487. Find the method of moments estimates for the unknown parameters $\lambda_{1}$ and $\lambda_{2}$ of the "double Poisson distribution" based on the sample $x_{1}, x_{2}, \ldots, x_{n}$

$$
P\left(X=x_{i}\right)=\frac{1}{2} \cdot \frac{\lambda_{1}^{x_{i}}-\lambda_{1}}{x_{i}!}+\frac{1}{2} \cdot \frac{\lambda_{2}^{x_{i}}-\lambda_{2}}{x_{i}!}
$$

where $x_{i}$ is the number of occurrences of the event in $n_{i}$ trials with $\lambda_{i}$ and $\lambda_{2}$ being positive numbers, and $\lambda_{2}>\lambda_{1}$.
Solution. If the random variable $Z$ is distributed according to the Poisson law with parameter $\lambda$, then its theoretical initial moments of the first and second order are respectively (see problems 207, 227):

$$
\begin{aligned}
& v_{1}=M(Z)=\lambda \\
& v_{2}=M\left(Z^{2}\right)=\lambda+\lambda^{2}
\end{aligned}
$$

Let's find the theoretical initial moments of the first and second order of the considered random variable $X$, taking into account the relationships ( ${ }^{2}$ ):

$$
\begin{gathered}
v_{1}=M(X)=\lambda_{1} / 2+\lambda_{2} / 2=\left(\lambda_{1}+\lambda_{2}\right) / 2 \\
v_{2}=M\left(X^{2}\right)=(1 / 2)\left(\lambda_{1}+\lambda_{1}^{2}\right)+(1 / 2)\left(\lambda_{2}+\lambda_{2}^{2}\right)=v_{1}+\left(\lambda_{1}^{2}+\lambda_{2}^{2}\right) / 2
\end{gathered}
$$

From this

$$
\left\{\begin{array}{l}
\lambda_{1}+\lambda_{2}=2 v_{1} \\
\lambda_{1}^{2}+\lambda_{2}^{2}=2 v_{2}-2 v_{1}
\end{array}\right.
$$

Solving this system for the unknown parameters, taking into account that $\lambda_{2}>\lambda_{1}$, we get:

$$
\lambda_{1}=v_{1}-\sqrt{v_{2}-v_{1}-v_{1}^{2}}, \lambda_{2}=v_{1}+\sqrt{v_{2}-v_{1}-v_{1}^{2}}
$$