# On likelihood functions and characteristic functions

Let me preface this by saying that if someone manages to provide a solution to my problem, I will forever be indebted to them, as this problem has driven me crazy. Let us first assume that the discrete variables $$S_1,S_2,...,S_n$$ are independent and can be explained by an information matrix, say $$n\times k$$ matrix $$X=[x_0,x_1,...,x_{n-1}]$$, where $$X\in\mathbb{R}^{n\times k}$$. That would imply that their log-likelihood function can be expressed by $$\begin{eqnarray} L(\mathbf{S_n};\theta)=\log P\left[S_1=s_1,...,S_n=s_n\mid X\right]&=&\log\prod_{t=1}^{n}P\left[S_t=s_t\mid X\right]\\ &=&\sum\limits_{t=1}^{n}\log P\left[S_t=s_t\mid X\right] \end{eqnarray}$$ Now let us assume that a test statistic, say $$S_{n,\text{independent}}$$ is based on the above log-likelihood function. Therefore, the characteristic function of the statistic $$S_{n,\text{independent}}$$ w.r.t $$X$$ can be written as $$\begin{eqnarray} \phi_{S_{n,\text{independent}}}(u)&=&\mathbb{E}_{X}\left[\exp{\left(iuS_{n,\text{independent}}\right)}\right]\\ &=&\mathbb{E}_{X}\left[\exp{\left(iu\left\{\sum\limits_{t=1}^{n}\log P\left[S_t=s_t\mid X\right]\right\}\right)}\right]\\ &=&\mathbb{E}_{X}\left[\prod\limits_{t=1}^{n}\exp{\left(iu\left\{\log P\left[S_t=s_t\mid X\right]\right\}\right)}\right]\\ \end{eqnarray}$$ and due to independence of $$S_1,...,S_n$$

$$$$\phi_{S_{n,\text{independent}}}(u)=\prod\limits_{t=1}^{n}\mathbb{E}_{X}\left[\exp{\left(iu\left\{\log P\left[S_t=s_t\mid X\right]\right\}\right)}\right]$$$$

Dependent case:

Now consider an identical problem to the one before, where now $$S_1,...,S_n$$ are dependent. Then from given the chain rule in probability, the log-likelihood function will change to $$\begin{eqnarray} L(\mathbf{S_n};\theta)=\log P\left[S_1=s_1,...,S_n=s_n\mid X\right]\\ =\log\prod_{t=1}^{n}P\left[S_t=s_t\mid S_1=s_1,...,S_{t-1}=s_{t-1},X\right]\\ =\sum\limits_{t=1}^{n}\log P\left[S_t=s_t\mid S_1=s_1,...,S_{t-1}=s_{t-1} ,X\right] \end{eqnarray}$$ where now $$\begin{eqnarray} \phi_{S_{n,\text{dependent}}}(u)&=&\mathbb{E}_{X}\left[\exp{\left(iuS_{n,\text{dependent}}\right)}\right]\\ &=&\mathbb{E}_{X}\left[\exp{\left(iu\left\{\sum\limits_{t=1}^{n}\log P\left[S_t=s_t\mid S_1=s_1,...,S_{t-1}=s_{t-1}, X\right]\right\}\right)}\right]\\ &=&\mathbb{E}_{X}\left[\prod\limits_{t=1}^{n}\exp{\left(iu\left\{\log P\left[S_t=s_t\mid X\right]\right\}\right)}\right]\\ \end{eqnarray}$$. Am I correct that since $$S_1,...S_n$$ are no longer independent, such decomposition does not exist? $$$$\phi_{S_{n,\text{dependent}}}(u)\neq\prod\limits_{t=1}^{n}\mathbb{E}_{X}\left[\exp{\left(iu\left\{\log P\left[S_t=s_t\mid S_1=s_1,...,S_{t-1}=s_{t-1},X\right]\right\}\right)}\right]$$$$ Or could we decompose as above due to conditional independence?