Being a programmer by nature I can never remember when looking at a math formula if the subscript that indicates which value in a series starts at 0 or 1. I know computer arrays start at 0 so that is my instinct without thinking but noticing 0 is never used in my formula my guess in this case is 1.
In a nutshell: For $X_i$ is the minimum $i$ 0 or 1?
Edit:
My context is that $X_i$ is the $i$th point in a recorded data series. Here is the formula I am using it in if it helps:
$$\sigma(\tau) = \sqrt{\frac{\sum_{j=1}^{N-3n-1} \left[\sum_{i = j}^{n + j + 1} (X_{i+2n} - 2X_{i+n} + X_{i})\right]^2 }{6n^2(N-3n-1)}}$$
Generally 1. A lot of times they tell you. For example, $\sum_{n = 1}^\infty \frac{\mu(n)}{3^n}$, it tells you right there you start with $n = 1$.
Besides $-\infty$, as the dark knight suggests, some formulas start with 2, but again, they give you enough information to know this is the case, e.g., $\sum_{p \leq N} \frac{1}{p}(p - 1) \ldots$ and then they say something like "where the sum runs over the primes."
There are also instances it might not make a difference, as for example, in the formula for triangular numbers, $T_n = \sum_{i = 0}^n i$, it would make no difference if you started with $i = 1$ instead, except of course it would kind of make $T_0$ undefined.
And for what it's worth, in Wolfram Mathematica, initial subscripts default to 1. If you were to tell it "Sum[MoebiusMu[n]/3^n, {n, Infinity}]," it would understand that n starts at infinity (although I think MoebiusMu[0] is defined in that program, and although that should not make any difference whatsoever to the final sum).
EDIT: Oh, I see. You've got nested sums, kind of like you're in Inception. At the first sum, your iterator $j$ should initialize at 1 (you could start it any value between 1 and $N - 3n - 1$ if you wanted to confuse yourself further; but seriously, start it at 1). Then, when you enter the inner sum, your iterator $i$ starts at the same value as $j$, which is 1. If it helps any, think of it as FOR loops in Basic or C++ or whatever.