# Standard deviation of means over two dimensions

(Apologies if the notations are "unusual", I'm not sure what the correct notations should be. I'm putting an example at the end of the question.)

Let's assume there was an initial dataset of an n by m matrix $M=(x_{ij})$ with $1<=i<=n$ and $1<=j<=m$, from which the following two vectors have been calculated:

• the vector of the column-wise means, that is, a vector of length $n$ where each element of index $j$ is the mean of the $m$ elements in column $j$ of the matrix: $(\bar{x}_{*j})$ with $1<=i<=n$, where $\bar{x}_{*j} = \frac{1}{m}\sum_{i=1}^m{x_{ij}}$.
• the corresponding vector of the columns-wise standard deviations ($s_{*j}$).

I'd like to get the mean and standard deviation of the row-wise vector of means, that is, the column vector made of the mean of each row: $(\bar{x}_{i*})$ with $1<=j<=m$, where $\bar{x}_{i*} = \frac{1}{n}\sum_{j=1}^n{x_{ij}}$, assuming that the initial matrix has been lost.

The mean of the elements in the row-wise vector is also the mean of the elements in the column-wise vector (which is also the mean of all the elements in the matrix, with equal weight): $\frac{1}{m}\sum_{i=1}^m{\bar{x}_{i*}} = \frac{1}{m}\sum_{i=1}^m({\frac{1}{n}\sum_{j=1}^n{x_{ij}}}) = \frac{1}{n\times m}\sum_{i=1}^m\sum_{j=1}^n{x_{ij}}$

Is there any way to get the standard deviation within the row-wise vector of means without having the original matrix?

For example:

$$M = \left(\begin{matrix} 1 & 2 & 3\\ 7 & 5 & 4\\ 8 & 2 & 3\\ 5 & 2 & 4 \end{matrix}\right)$$

$$(\bar{x}_{*j}) = \left(\begin{matrix}5.25 & 2.75 & 3.5\end{matrix}\right)$$

$$(s_{*j}) = \left(\begin{matrix} 2.68 & 1.29 & 0.5\end{matrix}\right)$$

$$(\bar{x}_{i*}) = \left(\begin{matrix} 2\\ 5.33 \\ 4.33 \\ 3.66\end{matrix}\right)$$

The mean of $\{2, 5.33, 4.33, 3.66\}$ is $3.83$ (also the mean of ${5.25, 2.75, 3.5}$). The standard deviation of $\{2, 5.33, 4.33, 3.66\}$ is $1.21$. Would there be any way of calculating this standard deviation, knowing $(\bar{x}_{*j})$, $(s_{*j})$, but without knowing $M$ nor $(\bar{x}_{i*})$?

Is there anything else that could have been calculated "column-wise" (that is, independently for each column, thus excluding the "row-wise" means themselves) that would help find out the (preferably sample) standard deviation of the "row-wise" means?

Thank you.

• I don't think the standard deviation of {2,5.33,4.33,3.66} is 1.21; I make it nearer 1.4 (or over 1.6 if you use a bias adjustment). – Henry May 13 '11 at 22:19
• The general flavor of the replies is that any set of column statistics that are invariant under permutations cannot supply information to deduce anything specific about individual rows (although certain bounds can be obtained). All conventional statistics are permutation invariant, including all moments and order statistics. If you were to view columns as if they were time series you could obtain some permutation-dependent statistics and then be able to recover information about individual rows. – whuber May 13 '11 at 22:53

In general, to recover the sum along the rows you'll need to recover the full matrix. Knowing the other sum (in your example, $(5.25 2.75 3.5)$ ) is not enough. This can be seen in matrix form. Essentially, you have $A U = M$ where A is the data matrix, U is a row column with ones (or 1/N) and M is the sums or means along the colums or rows. If you could augment U by adding other operations, so that U is square, you could invert U and recover A. For example, if you had not only the sum of the four rows, but the sum of the first three rows, and the first two rows ... etc... you'd have 4 vector of 3 components, and then you could recover your matrix.
$$M_2 = \left(\begin{matrix} 1 & 2 & 3\\ 5 & 2 & 3\\ 7 & 2 & 3\\ 8 & 5 & 4 \end{matrix}\right)$$