I have a 60-40 weighted distribution, of uniform(0,7.5) and uniform(7.5,10) respectively, i.e. $$f_X(x)=(0.6/7.5)1_{x∈[0,7.5)}+(0.4/2.5)1_{x∈[7.5,1]}$$
I have worked out that $$E(X) = 0.6(7.5/2) + 0.4((10+7.5)/2) = 5.75$$ $$Var(X) = 9.0208$$
Right now this distribution is continuous, and I would like to make it discrete via the method of moment matching, with number of moments p = 2 and span h = 1.25.
How do I go about this?
I sort of understand how the equations work for p=1 (matching $m_0^k$ and $m_1^k$), but I'm not sure how to work it out for p=2.
*Notes:
Method of moment matching for arithmetizing a continuous distribution
We construct an arithmetic distribution that matches p moments of the arithmetic and the true severity distributions. Consider an arbitrary interval of length $ph$, denoted by $[x_k ; x_{k+ph})$. We locate point masses $m^k_0$, $m^k_1$, $m^k_2$, ... , $m^k_p$ at points $x_k$, $x_k + h$, ... , $x_k + ph$ so that the first p moments are preserved.
The system of p + 1 equations reflecting these conditions is $$\sum_{j=0}^p (x_k + jh)^rm^k_j = \int_{x_k - 0}^{x_k + ph - 0} x^r dF_X(x) - (*)$$
where r = 0,1,2,...,p and the notation “- 0” at the limits of the integral indicates that discrete probability at $x_k$ is to be included but discrete probability at $x_k + ph$ is to be excluded.
Arrange the intervals so that $x_k+1 = x_k + ph$ and so the endpoints coincide. Then the point masses at the endpoints are added together.
With $x_0 = 0$, the resulting discrete distribution has successive probabilities:
$f_0 = m_0^0$, $f_1 = m_1^0$, $f_2 = m_2^0$, ...
$f_p = m_p^0 + m_0^1$, $f_{p+1} = m_1^1$, $f_{p+2} = m_2^1$, ...
We need to solve the system of equations defined by $(*)$.
The solution of $(*)$ is $$m_j^k = \int_{x_k - 0}^{x_k + ph - 0} \prod_{i \neq j}\frac{x - x_k - ih}{(j-i)h}dF_X(x)$$ where j = 0,1,...,p
