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Random Vectors

Module by: Nick Kingsbury

Summary: This module introduces random vectors.

Random Vectors are simply groups of random variables, arranged as vectors. E.g.:

X=X1…XnT X X1 … Xn (1)
where X1 X1 , …… Xn Xn are nn separate random variables.

In general, all of the previous results can be applied to random vectors as well as to random scalars, but vectors allow some interesting new results too.

Figure 1: pdfs of (a) a 2-D normal distribution and (b) a Rayleigh distribution, corresponding to the magnitude of the 2-D random vectors.
Figure 1 (figure1.png)

Example - Arrows on a target

Suppose that arrows are shot at a target and land at random distances from the target centre. The horizontal and vertical components of these distances are formed into a 2-D random error vector. If each component of this error vector is an independent variable with zero-mean Gaussian pdf of variance σ2 σ 2 , calculate the pdf's of the radial magnitude and the phase angle of the error vector.

Let the error vector be

X=X1X2T X X1 X2 (2)
X1 X1 and X2 X2 each have a zero-mean Gaussian pdf given by
fx=12πσ2ⅇ-x22σ2 f x 1 2 σ 2 x 2 2 σ 2 (3)
Since X1 X1 and X2 X2 are independent, the 2-D pdf of XX is
fXx1x2=fx1fx2=12πσ2ⅇ-x12+x222σ2 fX x1 x2 f x1 f x2 1 2 σ 2 x1 2 x2 2 2 σ 2 (4)
In polar coordinates x1=rcosθ x1 r θ and x2=rsinθ x2 r θ To calculate the radial pdf, we substitute r=x12+x22 r x1 2 x2 2 in the above 2-D pdf to get:
Prr<R<r+δr=∫rr+δr∫-ππfXx1x2RdθdR r R r δ r R r r δ r θ fX x1 x2 R (5)
where ∫rr+δr∫-ππfXx1x2RdθdR≈δr∫-ππ12πσ2ⅇ-r22σ2rdθ=1σ2rⅇ-r22σ2δr R r r δ r θ fX x1 x2 R δ r θ 1 2 σ 2 r 2 2 σ 2 r 1 σ 2 r r 2 2 σ 2 δ r Hence the radial pdf of the error vector is:
fRr=limδr→0Prr<R<r+δrδr=1σ2rⅇ-r22σ2 fR r δ r 0 r R r δ r δ r 1 σ 2 r r 2 2 σ 2 (6)
This is a Rayleigh distribution with variance = 2σ2 2 σ 2 (these are two components of XX, each with variance σ2 σ 2 ).

The 2-D pdf of XX depends only on rr and not on θθ, so the angular pdf of the error vector is constant over any 2π 2 interval and is therefore fΘθ=12π fΘ θ 1 2 so that ∫-ππfΘθdθ=1 θ fΘ θ 1

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