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.
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