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I would like a clarification for the following equivalence:

$$f(x_1,i_1,...,x_n,i_n|\theta, \pi)=$$ $$\prod_{j=1}^n[f(x_j;\theta))^{i_j}\pi^{i_j}(1-\pi)^{1-i_j}]=[\prod_{j=1}^n f(x_j;\theta)^{i_j}]\pi^m(1-\pi)^{n-m}$$

where $I_j\text{~}Bin(1,\pi)$ is a r.v. such that $I_j=1$ if observation $X_j$ has been made and $I_j=0$ if observation $X_j$ is missing.

An observed sample is $(X_1,I_1),...,(X_n,I_n)$ and the above is the joint distribution of such sample.

What I don't understand is the index change to $m$ and $n$, where $m$ becomes the number of observations and $n-m$ the number of missing observations.

mavavilj
  • 4,109

1 Answers1

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$$\begin{align*} \prod_{j=1}^n\left[f(x_j;\theta))^{i_j}\pi^{i_j}(1-\pi)^{1-i_j}\right] &=\prod_{j=1}^nf(x_j;\theta))^{i_j}\times\prod_{j=1}^n[\pi^{i_j}(1-\pi)^{1-i_j}]\\ &=\prod_{j=1}^nf(x_j;\theta))^{i_j}\times\pi^{\sum_{j=1}^ni_j}\times(1-\pi)^{\sum_{j=1}^n(1-i_j)}\\ &=\prod_{j=1}^nf(x_j;\theta))^{i_j}\times\pi^{\sum_{j=1}^ni_j}\times(1-\pi)^{n-\sum_{j=1}^ni_j}\\ &=\prod_{j=1}^n f(x_j;\theta)^{i_j}\times\pi^m\times(1-\pi)^{n-m} \end{align*}$$ when$$m=\sum_{j=1}^ni_j$$

Xi'an
  • 105,342