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I was wondering if the $3\sigma$ rule that holds for 1D normal distribution also holds for multivariate normal distribution?

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Related:… – cardinal May 13 '12 at 2:35
up vote 6 down vote accepted

In the general case (multivariate with arbitrary covariance matrix), the natural generalization of the "normalized distance from the mean", $d = |x -u|/\sigma$, is given by the Mahalanobis distance

$$d = \sqrt{ ({\bf x} - {\bf \mu})^t {\bf \Sigma}^{-1} ({\bf x} - {\bf \mu})}$$

Points of constant Mahalanobis distance lie on an ellipsoid.

If (and only if) the components are independent and with same variance, then $d=\frac{\|{\bf x} - {\bf \mu}\|}{\sigma}$.

The threshold value that contains most (say, 99%) of the distribution varies with the dimension. Or, put in other way, the probability that $x$ takes a (Mahalanobis) distance less than $d=3$ decreases with the dimension. This figure, taken from here (which explains all this in more detail), displays that probability as a function of the dimension, for $d=2$ and $d=3$ (2-sigma and 3 sigma).

enter image description here

For example, we see that in 5 dimensions the probability that $x$ lies 'under 3 sigmas' is about 0.9, and for '2 sigmas' is around $0.4$

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Very nice point! – bgins May 10 '12 at 11:49

Yes, in a sense. For the 1D case, you need to standardize the normal variate to a standard normal variate $Z\sim N(0,1)$. For the multivariate normal distribution, each covariate must be not only normally distributed, but independent. You also need to know something more, the correlation matrix. Then, your question can be elegantly answered using the Gaussian copula. For the $8\times8$ intervals between $\{0,\pm1,\pm2,\pm3,\pm\infty\}$ in each coordinate, the 2D discrete probability distribution of discretized independent (zero correlation) standard normal variates would be: $$ \matrix{1.82\times10^{-6}&0.0000289&0.000183&0.000461&0.000461&0.000183&0.0000289&1.82\times10^{-6}\\0.0000289&0.000458&0.00291&0.00730&0.00730&0.00291&0.000458&0.0000289\\0.000183&0.00291&0.0185&0.0464&0.0464&0.0185&0.00291&0.000183\\0.000461&0.00730&0.0464&0.117&0.117&0.0464&0.00730&0.000461\\0.000461&0.00730&0.0464&0.117&0.117&0.0464&0.00730&0.000461\\0.000183&0.00291&0.0185&0.0464&0.0464&0.0185&0.00291&0.000183\\0.0000289&0.000458&0.00291&0.00730&0.00730&0.00291&0.000458&0.0000289\\1.82\times10^{-6}&0.0000289&0.000183&0.000461&0.000461&0.000183&0.0000289&1.82\times10^{-6}\\}$$

The above was generated in sage with the code below: NormalCDF is the normal cumulative density function $\Phi$, $Z=\{z_i\}$ is the set of $9$ boundary values given above (as an array), $p=\{\Phi(z_i)\}$, and $P=\{\Phi(z_i)-\Phi(z_{i-1})\}_{i=1}^8$ are the 1D probabilities of lying within each of the $8$ intervals bounded by these $9$ points, $M=(P_iP_j)$ a matrix representing the CDF of the 2D discrete distribution of lying within interval $(i,j)$, $N$ is an array of array of strings representing each $M_{ij}$ numerically approximated to $3$ places, and $L$ is a string representing $N$ as an unbracketed matrix in LaTeX. The last command displays the LaTeX within sage, assuming you have your worksheet set to display mathematical typesetting.

NormalCDF = lambda z: (1+sign(z))/2 if abs(z)==infinity else ((1+erf(z/sqrt(2)))/2).n()
Z = [-infinity]; Z.extend(range(-3,4)); Z.append(infinity)
p = [NormalCDF(z) for z in Z]
P = [p[i]-p[i-1] for i in range(1,len(p))]
M = Matrix(RDF,8,8,[[P[i]*P[j] for j in range(8)] for i in range(8)])
N = [[latex((P[i]*P[j]).n(digits=3)) for j in range(8)] for i in range(8)]
L = '\\matrix{' + (''.join(['& '.join(N[k])+'\\\\' for k in range(8)])) + '}'
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From the table above it seems that the %age of values lie within one standard deviation is not 68% as is the case with 1D? Moreover, is it fine to say that for 3$\sigma$ rule to hold, the each dimension should be independent and uncorrelated? Can you copy the 2D Gaussian function you have used for obtaining the above values. – shaikh May 10 '12 at 9:35
To recover the 1D rule, you need to sum rows or columns. If you sum the $3$rd and $4$th (rows/columns), you'll get $68$%. – bgins May 10 '12 at 10:21
Thank you very much for your answer. I have one confusion if you can help please. I read at several places that when there is no correlation, [the n-dimension multivariate Gaussian is simply the multiplication of n 1-dimensional Gaussian distribution][1]. When I am integrating the 2D function in the [document][1], i am getting only 46.6% within 1 standard deviation. [1]… – shaikh May 10 '12 at 10:58
Well, there must be an error somewhere, as your reading suggests. The density of a multivariate with independent univariates factors into the product of its marginals. I suspect you are integrating from $-\sigma$ to $\sigma$ in each dimension. This corresponds to the categorical variables both being in one of the two middle columns/rows, and has probability about $4\times11.7$%. This is correct, but it is a central, not a marginal, probability. Indeed, 4*M[3,3] yields $0.466064942674$. Marginal probabilites must sum an entire row or column, to eliminate a variable. – bgins May 10 '12 at 11:43
Yes, I was integrating from $-\sigma$ to $\sigma$ in each dimension. Actually I am interested in probability of 2D Gaussian distribution within some radius from the mean. That is why I integrated from $-\sigma$ to $\sigma$ in each dimension. Now I understood the difference between the central and marginal probability. Actually I need central probability for my problem. Thank you very much for such a nice explanation. – shaikh May 10 '12 at 12:07

It is often said that in high dimensions the probability distribution is concentrated away from the center. So although in 1 D a 3 sigma interval will contain more than 99% of the distribution a three sigma circle for a 2D gaussian with iid components will contain less mass than the 1 D counterpart and the same for 3D compared to 2D etc.

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