Method of Moments for Mixture distribution 
The restaurant owner also wants to reconfigure her seating layout, and has asked you for help in modeling her clients.
She gives you a dataset of past reservations, and tells you that she gets a mix of single clients who come to sit at the bar, and people who come (alone or in groups) for dinner.
You decide to model her clients using a mixture distribution, with fraction $\alpha$ of single bar patrons, and the remaining $(1-\alpha)$ fraction of diners; for the latter, you want to model the size of each group as $1+N_p$, where $N_p$ is Poisson distributed with parameter $\lambda$.


Compute the first and second moments.

The first moment is defined as $\mu_1 = E[X]$ and the second moment $\mu_2 = E[X^2]$. However, I am not sure what $X$ is in this case. Is it simply $X=1+N_p$, and then I take the expectation of this and the square of that same expression? In this case, I presume I have two equations for two unknowns and I solve for $\alpha$ and $\lambda$? Is this the case?
 A: The probability mass distribution for a mixture distribution of discrete distributions is like a sum of the probability mass distributions for the distributions in the mixture.
$$p(x) = \alpha p_1(x) + (1-\alpha) p_2(x) $$
when you compute the mean you get
$$\begin{array}{rcl}
E[X] &=& \sum_{ x} x p(x) \\
&=&  \sum_{ x } x (\alpha p_1(x) + (1-\alpha) p_2(x))\\
& =& \alpha \sum_{ x } x p_1(x) + (1-\alpha) \sum_{ x } x p_2(x)
\\ &=& \alpha E[X_1] + (1-\alpha) E[X_2]  \end{array}$$
when you compute the variance then you get
$$\begin{array}{rcl}
Var[X] &=& \sum_{ x} (x-E[X])^2 p(x) \\
\\ &=& \alpha \sum_{ x } (x-E[X])^2 p_1(x) + (1-\alpha) \sum_{ x } (x-E[X])^2 p_2(x)  \end{array}$$
these are not the variance of the original variables but instead the moment about a point (the point $E[X]$) which is not the original mean of the distributions.
Moment about a point
$$\begin{array}{rcl}
\sum_{x} (x-c)^2 p(x) &=& \sum_{x}(x-\mu_x+\mu_x-c)^2 p(x)\\
 &=& \sum_{x}\left((x-\mu_x)^2 + 2 (x-\mu_x)(\mu_x-c)+ (\mu_x-c)^2 \right)p(x)\\
 &=& \sum_{x}\left((x-\mu_x)^2 + (\mu_x-c)^2 \right)p(x)\\
 &=& Var(x) + (\mu_x-c)^2\\
\end{array}$$
Wrap up
$$E[X] = \alpha E[X_1] + (1 -\alpha) E[X_2]$$
$$Var[X] = \alpha Var[X_1] + (1 -\alpha) Var[X_2] + \alpha (E[X_1]-E[X])^2+ (1-\alpha) (E[X_2]-E[X])^2$$
