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@psboonstra This is a valid point. After posting it, I too realized that the question is ill-posed. I attempted to oversimplify a problem that I had encountered in finding a characteristic function of a Neyman-Pearson type test-statistic, for testing the orthogonality hypothesis for a predictive regression of the form $y_t=\beta x_{t-1}+u_t$. Say the Neyman-Pearson type test statistic is of the form $S_T(\beta_0\mid\beta_1)=\sum_{t=1}^TS_t(\beta_0\mid\beta_1)$, where $S_i(\beta_0\mid\beta_1)$ are based on log-likelihood ratios that are expressed in terms of $X$. Then the characteristic function is \begin{equation} \mathbb{E}_X\left[e^{iuS_T(\beta_0\mid\beta_1)}\right]=\mathbb{E}_X\left[\prod\limits_{t=1}^Te^{iuS_t(\beta_0\mid\beta_1)}\right] \end{equation}. If $y_1,y_2,...,y_T$ are dependent and $S_1(\beta_0\mid\beta_1),S_2(\beta_0\mid\beta_1),...,S_T(\beta_0\mid\beta_1)$ are expressed as $\log P_1(Y_1\leq 0), \log P_2(Y_2\leq0\mid Y_1\leq 0),...$ which can alternatively be expressed as $\log P_1(u_1\leq -\beta x_0), \log P_2(u_2\leq -\beta x_1\mid u_1\leq -\beta x_0),...$ does \begin{equation} \mathbb{E}_X\left[\prod\limits_{t=1}^Te^{iuS_t(\beta_0\mid\beta_1)}\right]=\prod\limits_{t=1}^T\mathbb{E}_Xe^{iuS_t(\beta_0\mid\beta_1)} \end{equation}

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  • $\begingroup$ The question seems ill-posed. The quantity $P(Y_t = y_t)$ is a function of parameters, which are fixed constants. Thus, the answer to your question would seem to be 'yes, trivially so because the expectation of any constant with respect to any distribution is the constant itself'. Under what distribution are you taking your expectation in your second line of math? $\endgroup$ – psboonstra Nov 27 '19 at 13:44
  • $\begingroup$ @psboonstra Thank you for you reply. The question was indeed ill-posed and I have provided some clarity. I would ignore the first part of the question that had been written earlier and look at the edit as a new question, as the previous version may only cause confusion. $\endgroup$ – Carl Nov 27 '19 at 16:21
  • $\begingroup$ @psboonstra I infact made some more additional changes to make it clear, though I would not be surprised if it is still confusing. $\endgroup$ – Carl Nov 27 '19 at 16:37
  • $\begingroup$ Can you confirm that your second-to-last equation $E_X\left[e^{itS_T(\beta_0|\beta_1)}\right] = E_X\left[\prod_{t=1}^Te^{itS_t(\beta_0|\beta_1)}\right]$ is written as intended? Specifically, there is a variable $t$ in the left-hand expression that is used as an indexing variable in the product in the right-hand expression, which makes the comparison misleading. $\endgroup$ – psboonstra Nov 27 '19 at 16:51
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    $\begingroup$ You lost me at the outset, because the right hand side is not a random variable--it's a number--so what are you trying to take the expectation of?? $\endgroup$ – whuber Nov 27 '19 at 19:27

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