# Are these Bernoulli variables independent?

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

$$$$P[\varepsilon_t\geq0\mid\varepsilon_1,\cdots,\varepsilon_{t-1},X]=P[\varepsilon_t<0\mid\varepsilon_1,\cdots,\varepsilon_{t-1},X]=\frac{1}{2}$$$$ Then the signs $$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)$$ should also be independent. However, if we write the likelihood function, we would not observe this $$$$l(s(\varepsilon_1+\beta x_1),\cdots,s(\varepsilon_n+\beta x_n)\mid X)=\\ \prod\limits_{t=1}^{n}P[\varepsilon_t\geq-\beta x_{t}\mid\varepsilon_1,\cdots,\varepsilon_{t-1},X]^{s(\varepsilon_t+\beta x_t)}P[\varepsilon_t<-\beta x_t\mid\varepsilon_1,\cdots,\varepsilon_{t-1},X]^{1-s(\varepsilon_t+\beta x_t)}$$$$ and as before since no assumptions exists on the median of $$\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

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 $$\varepsilon_i+\beta x_i>0$$.

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