th raw moment \(m_j\) of the log return. Below is the first 5: \begin{align*} m_1 &= \Delta t\left(\alpha\lambda+\mu-\frac{\sigma^2}{2}\right),\\ m_2 &= \Delta t\left(\alpha^2\lambda+\beta^2\lambda+\sigma^2+\Delta t\left(\alpha\lambda+\mu-\frac{\sigma^2}{2}\right)^2\right),\\ m_3 &= \Delta t\left(\alpha^3\lambda+3\alpha\beta^2\lambda+3\Delta t\left(\alpha^2\lambda+\beta^2\lambda+\sigma^2\right)\left(\alpha\lambda+\mu-\frac{\sigma^2}{2}\right)+\Delta t^2\left(\alpha\lambda+\mu-\frac{\sigma^2}{2}\right)^3\right),\\ m_4 &= \Delta t\left(\alpha^4\lambda+6\alpha^2\beta^2\lambda+3\beta^4\lambda+6\Delta t^2\left(\alpha^2\lambda+\beta^2\lambda+\sigma^2\right)\left(\alpha\lambda+\mu-\frac{\sigma^2}{2}\right)^2+4\alpha\lambda\Delta t\left(\alpha^2+3\beta^2\right)\left(\alpha\lambda+\mu-\frac{\sigma^2}{2}\right)+3\Delta t\left(\alpha^2\lambda+\beta^2\lambda+\sigma^2\right)^2+\Delta t^3\left(\alpha\lambda+\mu-\frac{\sigma^2}{2}\right)^4\right),\\ m_5 &= \Delta t\left(\alpha^5\lambda+10\alpha^3\beta^2\lambda+15\alpha\beta^4\lambda+10\Delta t^3\left(\alpha^2\lambda+\beta^2\lambda+\sigma^2\right)\left(\alpha\lambda+\mu-\frac{\sigma^2}{2}\right)^3+\frac{5}{2}\alpha\lambda\Delta t^2\left(\alpha^2+3\beta^2\right)\left(-2\alpha\lambda-2\mu+\sigma^2\right)^2+15\Delta t^2\left(\alpha^2\lambda+\beta^2\lambda+\sigma^2\right)^2\left(\alpha\lambda+\mu-\frac{\sigma^2}{2}\right)+10\alpha\lambda\Delta t\left(\alpha^2+3\beta^2\right)\left(\alpha^2\lambda+\beta^2\lambda+\sigma^2\right)+5\lambda\Delta t\left(\alpha^4+6\alpha^2\beta^2+3\beta^4\right)\left(\alpha\lambda+\mu-\frac{\sigma^2}{2}\right)+\Delta t^4\left(\alpha\lambda+\mu-\frac{\sigma^2}{2}\right)^5\right). \end{align*} Matching these formulas to the sample moments of the empirical log return, we get a high order polynomial equation (for example the last term of \(m_5\) is of degree 10), and the solution could easily be complex numbers. I did an experiment years ago fitting this model to the daily log return of S&P 500. The result is lost now but it does get complex numbers everywhere.

Closed Form Estimators

It might not be obvious but the MOM estimators in this situation can actually be found in closed form. Below are the key steps to derive the estimators. First denote by \(\hat m_j\) the sample moments from the empirical data. Our goal is to solve the following system of equations for the 5 model parameters \(\mu, \sigma, \lambda, \alpha\) and \(\beta\): \begin{align*} \hat m_1 &= \Delta t\left(\alpha\lambda+\mu-\frac{\sigma^2}{2}\right),\\ \hat m_2 &= \Delta t\left(\alpha^2\lambda+\beta^2\lambda+\sigma^2+\Delta t\left(\alpha\lambda+\mu-\frac{\sigma^2}{2}\right)^2\right),\\ \hat m_3 &= \Delta t\left(\alpha^3\lambda+3\alpha\beta^2\lambda+3\Delta t\left(\alpha^2\lambda+\beta^2\lambda+\sigma^2\right)\left(\alpha\lambda+\mu-\frac{\sigma^2}{2}\right)+\Delta t^2\left(\alpha\lambda+\mu-\frac{\sigma^2}{2}\right)^3\right),\\ \hat m_4 &= \Delta t\left(\alpha^4\lambda+6\alpha^2\beta^2\lambda+3\beta^4\lambda+6\Delta t^2\left(\alpha^2\lambda+\beta^2\lambda+\sigma^2\right)\left(\alpha\lambda+\mu-\frac{\sigma^2}{2}\right)^2+4\alpha\lambda\Delta t\left(\alpha^2+3\beta^2\right)\left(\alpha\lambda+\mu-\frac{\sigma^2}{2}\right)+3\Delta t\left(\alpha^2\lambda+\beta^2\lambda+\sigma^2\right)^2+\Delta t^3\left(\alpha\lambda+\mu-\frac{\sigma^2}{2}\right)^4\right),\\ \hat m_5 &= \Delta t\left(\alpha^5\lambda+10\alpha^3\beta^2\lambda+15\alpha\beta^4\lambda+10\Delta t^3\left(\alpha^2\lambda+\beta^2\lambda+\sigma^2\right)\left(\alpha\lambda+\mu-\frac{\sigma^2}{2}\right)^3+\frac{5}{2}\alpha\lambda\Delta t^2\left(\alpha^2+3\beta^2\right)\left(-2\alpha\lambda-2\mu+\sigma^2\right)^2+15\Delta t^2\left(\alpha^2\lambda+\beta^2\lambda+\sigma^2\right)^2\left(\alpha\lambda+\mu-\frac{\sigma^2}{2}\right)+10\alpha\lambda\Delta t\left(\alpha^2+3\beta^2\right)\left(\alpha^2\lambda+\beta^2\lambda+\sigma^2\right)+5\lambda\Delta t\left(\alpha^4+6\alpha^2\beta^2+3\beta^4\right)\left(\alpha\lambda+\mu-\frac{\sigma^2}{2}\right)+\Delta t^4\left(\alpha\lambda+\mu-\frac{\sigma^2}{2}\right)^5\right), \end{align*} where \(\Delta t\) and \(\hat m_j\) are known constants.

First note that there are some common terms on the right hand side of the equations. By applying the substitutions \begin{align*} A &= \alpha \lambda +\mu -\frac{\sigma^2}{2}, \\ B &= \alpha^2 \lambda +\beta^2 \lambda +\sigma^2, \\ C &= \alpha^3\lambda+3\alpha\beta^2\lambda, \\ D &= \alpha^4\lambda+6\alpha^2\beta^2\lambda+3\beta^4\lambda, \\ E &= \alpha^5\lambda+10\alpha^3\beta^2\lambda+15\alpha\beta^4\lambda, \end{align*} the system can be rewritten in a much cleaner form: \begin{align*} \hat m_1 &= \Delta tA,\\ \hat m_2 &= \Delta t\left(B + A^2\Delta t\right),\\ \hat m_3 &= \Delta t\left(C + 3AB\Delta t + A^3\Delta t^2\right),\\ \hat m_4 &= \Delta t\left(D + 6A^2B\Delta t^2 + 4AC\Delta t + 3B^2\Delta t + A^4\Delta t^3\right),\\ \hat m_5 &= \Delta t\left(E + 10A^3B\Delta t^3 + 10A^2C\Delta t^2 + 15AB^2\Delta t^2 + 10BC\Delta t + 5AD\Delta t + A^5\Delta t^4\right). \end{align*} Now the first equation can be used to solve for \(A\), and then the second equation can be used to solve for \(B\) given the value of \(A\). Using the same procedure, \(C, D\) and \(E\) can also be backed out one by one. So the problem reduces to solving the system of equations \begin{align*} A &= \alpha \lambda +\mu -\frac{\sigma^2}{2}, \\ B &= \alpha^2 \lambda +\beta^2 \lambda +\sigma^2, \\ C &= \alpha^3\lambda+3\alpha\beta^2\lambda, \\ D &= \alpha^4\lambda+6\alpha^2\beta^2\lambda+3\beta^4\lambda, \\ E &= \alpha^5\lambda+10\alpha^3\beta^2\lambda+15\alpha\beta^4\lambda, \end{align*} where \(A, B, C, D\) and \(E\) are known constants. We will focus on the last 3 equations now. Observe that \(E = \alpha D + 4\beta^2 C\) and hence \begin{align*} \beta^2 = \frac{E-\alpha D}{4C}. \end{align*} Substitute this into the \(C\) equation, isolate \(\lambda\) and simplify to get \begin{align*} \lambda = \frac{4C^2}{3E\alpha - 3D\alpha^2 + 4C\alpha^3}. \end{align*} Now in the \(D\) equation write all \(\beta^2\) and \(\lambda\) in terms of \(\alpha\) to get \begin{align*} D &= \lambda (\alpha^4+6\alpha^2\beta^2+3\beta^4)\\ &= \frac{4C^2\left(\alpha^4+6\alpha^2\left(\frac{E-\alpha D}{4C}\right)+3\left(\frac{E-\alpha D}{4C}\right)^2\right)}{3E\alpha - 3D\alpha^2 + 4C\alpha^3}\\ &= \frac{4C^2\alpha^4+6C\alpha^2\left(E-\alpha D\right)+\frac34\left(E-\alpha D\right)^2}{3E\alpha - 3D\alpha^2 + 4C\alpha^3}. \end{align*} Multiplying the denominator on both sides, we obtain \begin{align*} 4C^2\alpha^4+6C\alpha^2\left(E-\alpha D\right)+\frac34\left(E-\alpha D\right)^2 - D (3E\alpha - 3D\alpha^2 + 4C\alpha^3) = 0. \end{align*} This is a quartic equation of \(\alpha\) and the solution is known in closed form. The discriminant can give us some ideas whether the equation has real roots.

Once the value of \(\alpha\) is found, since \(\lambda\) and \(\beta\) can be written in terms of \(\alpha\), their values can also be found. Finally, these numbers can be used to back out \(\mu\) and \(\sigma\) using the \(A\) and \(B\) equations.