Thank you for the response Ackbach! It's a problem which I have to solve for my thesis in the area of financial engineering (calibration method: matching moments). If you want, I state the complete system. It consist of four equations and four unknowns $\lambda, \mu, \delta$ and $\sigma$:
$$\left \{ \begin{array}{rcl} \displaystyle v & = & T(\sigma^2+\lambda(\mu^2+\delta^2)) \\ s & = & \displaystyle \frac{\lambda (\mu^3+3\mu \delta^2)}{\sqrt{T}(\sigma^2+\lambda(\sigma^2+\delta^2))^{3/2}} \\
k & = & \displaystyle 3+\frac{\lambda(3\delta^4+6\delta^2\mu^4+\mu^4)}{T(\sigma^2+\lambda(\mu^4+\delta^2))^2} \\
w & = & \displaystyle \frac{\lambda (15 \delta^4 \mu + 10 \delta^2 \mu^3 + \mu^5)+10T\lambda(\mu^3+3\delta^2 \mu)(\sigma^2+\lambda(\mu^2+\sigma^2)}{T^{3/2}(\lambda(\delta^2+\mu^2)+\sigma^2)^{5/2}}
\end{array}\right. $$
where $v,s,k$ and $w$ are constants, more precisely they are measurements which I have. The unknowns $\mu \in \mathbb{R}$ and $\delta > 0$ are the parameters of a Normal distribution, $N(\mu, \delta^2)$. So as you can see in the second equation, the numerator is the third central moment of $N(\mu,\delta^2)$. Similarly, in the third equation the numerator is the fourth central moment. In the last equation the numerator also consist of the fifth central moment, a third central moment and an additional second central moment where $\sigma^2$ is added.
Moreover, all numerators are of the same form, using the first equation of the system they all can be written in function of $v/T$ (which is known). In that case my system reduces to:
$$\left \{ \begin{array}{rcl} \displaystyle v & = & T(\sigma^2+\lambda(\mu^2+\delta^2)) \\ \frac{s}{T} v^{3/2} & = & \displaystyle \lambda (\mu^3+3\mu \delta^2) \\
\frac{(k-3)v^2}{T} & = & \lambda(3\delta^4+6\delta^2\mu^4+\mu^4)\\
\frac{v^{5/2}}{T}(w-10s) & = & \displaystyle \lambda (15 \delta^4 \mu + 10 \delta^2 \mu^3 + \mu^5)
\end{array}\right. $$
and where I also used the first and second equation to further simplify the last equation. The system is now simplified. To summarize: $\mu \in \mathbb{R}, \sigma>0, \delta>0, \lambda>0$.
I know it's pretty messed up to solve. My idea was to solve the last two equations simultaneously so that I get an expression for $\mu$ and $\sigma$. From there I can use equation 2 to obtain an expression for $\lambda$. Finally equation one will then give me an expression for $\sigma$.
I will look for a free download/trial of Mathematica to try and solve with the given constraints on the parameters.
