Download FUNCTIONS OF ONE RANDOM VARIABLE

FUNCTIONS OF ONE RANDOM VARIABLE∗
Constantin VERTAN
April 9th, 2015
As shown before, any random variable is a function, ξ : Ω −→ R; this function can be composed with
any real-valued, real argument function (g : R −→ R), resulting another random variable, denoted, for
instance as η, which is:
η = g ◦ ξ = g(ξ), η : Ω −→ R.
The following problem occurs?: how to characterize the new random variable η (more specifically, how
to find the probability density function of the random variable η) knowing the statistical properties of
the random variable ξ.
Let’s consider a given, fixed value x. The probability that the value of a particular realization of the
random variable ξ s’a is equal to x equals the probability that the value of that particular realization of
the random variable ξ is within the infinitely small interval [x, x + dx] (dx −→ 0). But this is also:
∫
x+dx
Prob{ξ ∈ [x; x + dx]} = Fξ (x + dx) − Fξ (x) =
wξ (t)dt = wξ (x) |dx| .
(1)
x
If the transform function g is bijective, the value x is uniquely mapped to the value y = g(x). in the
same as we deduced (1), we can write that:
Prob{η ∈ [y; y + dy]} = wη (y) |dy| .
(2)
But since y is uniquely obtained out of x, it follows that the probabilities given by (1) and (2) are equal,
thus:
wξ (x) |dx| = wη (y) |dy| .
We are interested in wη (y), so we will write:
dx ( −1 )
1 1
wη (y) = wξ (x) = wξ (x) ′
(y)
.
x=g −1 (y) = wξ g
′
−1
dy
|g (x)|
|g (g (y))|
(3)
This formula (3) is valed only in the case of a transform function g which is bijective over its entire
domain, or, in some other words, only if the equation y = g(x) in which x is unknown and y is a
parameter, has an unique solution. If the function g is not bijective, its domain must be decomposed in
intervals of bijectivity. Within each such interval, the equation y = g(x), in which x is unknown and y is
a parameter, has an unique solution (denoted by xk ). In this case, the formula (3) becomes:
wη (y) =
∑
k
wξ (xk )
1
−1
|g ′ (xk )| xk =g (y)
.
(4)
The essential condition for the above relation to hold is that the number of intervals of bijectivity is finite
or countable.
∗ This document is intended to be used in the framework of the course Decision and estimation for information processing,
held for the students of the English stream at the Faculty of Electronics, Telecommunications and Information Technology
of the Politehnica University of Bucharest. This material discusses applications of the fundamental theoretical definitions
regarding the concept of random variable, is intended for use at the seminars and does not replaces the course notes.
1
For the pair of random variables ξ and η one can verify the theorem of the mean:
∫ ∞
∫ ∞
∫ ∞
η=
ywη (y)dy =
ywξ (x)dx =
g(x)wξ (x)dx.
−∞
−∞
(5)
−∞
(
)
Ex. 1 Let ξ be a random variable uniformly distributed within − π2 , π2 and consider the function
)
(
˙ Compute the
g : − π2 , π2 −→ (−1; 1), with g(x) = sin(x). The random variable η is given by η = g(ξ ).
pdf of the random variable η.
First, one has to check the bijectivity of the transform function g(x) within its domain, and, if this
does not hold, one has to divide this interval into subintervals for which the function is bijective. The
bijectivity can be studied in a very simple manner, by solving the equation y = g(x), with the unknown
(
)
x and the parameter y. In this case, the equation y = sin(x) has (one unique
solution for x ∈ − π2 , π2 ,
)
which is x = arcsin(y). Thus g −1 (y) = arcsin(y), g −1 : (−1; 1) −→ − π2 , π2 .
The derivative of the function g(x) is g ′ (x) = cos(x).
According to the formula of computing the new pdf (3) we have:
(
)
fη (y) = wξ g −1 (y)
1
1
fξ (arcsin(y))
= wξ (arcsin(y))
= √
.
|g ′ (g −1 (y))|
|cos(arcsin(y))|
1 − y2
(
)
The random variable ξ is uniformly distributed within − π2 , π2 ; then it follows that
(
)
{ 1
, if x ∈ − π2 , π2 ,
π
wξ (x) =
0, otherwise.
Then:
1
{
wη (y) = √
1 − y2
{
(
)
√ 1 2 π1 , if y ∈ (−1, 1)
if arcsin(y) ∈ − π2 , π2
1−y
=
0, otherwise
0, otherwise
1
π,
Ex. 2 A random variable ξ is transformed via a linear function g(x) = αx + β (α ̸= 0), into the random
variable η. Compute the pdf, the mean and the variance of the random variable η, considering that ξ is
distributed according to a normal distribution, and respectively according to an uniform distribution.
Any linear function is bijective; the equation y = g(x) with the unknown x has the solution x = y−β
α ,
′
and thus g −1 (y) = y−β
.
The
derivative
of
the
linear
function
is
g
(x)
=
α.
Under
these
circumstances,
α
the pdf of the random variable η is given by the formula (3), being:
)
(
( −1 )
y−β
(
)
w
ξ
wξ g (y)
α
1
y−β
wη (y) = ′ −1
=
=
wξ
.
(6)
|g (g (y))|
|α|
|α|
α
If ξ is normally distributed (with mean µ and variance σ 2 ) then:
)
(
1
(x − µ)2
2
wξ (x) = N (µ, σ ) = √
.
exp −
2σ 2
2πσ 2
By replacing in (6) we get:
 (
wη (y) =
1
1
1

√
exp −
|α| 2πσ 2
=√
exp
2πα2 σ 2
(
y−β
α
−µ
2σ 2
(y − (αµ + β))
2πα2 σ 2
2
)
)2 

=
(
)
= N αµ + β, (ασ)2 .
This means that the distribution of the random variable obtained by the linear transformation is still
normal, having a mean transformed via the same linear function and a variance scaled α2 times.
2
If the random variable ξ is uniformly distributed (within [a; b], for instance), its pdf is:
{ 1
b−a , if x ∈ [a, b],
wξ (x) =
0, otherwise.
Then:
1
wη (y) =
wξ
|α|
(
y−β
α
)
1
=
|α|
{
if y−β
α ∈ [a, b],
0, otherwise.
1
b−a ,
One has to consider two cases, according to the sign of α; case 1, α > 0:
{
1
α(b−a) , if y ∈ [αa + β, αb + β],
wη (y) =
0, otherwise.
case 2, α < 0:
{
wη (y) =
1
− α(b−a)
, if y ∈ [αb + β, αa + β],
0, otherwise.
In both cases one can notice that the distribution is still uniform after the transform and according to
the previous proof, the mean is the middle of the definition interval, and the variance is 1/12 from the
squared interval length:
η=α
α2 (b − a)2
a+b
+ β = αξ + β and ση2 =
= α2 σξ2 .
2
12
The new mean is the original mean transformed via the same function (as the random variable) and the
new variance is the α2 scaled version of the original variance.
Ex. 3 One considers the random variable ξ, uniformly distributed in [−c, c]. Compute the pdf and the
cumulative density function of the random variable η = 1/ξ 2 .
The transform function is g(x) = 1/x2 . The derivative is g ′ (x) = −2/x3 . The function is not bijective
over the entire real axis, but it is bijective within the intervals (−∞, 0) and (0, ∞). The solutions of the
√
√
equation y = g(x) are: x1 = 1/ y and x2 = −1/ y, for y > 0. If y < 0 the equation has no solutions
and fη (y) = 0. Then we can apply the formula (4) for y > 0:
wη (y) =
∑
wξ (xk )
k
+wξ (x2 )
1
|g ′ (x
2 )|
x
1
x =g−1 (y) = wξ (x1 ) 1
−1
+
k
|g ′ (xk )|
|g ′ (x1 )| x1 =g (y)
2 =g
−1 (y)
√
√
wξ (1/ y)
wξ (−1/ y)
1 −3
√
√
2
wη (y) = +
( √ )3 (
√ )3 = 2 y (wξ (1/ y) + wξ (−1/ y)) .
−2/ 1/ y −2/ −1/ y The random variable ξ is uniformly distributed; then:
{ 1
2c , if x ∈ [−c, c],
wξ (x) =
0, otherwise.
It follows that:
{ 1
√
√
y ∈ (−∞, −1/c] ∪ [1/c, ∞),
2c , if
wξ (1/ y) =
=
0, otherwise.
{ 1
2
2c , if y ∈ [1/c , ∞),
0, otherwise.
{ 1
√
√
2c , if − y ∈ (−∞, −1/c] ∪ [1/c, ∞), =
wξ (−1/ y) =
0, otherwise.
{ 1
2
2c , if y ∈ [1/c , ∞),
=
0, otherwise.
3
{
Then:
wη (y) =
1 − 32
,
2c y
if y ∈ [1/c2 , ∞),
0, otherwise.
The cumulative density function of random variable η is:
{
∫ y
0, if y < 1/c2 ,
wη (t)dt =
Fη (y) =
1
1 − 12 y − 2 , if y ∈ [1/c2 , ∞).
−∞
Ex. 4 A signal with values distributed normally according to N (0, σ 2 ) (zero mean and variance σ 2 ) is
rectified by a ideal diode. Compute the pdf of the values of the resulting signal.
The transform function of the rectifier circuit containing a single diode (half-wave rectifier) is:
{
x, if x > 0,
g(x) =
0, if x < 0.
The function g(x) is not bijective; in this case one cannot simply apply formula (4) sice the solution set
for the equation y = g(x) with x < 0 is not countable; more precisely {x ∈ R|g(x) = 0} = (−∞, 0]. The
problem will be solved by computing the cumulative density function Fη (y) = Prob{η 6 y}.
If y < 0, Fη (y) = Prob{η 6 y} = Prob{η < 0} = 0.
If y = 0, Fη (y) = Prob{η 6 0} = Prob{η = 0} = Prob{ξ 6 0} = Fξ (0) = 0.5.
If y > 0, Fη (y) = Prob{η 6 y} = Prob{ξ 6 x} = Fξ (x). ’In concluzie,
{
Fξ (y), if y > 0,
Fη (y) =
0, otherwise.
The desired pdf is the derivative of Fη (y), that is:
{
dFη (y)
0, if y < 0,
wη (y) =
=
,
0.5δ(y) + N (0, σ 2 )U (y), for y > 0.
dy
where U is the unitary step function. One has to notice that the new random variable η has a concentrated
probabiliy at value zero, that is the probability of obtaining exactly the value zero is non-zero:
∫ ε
∫ ε
∫ ε
1
1
P {η = 0} = lim
wη (y)dy =
lim
δ(y)dy + lim
wξ (x)dx = .
ε−→0 −ε
ε−→0 0
2 ε−→0 −ε
2
Ex. 5 Find the transform function that maps an uniform distribution within [0; 1] into a Rayleigh
distribution.
Let ξ be the uniformly distributed random variable and η be the Rayleigh distributed random variable.
Then:
{
1, if x ∈ [0, 1],
wξ (x) =
,
0, otherwise.
{
y2
y − 2α
2
e
, if y > 0,
2
α
wη (y) =
0, otherwise.
The supports of the two pdf’s are [0, 1], respectivly [0, ∞), and then the unknown transform function
must satisfy g : [0, 1] −→ [0, ∞).
Let assume that the function g is bijective; then, according to (3) we have:
(
)
wη (y) = wξ g −1 (y)
4
1
.
|g ′ (g −1 (y))|
(
)
The inverse of the transform function exists and it is g −1 : [0, ∞) −→ [0, 1]. This means that fξ g −1 (y) =
1 and thus:
′ ( −1 )
g g (y) = wη (y).
But, since g is bijective, then it is strictly monotonic. By imposing g(0) = 0, it follows that the function
cannot be otherwise than increasing, and its derivative is positive.
( −1 )′
g (y) = wη (y),
∫y
∫y
2
t − t22
− y
2α dt = 1 − e 2α2 = x.
g −1 (y) =
wη (t)dt =
e
α2
−∞
0
From here one evaluates y as a function of x and then:
√
y = −2α2 ln(1 − x).
√
Thus g(x) = −2α2 ln(1 − x).
Ex. 6 Prove that in the case when ξ is a random variable, its cumulative density function will transform
it into an uniformly distributed random variable within [0, 1].
If the pdf of random variable ξ is wξ (x), then its associated cumulative density function is Fξ (x) =
∫x
wξ (t)dt. If this is also the transform function of the random variable, then g(x) = Fξ (x), with
−∞
g : R −→ [0, 1]. The derivative of the transform function is:
g ′ (x) =
dFξ (x)
= wξ (x),
dx
and |g ′ (x)| = |wξ (x)| = wξ (x) (since the pdf is positive). then, according to (3) we have:
1 1 −1
= 1 for y ∈ [0; 1].
wη (y) = wξ (x) ′
x=g −1 (y) = wξ (x)
|g (x)|
wξ (x) x=Fξ (y)
This is indeed an uniform distribution within [0, 1].
Ex. 7 The dissipated electrical power into a resistor R = 1kΩ is modelled as a random variable having
an uniform distribution within [Pmin , Pmax ] = [1W, 10W ]. Find the distribution of the current values
within the resistor.
Ex. 8 A constant valued resistor R is conected at a current generator. The current generated by the
generator can be considered a random variable uniformly distributed within [Imin , Imax ]. Compute the
mean power dispersed by the resistor and the pdf of that power.
Ex. 9 Prove (using the theorem of the mean (5)) that for a linear transform function (g(x) = αx + β)
the variance of the resulting random variable is the original variance scaled α2 times, and the resulting
mean is original mean transformed via g(x).
Ex. 10 Compute the mean information obtained following the realization of an event which probability
is distributed according to a 1/x function within [α, 1].
Ex. √11 By measuring the anodic current of a vacuum diode one notices a linear distribution within 0
and 2 mA. The anodic voltage of the vacuum diode is generated
√ by a voltage source that has to be tested
(the dispersion of the voltage must not exceed 20 V ). If A = 2/1000 mA/V 3/2 is the tested generator
fit?
Ex. 12 The transfer function of an ideal bi-phase rectifier is described by the function g(x) = |x|. Compute the pdf, the mean voltage and the mean power of the random signal ξ(t) obtained after rectification,
if ξ(t) is a) normally distributed N (0, σ 2 ), b) uniformly distributed in [−1; 1] and [0; 1].
Ex. 13 A non-linear limiter is defined by the transfer function:

 0, if x 6 0,
x, if x ∈ (0, 1],
y = g(x) =

1, if x > 1.
5
At the input of the circuit is applied a signal with values following the pdf wX (x) =
Compute the pdf of the output signal.
1
1 −2|x|
2e
+ 12 δ(x).
Acknowledgement
This work has been funded by the Sectoral Operational Programme Human Resources Development 20072013 of the Romanian Ministry of Education and Scientific Research through the Financial Agreement
POSDRU/174/1.3/S/149155.
References
[1] M. Ciuc, C. Vertan: Statistical signal processing, (in Romanian: Prelucrarea statistică a semnalelor),
Ed. MatrixROM, Bucharest, 2005.
[2] A. T. Murgan, I. Spânu, I. Gavăt, I. Sztojanov, V. E. Neagoe, A. Vlad: Exercises for the Theory of
Information Transmission (in Romanian: Teoria Transmisiunii Informaţiei - probleme), Ed. Didactică
şi Pedagogică, Bucharest, 1983.
[3] C. Vertan, I. Gavăt, R. Stoian: Random variables and processes: principles and applications (in
Romanian: Variabile şi procese aleatoare: principii şi aplicaţii), Ed. Printech, Bucharest, 1999.
[4] A. Papoulis: Probability, random variables and stochastic processes, McGraw Hill Inc., 1991.
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