Given $n$ random variables of variance 1 with pairwise correlation $-1/(n-1)$ and the value of $n-1$ of the variables, can we recover the $n$th value? Let $X_i$ be $i=1$ to $n$ be random variables of variance 1 with pairwise correlation $\frac{-1}{n-1}$. Suppose we know the value of any $n-1$ of the $X_i$, can we recover the $n$th value?
For the case that $n=2$, this would be asking if when two random variables $X$ and $Y$ (both of which has variance 1) have correlation $-1$, can we determine the value of one of the variables given the other?
I understand how to construct random variables that have the specified correlation and satisfy the property that given $n-1$ of the values, we can deduce the $n$th value, but I'm not sure how to prove if this always holds (if it does).
For context, the quantity $\frac{-1}{n-1}$ is the minimum pairwise correlation of $n$ random variables.
 A: Since $$\text{Var}(X_1 + \cdots + X_n) = n + n(n-1) \left(-\frac{1}{n-1}\right) = 0$$
the sum of the $n$ random variables is a constant. If you know this constant, then yes knowing the realizations of $n-1$ of the random variables will give you the last one.
However, if we don't know this constant, without further information beyond the variances and correlations given in your question, I'm not sure we can do much. For instance, I believe the mean of each distribution can be freely set to be anything. Seeing a single realization of $n-1$ of the random variables would not tell you much about the last one at all, since variance and correlation are invariant under additive shifts.
A: As a concrete example of @angryavian's point, with $n=2$
If you take $X_1\sim N(0,1)$, you could have  $X_2=42-X_1$ or $X_2=69-X_1$ or $X_2=17-X_1$ and these would all have unit variance and correlation $-1$.  In general, $X_n = A-\sum_{i=1}^{n-1} X_i$, and $A$ can be chosen freely.
If you also required $E[X_i]=0$ there would be a unique answer
