Functions of Random Variables
Let’s consider a continuous r.v., Y , which
is a differentiable, increasing function
of a second continuous r.v., X:
Y = g(X).
Because g is differentiable and increasing, g0 and g−1 are guaranteed to exist.
Because g maps all x ≤ s ≤ x + ∆x to
y ≤ t ≤ y + ∆y:
Z
x+∆x
Z
y+∆y
fY (t)dt.
fX (s)ds =
x
y
It follows that for small ∆x:
fY (y)∆y ≈ fX (x)∆x.
fY (y)
Y = g(X)
∆y
∆x
fX (x)
Dividing by ∆y, we get an approximate
expression for fY in terms of fX :
∆x
fY (y) ≈ fX (x) .
∆y
Functions of Random Variables (contd.)
This is exact in the limit ∆x → 0:
∆x
fY (y) = lim fX (x)
∆x→0
∆y
1
= lim fX (x)
∆x→0
∆y/∆x
fX (x)
= lim
.
∆x→0 ∆y/∆x
From calculus we know that:
∆y
g(x + ∆x) − g(x)
lim
= lim
= g0(x).
∆x→0 ∆x
∆x→0
∆x
Consequently
fX (x)
fY (y) = 0 .
g (x)
Functions of Random Variables (contd.)
Substituting g−1(y) for x:
fX (g−1(y))
fY (y) = 0 −1
g (g (y))
yields an expression for fY in terms of
g0, g−1, and fX .
Linear Example
Consider a continuous r.v., Y , which is
a linear function of a continuous r.v., X.
Specifically, Y = aX + b. It follows that
g(x) = ax + b
g0(x) = a
g−1(y) = (y − b)/a.
Substituting the above into
fX (g−1(y))
fY (y) = 0 −1
g (g (y))
yields
1
y−b
fY (y) = fX
.
a
a
Linear Example (contd.)
Let fX (x) =
2
√1 e−(x−µ) /2, then
2π
2
y−b
1 − a −µ /2
fY (y) = √ e
a 2π
.
Linear Example (contd.)
Unfortunately, there is a problem. When
a = −1, the function g is not increasing
(it is decreasing). Consequently,
− fY (y)∆y ≈ fX (x)∆x.
It follows that for decreasing functions,
fX (g−1(y))
.
fY (y) =
0
−1
−g (g (y))
However, we can derive a function which
is correct in both cases by replacing g0(.)
with |g0(.)| in the expression relating fY (.)
and fX (.):
fX (g−1(y))
fY (y) = 0 −1
.
|g (g (y))|
Quadratic Example
Consider a continuous r.v., Y , which is
a quadratic function of a continuous r.v.,
X. Specifically, Y = X 2. It follows that
g(x) = x2
g0(x) = 2x
√
−1
g (y) = y.
Substituting the above into
fX (g−1(y))
fY (y) = 0 −1
|g (g (y))|
yields
√
fX ( y)
fY (y) = √ .
|2 y|
Quadratic Example (contd.)
Let fX (x) =
2
√1 e−x /2,
2π
then
1 e−y/2
fY (y) = √
√ .
2π |2 y|
Unfortunately, there is a problem:
1 e−y/2
1
√
fY (y)dy =
√ dy = 6= 1.
2
0
0
2π |2 y|
Can anyone see the mistake?
Z
∞
Z
∞
Quadratic Example (contd.)
The mistake is that two different values
of Y satisfy Y = X 2:
√
−1
g (y) = ± y.
We decided to use the positive square
root arbitrarily and ignored the negative square root. Hence the factor of
two error. In general, if a function does
not have a unique inverse, we must sum
over all possible inverse values:
fX (g−1
i (y))
fY (y) = ∑ 0 −1
.
i=1 |g (gi (y))|
n
Quadratic Example (contd.)
√
√
−1
−1
Let g1 (y) = y and g2 (y) = − y,
then
−1
fX (g−1
(y))
f
(g
X
1
2 (y))
fY (y) = 0 −1
+ 0 −1
|g (g1 (y))| |g (g2 (y))|
1 e−y/2
= √
√ .
2π y
This is an example of the χ2 distribution.
Kinetic Energy
Recall from physics, that the kinetic energy, K, of a moving body is given by
1 2
K = mV
2
where m is mass and V is velocity. Let
V be a normally distributed random variable with mean, µ, and variance, σ2:
1 −(v−µ)2/2σ2
fV (v) = √ e
.
σ 2π
We would like to compute fK (k), the
p.d.f for the continuous random variable, K.
Kinetic Energy (contd.)
−1
We start by computing, g0, g−1
,
and
g
1
2 :
1 2
g(v) = mv
2
0
g (v) = mv
p
−1
g1 (k) = 2k/m
p
−1
g2 (k) = − 2k/m.
Substituting fV (v) and the above into
−1
(k))
fV (g−1
f
(g
V
1
2 (k))
fK (k) = 0 −1
+ 0 −1
|g (g1 (k))| |g (g2 (k))|
yields
√
2
−
2k/m−µ /2σ2
√
2
− − 2k/m−µ /2σ2
+e
p
√
fK (k) =
σ 2π m 2k/m
√
2
√
2
2
−
2k/m−µ /2σ
−
2k/m+µ /2σ2
e
+e
√
=
.
2σ π k m
e