Let $\textbf{x} = (x_1, \dots, x_n)$ a real vector,
We define the vector $\textbf{y} = (y_1, \dots, y_n)$ following this protocol:
- We take a subset of $\{1, \dots, n\}$ of cardinal $n-l$, we denote this subset by $S_l$, $\forall i \in S_l: y_i = x_i$.
For the remaining $l$ elements of $\{1, \dots, n\}$: Let $a \in [-1,1]$, independently:
$y_i = x_i$ with probability $\frac{1+a}{2}$
$y_i = - x_i$ with probability $\frac{1-a}{2}$
We will say that $\textbf{y} \sim \mathcal{N}_{a,l}(\textbf{x})$ if the vector $ \textbf{y}$ was generated following the protocol above.
My question is: What is $\underset{\textbf{y} \sim \mathcal{N}_{a,l}(\textbf{x})}{\mathbb{E}}[ \underset{i \in S}{\prod} y_i ]$ for some subset $S \subseteq [n]$
My attempt:
$\underset{\textbf{y} \sim \mathcal{N}_{a,l}(\textbf{x})}{\mathbb{E}}[ \underset{i \in S}{\prod} y_i ] = \underset{i \in S}{\prod} \underset{\textbf{y} \sim \mathcal{N}_{a,l}(\textbf{x})}{\mathbb{E}}[ y_i ]$ (independence)
There's $\binom{l}{|S|}$ ways to choose $|S| -l$ elements from the set $|S|$
Hence, the expectation is: $\underset{\textbf{y} \sim \mathcal{N}_{a,l}(\textbf{x})}{\mathbb{E}}[ \underset{i \in S}{\prod} y_i ] = \underset{i \in S}{\prod} ( \binom{l}{|S|} x_i + \frac{1+a}{2} x_i - \frac{1-a}{2} x_i )$
And this gives: $\underset{\textbf{y} \sim \mathcal{N}_{a,l}(\textbf{x})}{\mathbb{E}}[ \underset{i \in S}{\prod} y_i ] = ( \binom{l}{|S| + a} + a)^{|S|} \underset{i \in S}{\prod} x_i $
However, I'm not sure about the first part of the expectation ($\binom{l}{|S|} x_i$). It seems like the protocol is divided into a deterministic way of fixing $n-l$ elements of the vector $\textbf{x}$ and the other part is random. How would the expectation be expressed in this case ?