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-1

product of marginal density functions and hence

the components of X are jointly independent.

-2

Furthermore, it can be shown that if some

non-diagonal element, i j , of Î£ is zero, the

-2 -1 0 1 2

corresponding random variables Xi and X j are

x1

pairwise independent.

2.8.11 Computing Probabilities. If is any

0.2

subset of R m , then the probability that a random

0.15

outcome of the random vector X occurs in is

0.1

Z

given as the integral of the density f X over the area

0.05

:

0

2

P Xâˆˆ = f X (x) d x.

1

2

0 1

Y

0

-1

X

Of particular interest are those regions p chosen

-1

-2

-2

so that p is the smallest region for which

Figure 2.9: A bivariate normal density function

with variances Ïƒ1 = Ïƒ2 = 1 and covariances P X âˆˆ p = p.

2 2

Ïƒ1,2 = Ïƒ2,1 = 0.5.

These areas turn out to be the interior regions

Top: Contours of constant density;

bounded by contours of constant probability

Bottom: three-dimensional representation.

density. That is, for any given p, there is a constant

Îº p such that p is given by

X are multivariate normal. The reverse is not true

in general. = {x : f (x) â‰¥ Îº }. p p

X

2.8.10 Independence. The covariance matrix For multivariate normal distributions, the region

plays much the same role in the multivariate p is the interior of an ellipsoid (such as those

case as does the variance in the scalar case: it shown in Figure 2.9).

determines the spread of the distribution and the The contours of constant density in a multivari-

shape of the region occupied by the main body of ate normal distribution are given by the contours

the distribution. of the Mahalanobis distance

Suppose that Î£ = Ïƒ 2 I, where I is the identity

D2 (x) = (x âˆ’ Âµ)T Î£âˆ’1 (x âˆ’ Âµ).

matrix. Then the contours of constant density are

circular.

If Î£ = diag(Ïƒ1 , Â· Â· Â· , Ïƒm ), the contours of the

2 2

Assuming X âˆ¼ N (Âµ, Î£), it can be shown that

scaled random vector Î£âˆ’1/2 X are circular where D2 âˆ¼ Ï‡ 2 (m), so that

âˆ’1

Î£âˆ’1/2 = (Ïƒ1 , Â· Â· Â· , Ïƒm ). The scaled random

âˆ’1

Îºp

vector is identical to X except that each element

P D (X) > Îº p = Ï‡m (u) du = p.

2 2

has been divided by its standard deviation. With 0

such a diagonal covariance matrix, the density

Thus, the problem of calculating probabilities

function of X is given by

reduces to the inversion of the Ï‡ 2 distribution.

(xi âˆ’Âµi )2

m

âˆ’

1 We will return to this concept when introducing

i=1

2Ïƒi 2

f X (x) = .

e

m

(2Ï€ )m/2 i=1 Ïƒi

statistical tests in [6.2.2].

2.8: Random Vectors 43

Suppose now that we wish characterize the

10

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â€˜normalâ€™ winds at our location by identifying the

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â€¢ â€¢ â€¢â€¢ â€¢

95% of possible wind vectors that are closest to

â€¢ â€¢â€¢

â€¢

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8

â€¢

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â€¢

the mean. That is, we wish to exclude from our

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