$n$ boolean coordinates given parity equivalent to $n-1$ boolean coordinates For a given $x\in\left\{ 0,1\right\} ^{n}$, define $x_{\setminus k}=\left(x_{1},\dots,x_{k-1},x_{k+1},\dots,x_{n}\right)$. [Is there standard notation for this?]
Prove or disprove the following claim:
Let $X$ be some random variable with values in $\left\{ 0,1\right\} ^{n}$. There exists $k\in\left[n\right]$ s.t for all $y\in\left\{ 0,1\right\} ^{n}$,
$$Pr\left(X=y\vert\bigoplus_{i=1}^{n}X_{i}=\bigoplus_{i=1}^{n}y_{i}\right)=Pr\left(X_{\setminus k}=y_{\setminus k}\right)$$
Where $\oplus$ is the XOR function.
Intuitively the claim sounds true, since for $n-1$ boolean coordinates and a parity of 1's there exists a unique $n$-coordinate boolean vector.
 A: The question is not well posed since it uses a conditional probability.
Let $\pi(x) = \bigoplus_{i=1} x_i$
If $n=2$, $X \in \{ (0,0), (1,1) \}$
with equal probability and $y=(0,1)$, then
$P[\pi(X) = \pi(y)] = 0$
A: It's not really an issue of conditional probability being defined or not. I mean, that's a problem, but it's not the only problem.
As in copper.hat's answer, define $\pi(x) = \bigoplus_{i=1}^n x_i$. Then we have, for any $k$, that $X=y$ holds if and only if $X_{\setminus k} = y_{\setminus k}$ and $\pi(X) = \pi(y)$, because $\pi$ allows us to reconstruct $x_k$ given $y_k$. Therefore, if $\pi(X) =\pi(y)$ happens with positive probability, we have
$$\Pr[X=y \mid \pi(x)=\pi(y)] = \Pr[X_{\setminus k} = y_{\setminus k} \mid \pi(x) = \pi(y)].$$But there is no reason to expect the second probability to equal $\Pr[X_{\setminus k} = y_{\setminus k}]$, without that conditional probability.

For a simple counterexample, again take $n=2$, and suppose that $X$ is chosen uniformly at random from $\{(0,0), (0,1), (1,0)\}$. Let $y=(0,0)$.
Then $\Pr[X = y \mid \pi(x) = \pi(y)] = 1$, because $\pi(y)=0$ and the only possible value of $X$ with $\pi(X) = 0$ is $(0,0)$. However, neither $k=1$ nor $k=2$ satisfy the condition we want:


*

*$\Pr[X_{\setminus 1} = y_{\setminus 1}] = \Pr[X_2 = 0] = \frac23$.

*$\Pr[X_{\setminus 2} = y_{\setminus 2}] = \Pr[X_1 = 0] = \frac23$.

