variance of k element average of sequence I want to know how to theoretically find the variance of a sequence obtained by taking k element averages of another known sequence.
Lets say Sequence S1 has some N elements that are say uniformly distributed(for simplicity) between -1.0 and 1.0. Assuming N to be very large, the mean of the sequence will be 0(approximately) and variance will be $\frac{2^2}{12}$.
S1 -> ${a_1,a_2,a_3,a_4,.......,a_N}$
A new sequence S2 is formed by taking k element average of sequence S1.
S2 -> ${b_1,b_2,b_3,....,b_{N-k}}$
where $b_i = \frac{a_i + a_{i+1} + a_{i+2} + .... + a_{i+k}}{k}$
Intuitively I can say that the mean of S2 will also be 0(approximately). But what about the variance of S2. I remember there is some formula/method in probability and statistics that directly gives the mean and variance of the new sequence(S2) given the mean and variance of S1. But I don't remember it.
 A: When you add terms, the means and variances add.  Since the mean of each element is $0$, the mean of a sum is also $0$.  As you say, the variance of each element is $\frac {2^2}{12}=\frac 13$  The variance of the sum of $k$ terms is then $\frac k3$, so the variance of the average of $k$ terms is $\frac 1{3k}$.
A: First of all I think we need to clarify a little terminology. Assuming OP is concerning about the sample mean of the sequence:
$$ \bar{A}_N = \frac {1} {N} \sum_{i=1}^N A_i $$
Then as what OP said, the theoretical mean of this estimator is equal to the true population mean (i.e. it is an unbiased estimator).
$$ E[\bar{A}_N] = E[A_1]$$
When OP mentioned 

the mean of sequence will be $0$ (approximately)

it seems that OP is stating the sample mean of the sequence will converge to the theoretical mean in probability / a.s. as $N \to \infty$, by LLN
$$\bar{A}_N \stackrel {p} {\to} E[A_1]$$
From all the answers/comments, I think we just stick to the first statement, and thus the expectation is exact for every $N$, not only approximately. Please correct me if I misinterpret your question intention.
Now begins. This question reminds me of a $MA(k)$ model, and we are dealing with the sample mean
$$ \bar{B} = \frac {1} {N} \sum_{i=1}^N B_i $$
and
$$ B_i = \frac {1} {k} \sum_{j=0}^{k-1} A_{i+j}, i = 1, 2, \ldots, N - k + 1$$
As a $MA(k)$ model, those $B_i$ which within $k$-lags are correlated, so the variance of this sample mean $\bar{B}_N$ differs from the first example $\bar{A}_N$. Assuming $N > 2(k-1)$, then by simple counting,
$$ \bar{B} = \frac {1} {kN} \left[\sum_{i=1}^{k-1} iA_i + \sum_{i=k}^{N-k+1} kA_i + \sum_{i=N-k+2}^{N} (N-i+1)A_i\right]$$
As a result
$$ \begin{align} 
Var[\bar{B}] & = \frac {Var[A_1]} {k^2N^2}  \left[\sum_{i=1}^{k-1}i^2
+ (N-2k+2)k^2 + \sum_{i=N-k+2}^{N} (N-i+1)^2\right] \\
& = \frac {Var[A_1]} {k^2N^2}\left[k^2N - 2(k-1)k^2 + \frac {(k-1)k(2k-1)} {6} \times 2 \right] \\
& = \frac {Var[A_1]} {kN^2}\left[kN - 2(k-1)k + \frac {(k-1)(2k-1)} {3}\right] \\
& = \frac {Var[A_1]} {kN^2}\left[kN + \frac {(k-1)(2k-1) - 6(k-1)k} {3} \right] \\
& = \frac {Var[A_1]} {N} \left[1 - \frac{(k-1)(4k+1)} {3kN}\right]
\end{align}$$ 
