I was reading a paper in which it was assumed that $\varepsilon_1,\cdots,\varepsilon_n$ conditional on $X$ possess serial (non-linear) dependence, such that

\begin{equation} P[\varepsilon_t\geq0\mid\varepsilon_1,\cdots,\varepsilon_{t-1},X]=P[\varepsilon_t<0\mid\varepsilon_1,\cdots,\varepsilon_{t-1},X]=\frac{1}{2} \end{equation} Then the signs $s(\varepsilon_1),\cdots,s(\varepsilon_n)$ are i.i.d and distributed as $Bi(1,0.5)$. Proof: We can write the likelihood function of the signs conditional on X as \begin{eqnarray} l(s(\varepsilon_1),\cdots,s(\varepsilon_n)\mid X)&=&\prod\limits_{t=1}^{n}P[\varepsilon_t\geq0\mid\varepsilon_1,\cdots,\varepsilon_{t-1},X]^{s(\varepsilon_t)}P[\varepsilon_t<0\mid\varepsilon_1,\cdots,\varepsilon_{t-1},X]^{1-s(\varepsilon_t)}\\ &=&\left(\frac{1}{2}\right)^{s(\varepsilon_t)}\left(\frac{1}{2}\right)^{1-s(\varepsilon_t)}=\left(\frac{1}{2}\right)^n \end{eqnarray} This holds for any combination of $t=1,\cdots,n$, if there is a permutation $\pi:i\rightarrow j$ such that the earlier assumption on the conditional median holds. Now instead lets assume we are interested in the signs $s(\varepsilon_1+\beta x_1),\cdots,s(\varepsilon_n+\beta x_n)$. Intuitively, conditional on $X$, as $\beta x_1,\cdots,\beta x_n$ are constant, and since $s(\varepsilon_1),\cdots,s(\varepsilon_n)$ are independent, then the signs $s(\varepsilon_1+\beta x_1),\cdots,s(\varepsilon_n+\beta x_n)$ should also be independent. However, if we write the likelihood function, we would not observe this \begin{equation} l(s(\varepsilon_1+\beta x_1),\cdots,s(\varepsilon_n+\beta x_n)\mid X)=\\ \prod\limits_{t=1}^{n}P[\varepsilon_t\geq-\beta x_{t}\mid\varepsilon_1,\cdots,\varepsilon_{t-1},X]^{s(\varepsilon_t+\beta x_t)}P[\varepsilon_t<-\beta x_t\mid\varepsilon_1,\cdots,\varepsilon_{t-1},X]^{1-s(\varepsilon_t+\beta x_t)} \end{equation} and as before since no assumptions exists on the median of $\varepsilon_t+\beta x_t$ conditional on its own past and $X$, then the joint probabilities $P[\varepsilon_t\geq-\beta x_{t}\mid\varepsilon_1,\cdots,\varepsilon_{t-1},X]$ vary across time. Is my conclusion correct that thus, the signs $s(\varepsilon_1+\beta x_1),\cdots,s(\varepsilon_n+\beta x_n)$ cannot be concluded to be independent? Or am I missing something here?

Thanks in advance.


1 Answer 1


Your conclusion is correct. If signs $s(\varepsilon_1),\cdots,s(\varepsilon_n)$ are i.i.d and distributed as $Bi(1,0.5)$, you cannot conclude that signs $s(\varepsilon_1+\beta x_1),\cdots,s(\varepsilon_n+\beta x_n)$ are independend (or identically distributed). For example, if $\beta x_i > 0$, it is possible that $\varepsilon_n+\beta x_i$ is always positive, whatever $s(\varepsilon_i)$ is. If all elements of $\beta x$ are positive, you can randomly select signs of $s(\varepsilon_1),\cdots,s(\varepsilon_n)$, and then select these variables in such way that if $\varepsilon_{i-1}+\beta x_{i-1}<0$ then $\varepsilon_i+\beta x_i>0$.

  • $\begingroup$ I couldn't follow the relevance of your final statement to the question. Could there be some typos in it? $\endgroup$
    – whuber
    Commented May 26, 2020 at 13:15
  • $\begingroup$ @whuber - yes, fixed. $\endgroup$
    – user31264
    Commented May 26, 2020 at 17:14

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