Moment estimator for 4 parameters in some distribution function From Mathematische statistiek by A. van der Vaart:
"Let $X_1,X_2,\ldots,X_n$ be a random sample from the distribution function $x\rightarrow p\Phi(x-\mu)+(1-p)\Phi\bigl( (x-\nu)/\sigma\bigl)$. The parameters $p\in [0,1],\mu,\nu\in\mathbb{R}$ and $\sigma\in (0,\infty)$ are unknown. Construct a moment estimator for $(p,\mu,\nu,\sigma)$ and show that it is asymptotically normal."
We have to have an expectation, so what I did is taking the derivative of this distribution function to get a density. Then $E(X)$ is equal to $\int_{-\infty}^{\infty} x f(x) dx$ with $f$ the found density. It turns out this is equal to $\mu p + \frac{(1-p)\nu}{\sigma^2}$. But then I get stuck because this function in $(p,\mu,\nu,\sigma)$ is not injective so it is not possible to get a moment estimator. Then I tried $E(X^2)$ but that gives nothing better. I also don't see a link between $E(X)$ and $E(X^2)$, so interpreting it as a system of equations and eliminating also doesn't work.
Can anyone see a solution? I thank in advance, because I can't thank in the comments.
 A: $$\newcommand{\e}{\operatorname{E}}$$
Suppose $\Pr(Y=1) = p$ and $\Pr(Y=0) = 1-p.$
Suppose $X\mid (Y=1) \sim N(0,\mu)$ and $X\mid (Y=0) \sim N(\nu,\sigma^2).$
Then $X$ has just the distribution that was given.
$$
\e(X^n) = \e(X^n\mid Y=1)\Pr(Y=1) + \e(X^n\mid Y=0)\Pr(Y=0).
$$
Use that to find the first four moments.
The first moment should be $p\mu+(1-p)\nu.$
If the whole system of four functions of those four parameters is injective, then the estimator exists. You shouldn't expect any one of those four to be injective; only the whole system.
Postscript:
Let's look at $\e(X^n\mid Y=0).$
\begin{align}
\e(X^n\mid Y=0) & = \e( (\nu + \sigma Z)^n) \text{ where } Z\sim N(0,1).
\end{align}
One can expand $(\nu+\sigma Z)^n$ by the binomial theorem. If $n=4,$ this is $\nu^4 + 4\nu^3\sigma Z + 6\nu^2\sigma^2 Z^2 + 4\nu\sigma^3 Z^3 + \sigma^4 Z^4.$
So the problem is reduced to that of finding $\e(Z^k).$
$$
\e(Z^k) = \frac1 {\sqrt{2\pi}} \int_{-\infty}^\infty z^k e^{-z^2/2} \, dz = 0 \text{ if $k$ is odd, by symmetry.}
$$
So now assume $k$ is even. We will need only need $k=2$ and $k=4.$
\begin{align}
\e(Z^k) & = \frac1 {\sqrt{2\pi}} \int_{-\infty}^\infty z^k e^{-z^2/2} \, dz \\[10pt]
& = \sqrt{\frac 2 \pi} \int_0^\infty z^{k-1} e^{-z^2/2} (z\,dz) = \sqrt{\frac 2 \pi} \int_0^\infty \left(\sqrt{2u} \, \right)^{k-1} e^{-u} \, du \\[10pt]
& = \frac{2^{k/2}}{\sqrt{\pi}} \int_0^\infty u^{((k+1)/2) \, - \, 1} e^{-u} \, du = \frac{2^{k/2}}{\sqrt{\pi}} \Gamma\left( \frac{k+1} 2 \right).
\end{align}
If $k=4$ then we have
$$
\frac 4 {\sqrt \pi} \Gamma\left( \frac 5 2 \right) = \frac 4 {\sqrt \pi} \cdot \frac 1 2 \cdot \frac 3 2 \Gamma\left( \frac 1 2 \right) = 3 \qquad \text{ since } \Gamma\left( \frac 1 2 \right) = \sqrt \pi.
$$
Thus we have $\e(Z^4) = 3.$
It is widely known that $\e(Z^2) = 1.$ So
$$
\e((\nu+\sigma Z)^4) = \nu^4 + 6 \nu^2 \sigma^2 \e(Z^2) + \sigma^4 \e(Z^4) = \nu^4 + 6\nu^2\sigma^2 + 3\sigma^4.
$$
