The Multivariate Gaussian Distribution
Chuong B. Do October 10, 2008
A vector-valued random variable X = X1 · · · Xn T is said to have a multivariate normal (or Gaussian) distribution with mean μ ∈ Rn and covariance matrix Σ ∈ Sn++1 if its probability density function2 is given by
p(x;μ,Σ) = 1 exp−1(x−μ)TΣ−1(x−μ). (2π)n/2|Σ|1/2 2
We write this as X ∼ N (μ, Σ). In these notes, we describe multivariate Gaussians and some of their basic properties.
1 Relationship to univariate Gaussians
Recall that the density function of a univariate normal (or Gaussian) distribution is
given by
p(x;μ,σ2)= √1 exp− 1 (x−μ)2. 2πσ 2σ2
Here, the argument of the exponential function, − 1 (x − μ)2, is a quadratic function of the 2σ2
variable x. Furthermore, the parabola points downwards, as the coefficient of the quadratic term is negative. The coefficient in front, √ 1 , is a constant that does not depend on x;
2πσ
hence, we can think of it as simply a “normalization factor” used to ensure that
√1 ∞exp−1(x−μ)2=1. 2πσ −∞ 2σ2
1Recall from the section notes on linear algebra that Sn++ is the space of symmetric positive definite n×n matrices, defined as
Sn++ =A∈Rn×n :A=AT andxTAx>0forallx∈Rn suchthatx̸=0.
2In these notes, we use the notation p(•) to denote density functions, instead of fX(•) (as in the section notes on probability theory).
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0 1 2 3 4 5 6 7 8 9 10
0.02
0.015
0.01
0.005
0 10
5 10
0
−5
5
−10 −10
−5
0
Figure 1: The figure on the left shows a univariate Gaussian density for a single variable X. The figure on the right shows a multivariate Gaussian density over two variables X1 and X2.
In the case of the multivariate Gaussian density, the argument of the exponential function,
−21(x−μ)TΣ−1(x−μ), is a quadratic form in the vector variable x. Since Σ is positive definite, and since the inverse of any positive definite matrix is also positive definite, then for any non-zero vector z, zT Σ−1z > 0. This implies that for any vector x ̸= μ,
(x − μ)T Σ−1(x − μ) > 0 − 21 ( x − μ ) T Σ − 1 ( x − μ ) < 0 .
Like in the univariate case, you can think of the argument of the exponential function as being a downward opening quadratic bowl. The coefficient in front (i.e., 1 ) has an
(2π)n/2|Σ|1/2
even more complicated form than in the univariate case. However, it still does not depend
on x, and hence it is again simply a normalization factor used to ensure that
1 ∞ ∞ ··· ∞ exp−1(x−μ)TΣ−1(x−μ)dx1dx2···dxn =1.
(2π)n/2|Σ|1/2 −∞ −∞ −∞ 2 2 The covariance matrix
The concept of the covariance matrix is vital to understanding multivariate Gaussian distributions. Recall that for a pair of random variables X and Y , their covariance is defined as
Cov[X,Y]=E[(X−E[X])(Y −E[Y])]=E[XY]−E[X]E[Y].
When working with multiple variables, the covariance matrix provides a succinct way to summarize the covariances of all pairs of variables. In particular, the covariance matrix, which we usually denote as Σ, is the n×n matrix whose (i,j)th entry is Cov[Xi,Xj].
2
The following proposition (whose proof is provided in the Appendix A.1) gives an alter- native way to characterize the covariance matrix of a random vector X:
Proposition 1. For any random vector X with mean μ and covariance matrix Σ,
Σ = E[(X − μ)(X − μ)T ] = E[XXT ] − μμT . (1)
In the definition of multivariate Gaussians, we required that the covariance matrix Σ be symmetric positive definite (i.e., Σ ∈ Sn++). Why does this restriction exist? As seen in the following proposition, the covariance matrix of any random vector must always be symmetric positive semidefinite:
Proposition 2. Suppose that Σ is the covariance matrix corresponding to some random vector X. Then Σ is symmetric positive semidefinite.
Proof. The symmetry of Σ follows immediately from its definition. Next, for any vector z ∈ Rn, observe that
nn
zTΣz = (Σijzizj) (2)
i=1 j=1 nn
= (Cov[Xi,Xj]·zizj) i=1 j=1
nn
=(E[(Xi −E[Xi])(Xj −E[Xj])]·zizj)
i=1 j=1
nn
=E (Xi −E[Xi])(Xj −E[Xj])·zizj . (3) i=1 j=1
Here, (2) follows from the formula for expanding a quadratic form (see section notes on linear algebra), and (3) follows by linearity of expectations (see probability notes).
To complete the proof, observe that the quantity inside the brackets is of the form i j xixj zizj = (xT z)2 ≥ 0 (see problem set #1). Therefore, the quantity inside the expectation is always nonnegative, and hence the expectation itself must be nonnegative. We conclude that zT Σz ≥ 0.
From the above proposition it follows that Σ must be symmetric positive semidefinite in order for it to be a valid covariance matrix. However, in order for Σ−1 to exist (as required in the definition of the multivariate Gaussian density), then Σ must be invertible and hence full rank. Since any full rank symmetric positive semidefinite matrix is necessarily symmetric positive definite, it follows that Σ must be symmetric positive definite.
3
3 The diagonal covariance matrix case
To get an intuition for what a multivariate Gaussian is, consider the simple case where n = 2, and where the covariance matrix Σ is diagonal, i.e.,
x = x 1 μ = μ 1 x2 μ2
In this case, the multivariate Gaussian density has the form,
p ( x ; μ , Σ ) =
=
where we have relied on the explicit formula for the determinant of a 2 × 2 matrix3, and the fact that the inverse of a diagonal matrix is simply found by taking the reciprocal of each diagonal entry. Continuing,
p ( x ; μ , Σ ) = =
1 2πσ1σ2
1
2 π σ 1 σ 2
1 e x p − 1 x 1 − μ 1 T σ 12
σ201/2 2x−μ0σ2 x−μ
2π1 2 2 2 2 2
0 σ2
T1 0
1 e x p − 1 x 1 − μ 1 σ 12
x 1 − μ 1 , 2π(σ12·σ2−0·0)1/2 2 x2−μ2 0 σ2 x2−μ2
e x p
exp− 1 (x1 −μ1)2 − 1 (x2 −μ2)2
1
2 x2 −μ2 σ2(x2 −μ2)
Σ = σ 12 0 0 σ2
0 − 1 x 1 − μ 1
1
T 1 (x − μ ) − 1 x 1 − μ 1 σ 12 1 1
2 σ 12 2 σ 2 2
exp− 1 (x1 −μ1)2· √ 1 exp− 1 (x2 −μ2)2.
= √ 1 2πσ1
The last equation we recognize to simply be the product of two independent Gaussian den- sities, one with mean μ1 and variance σ12, and the other with mean μ2 and variance σ2.
More generally, one can show that an n-dimensional Gaussian with mean μ ∈ Rn and diagonal covariance matrix Σ = diag(σ12, σ2, . . . , σn2 ) is the same as a collection of n indepen- dent Gaussian random variables with mean μi and variance σi2, respectively.
4 Isocontours
Another way to understand a multivariate Gaussian conceptually is to understand the shape of its isocontours. For a function f : R2 → R, an isocontour is a set of the form
→ R,
x∈R2 :f(x)=c. a b
for some c ∈ R.4
3Namely, c d = ad − bc.
2σ12 2πσ2 2σ2
4 n Isocontours are often also known as level curves. More generally, a level set of a function f : R
isasetoftheformx∈R2 :f(x)=cforsomec∈R. 4
4.1 Shape of isocontours
What do the isocontours of a multivariate Gaussian look like? As before, let’s consider the case where n = 2, and Σ is diagonal, i.e.,
x = x 1 x2
As we showed in the last section, p(x;μ,Σ)= 1
μ = μ 1 Σ = σ 12 0 μ2 0 σ2
exp− 1 (x1 −μ1)2 − 1 (x2 −μ2)2. (4) 2 σ 12 2 σ 2 2
2 π σ 1 σ 2
Now, let’s consider the level set consisting of all points where p(x; μ, Σ) = c for some constant
c ∈ R. In particular, consider the set of all x1, x2 ∈ R such that
c= 1 exp− 1 (x1 −μ1)2 − 1 (x2 −μ2)2
2 π σ 1 σ 2 2 σ 12 2 σ 2 2 2πcσ1σ2 =exp− 1 (x1 −μ1)2 − 1 (x2 −μ2)2
2σ12 2σ2 log(2πcσ1σ2)=−1(x1−μ1)2− 1(x2−μ2)2
2σ12 2σ2
log 1 = 1 (x1−μ1)2+ 1 (x2−μ2)2
2 π c σ 1 σ 2 2 σ 12 2 σ 2 2
(x1 − μ1)2 (x2 − μ2)2
1= 2 1 + 2 1 . 2σ1 log 2πcσ1σ2 2σ2 log 2πcσ1σ2
Defining
r1 =2σ12log it follows that
1 r2 =2σ2log 1 2πcσ1σ2 2πcσ1σ2
x1 −μ12 x2 −μ22 1= r + r .
12
,
(5)
Equation (5) should be familiar to you from high school analytic geometry: it is the equation of an axis-aligned ellipse, with center (μ1,μ2), where the x1 axis has length 2r1 and the x2 axis has length 2r2!
4.2 Length of axes
To get a better understanding of how the shape of the level curves vary as a function of the variances of the multivariate Gaussian distribution, suppose that we are interested in
5
8
6
4
2
0
−2
−4
−6
−6 −4 −2 0 2 4 6 8 10 12
8
6
4
2
0
−2
−4
−4 −2 0 2 4 6 8 10
Figure 2:
The figure on the left shows a heatmap indicating values of the density function for an
axis-aligned multivariate Gaussian with mean μ = 32 and diagonal covariance matrix Σ = 25 0. Notice that the Gaussian is centered at (3,2), and that the isocontours are all
09
elliptically shaped with major/minor axis lengths in a 5:3 ratio. The figure on the right shows a heatmap indicating values of the density function for a non axis-aligned multivariate
Gaussian with mean μ = 3 and covariance matrix Σ = 10 5. Here, the ellipses are 255
again centered at (3,2), but now the major and minor axes have been rotated via a linear transformation.
6
the values of r1 and r2 at which c is equal to a fraction 1/e of the peak height of Gaussian density.
First, observe that maximum of Equation (4) occurs where x1 = μ1 and x2 = μ2. Substi-
tuting these values into Equation (4), we see that the peak height of the Gaussian density
is 1 . 2πσ1σ2
Second, we substitute c = 1 1 into the equations for r1 and r2 to obtain e 2πσ1σ2
21√ r 1 = 2 σ 1 l o g 2 π σ 1 σ 2 · 1 1 = σ 1 2
e 2πσ1σ2 1
√ r2=2σ2log2πσ1σ2·1 1 =σ2 2.
e 2πσ1σ2
From this, it follows that the axis length needed to reach a fraction 1/e of the peak height of the Gaussian density in the ith dimension grows in proportion to the standard deviation σi. Intuitively, this again makes sense: the smaller the variance of some random variable xi, the more “tightly” peaked the Gaussian distribution in that dimension, and hence the smaller the radius ri.
4.3 Non-diagonal case, higher dimensions
Clearly, the above derivations rely on the assumption that Σ is a diagonal matrix. However, in the non-diagonal case, it turns out that the picture is not all that different. Instead of being an axis-aligned ellipse, the isocontours turn out to be simply rotated ellipses. Furthermore, in the n-dimensional case, the level sets form geometrical structures known as ellipsoids in Rn.
5 Linear transformation interpretation
In the last few sections, we focused primarily on providing an intuition for how multivariate Gaussians with diagonal covariance matrices behaved. In particular, we found that an n- dimensional multivariate Gaussian with diagonal covariance matrix could be viewed simply as a collection of n independent Gaussian-distributed random variables with means and vari- ances μi and σi2, respectvely. In this section, we dig a little deeper and provide a quantitative interpretation of multivariate Gaussians when the covariance matrix is not diagonal.
The key result of this section is the following theorem (see proof in Appendix A.2). Theorem 1. Let X ∼ N(μ,Σ) for some μ ∈ Rn and Σ ∈ Sn++. Then, there exists a matrix
B∈Rn×n suchthatifwedefineZ=B−1(X−μ),thenZ∼N(0,I). 7
To understand the meaning of this theorem, note that if Z ∼ N(0,I), then using the analysis from Section 4, Z can be thought of as a collection of n independent standard normal random variables (i.e., Zi ∼ N (0, 1)). Furthermore, if Z = B−1(X − μ) then X = BZ + μ follows from simple algebra.
Consequently, the theorem states that any random variable X with a multivariate Gaus- sian distribution can be interpreted as the result of applying a linear transformation (X = BZ + μ) to some collection of n independent standard normal random variables (Z).
8
Appendix A.1
Proof. We prove the first of the two equalities in (1); the proof of the other equality is similar. Cov[X1, X1] · · · Cov[X1, Xn]
. .. . Σ=...
Cov[Xn, X1] · · · Cov[Xn, Xn]
E[(X1 − μ1)2] · · · E[(X1 − μ1)(Xn − μn)]
. .. . =...
E[(Xn − μn)(X1 − μ1)] · · · E[(Xn − μn)2] (X1 −μ1)2 ··· (X1 −μ1)(Xn −μn)
. .. .
=E . . . (6)
(Xn −μn)(X1 −μ1) ··· (Xn −μn)2 X1 −μ1
.
= E . X1 − μ1 · · · Xn − μn (7)
Xn − μn
= E (X − μ)(X − μ)T .
Here, (6) follows from the fact that the expectation of a matrix is simply the matrix found by taking the componentwise expectation of each entry. Also, (7) follows from the fact that for any vector z ∈ Rn,
z1 z1z1 z1z2 ··· z1zn T z2 z2z1 z2z2 ··· z2zn
zz =. z1 z2 ···zn = . . .. . . . ....
zn znz1 znz2 · · · znzn
Appendix A.2
We restate the theorem below:
Theorem 1. Let X ∼ N(μ,Σ) for some μ ∈ Rn and Σ ∈ Sn++. Then, there exists a matrix
B∈Rn×n suchthatifwedefineZ=B−1(X−μ),thenZ∼N(0,I).
The derivation of this theorem requires some advanced linear algebra and probability theory and can be skipped for the purposes of this class. Our argument will consist of two parts. First, we will show that the covariance matrix Σ can be factorized as Σ = BBT for some invertible matrix B. Second, we will perform a “change-of-variable” from X to a different vector valued random variable Z using the relation Z = B−1(X − μ).
9
Step 1: Factorizing the covariance matrix. Recall the following two properties of symmetric matrices from the notes on linear algebra5:
1. Any real symmetric matrix A ∈ Rn×n can always be represented as A = U ΛU T , where U is a full rank orthogonal matrix containing of the eigenvectors of A as its columns, and Λ is a diagonal matrix containing A’s eigenvalues.
2. If A is symmetric positive definite, all its eigenvalues are positive.
Since the covariance matrix Σ is positive definite, using the first fact, we can write Σ = UΛUT for some appropriately defined matrices U and Λ. Using the second fact, we can define Λ1/2 ∈ Rn×n to be the diagonal matrix whose entries are the square roots of the corresponding entries from Λ. Since Λ = Λ1/2(Λ1/2)T , we have
Σ=UΛUT =UΛ1/2(Λ1/2)TUT =UΛ1/2(UΛ1/2)T =BBT,
where B = UΛ1/2.6 In this case, then Σ−1 = B−TB−1, so we can rewrite the standard
formula for the density of a multivariate Gaussian as
p(x; μ, Σ) = 1 exp −1(x − μ)T B−T B−1(x − μ). (8)
Step 2: Change of variables. Now, define the vector-valued random variable Z = B−1(X −μ). A basic formula of probability theory, which we did not introduce in the section notes on probability theory, is the “change-of-variables” formula for relating vector-valued random variables:
Suppose that X = X1 · · · XnT ∈ Rn is a vector-valued random variable with joint density function fX : Rn → R. If Z = H(X) ∈ Rn where H is a bijective, differentiable function, then Z has joint density fZ : Rn → R, where
∂x1 ··· ∂x1
(2π)n/2|BBT |1/2 2
∂z1 ∂zn
. .. . fZ(z)=fX(x)·det . . . .
∂xn···∂xn ∂z1 ∂zn
Using the change-of-variable formula, one can show (after some algebra, which we’ll skip) that the vector variable Z has the following joint density:
pZ(z) = 1 exp−1zTz. (9) (2π)n/2 2
The claim follows immediately.
5See section on “Eigenvalues and Eigenvectors of Symmetric Matrices.”
6To show that B is invertible, it suffices to observe that U is an invertible matrix, and right-multiplying U by a diagonal matrix (with no zero diagonal entries) will rescale its columns but will not change its rank.
10