Extra Topic: DISTRIBUTIONS OF
FUNCTIONS OF
RANDOM VARIABLES
A little in Montgomery and Runger text in Section 5-5.
• Transformations (Continuous r.v.’s)
We have a continuous random variable X
and we know its distribution. We are interested, though, in a random variable Y
which is a transformation of X. For example, Y = X 2.
We wish to determine the distribution of Y .
The two methods we will discuss for continuous random variables:
(1) Distribution function (cdf) technique
(2) Change of variable (Jacobian) technique
1
• (2) Change of variable (Jacobian)
technique
Suppose X has pdf fX (x) with
support c1 < x < c2.
We want to find the pdf of Y = g(X) where
g is a one-to-one mapping.
Procedure:
1) Find the inverse function x = g −1(y).
d −1 g (y)
2) Then, fY (y) = fX (g −1(y)) · dy
| {z }
↑
Jacobian
of the transformation
for g(c1) < y < g(c2).
2
– Example 1 (Jacobian technique):
Suppose X has the pdf below...
3
4x
0<x<1
fX (x) =
0
otherwise
This pdf allows us to calculate the probability that X is in any particular interval.
Let Y = X 2 (one-to-one on X support).
0.0
0.4
Y
0.8
1.2
Y=X2
0.0
0.5
1.0
1.5
X
Find the pdf of Y using the Jacobian technique.
3
1) Y = X 2 ⇒ g(X) = X 2
x=
√
y (must be positive square root
since support of X is (0,1))
√
−1
⇒ g (y) = y
d −1 2) fY (y) = fX (g −1(y)) · dy
g (y)
√
d √ = fX ( y) · dy y √ 3 1 −1/2
= 4( y) · 2 y
3/2
−1/2
= 2y · y
= 2y
Thus, fY (y) =
2y
0
0<y<1
otherwise
NOTE: This is exactly what we got from
the cdf technique last time.
4
Where does this Jacobian term come from?
Recall the first technique we used to find
FY (y) for a continuous r.v., which was the
distribution function (cdf) technique.
We saw that we could find FY (y)=P (Y ≤ y)
using the graph of Y vs. X, the inverse of y,
and probabilities for X...
0.8
1.2
Y=X2
0.0
0.4
Y
y
y1
0.0
0.5
2
1.0
1.5
X
In the above example for Y =X 2 on X∈ (0, 1),
we found
√
P (Y ≤ y) = P (0 ≤ X ≤ y).
5
Thus, for a one-to-one increasing function
Y = g(X), we essentially used...
FY (y) =P (Y ≤ y)
=P (X ≤ g −1(y))
=FX (g −1(y)).
Taking the derivative to get the pdf...
d
d
fY (y) = FY (y) = FX (g −1(y))
dy
dy
d −1
−1
=fX (g (y)) · g (y)
dy
NOTE: Because g is an increasing
function, the derivative at the end is > 0.
When we have a decreasing function g(X),
the derivative on the end is < 0, but FY (y) =
P (Y ≤ y) = P (X > g −1(y)) = 1−FX (g −1(y)),
so the two negatives cancel which leads to one
equation for both increasing and decreasing
functions g(X)...
d −1 fY (y) = fX (g −1(y)) · dy
g (y)
6
– Example 2 (Jacobian technique):
Suppose X has the
pdf
1
0<x<1
fX (x) =
0
otherwise
Let Y = −2 loge(X)
(a decreasing function in X).
Find fY (y) using the Jacobian technique.
4
2
0
Y
6
8
Y=− 2ln(X)
0.0
0.2
0.4
0.6
X
7
0.8
1.0
1) Find the inverse function x = g −1(y).
x = e−y/2
d −1 −1
2) Then, fY (y) = fX (g (y)) · dy g (y).
d
−y/2
−y/2
fY (y) =fX (e
) · e
dy
1 −y/2
= 1 · − e
2
1 −y/2
= e
2
Thus,
(
fY (y) =
1 e−y/2
2
0<y<∞
0
otherwise
⇒ Y ∼ exponential(1/2)
8
– Example 3 (Jacobian technique):
Suppose X is a r.v. with
fX (x) = x8 for 0 ≤ x < 4.
Let Y = 2X + 4. Find fY (y).
1)
x = y−4
2
y−4
d
y
−
4
2)fY (y) =fX
· 2
dy
2
y−4
1
2
=
· 8
2
1
= (y − 4)
32
Thus,
( 1
4 ≤ y < 12
32 (y − 4)
fY (y) =
0
otherwise
NOTE: Same as in the cdf technique.
9
• Jacobian technique in a non one-toone function
As long as g(X) is one-to-one (increasing or
decreasing) then we can use the general rule
d −1 fY (y) = fX (g −1(y)) · dy
g (y) .
Can we still use the Jacobian technique if we
don’t have one-to-one correspondence or the
function is not monotonic?
Yes, but we will first need to partition the
support for X into intervals where the function IS monotone (either increasing or decreasing).
We will limit our discussion to functions whose
one-to-one partitions all coincide with the
same support for Y , such as...
10
Y
Y as a function of X
0
2
4
6
8
X
In the above, we can partition the support of
X into 4 intervals or ranges such that gi(X)
in each region i=1,2,3,4, is monotonic.
This gets us what we needed, which is for
gi−1(y) to be continuous and have a unique
value on the support of Y for each i.
Then, the pdf of Y is given by
4
X
d
−1
−1
fY (y) =
fX (gi (y)) · gi (y)
dy
i=1
which can be generalized to any number of k
partitions.
11
– Example 4 (Jacobian technique):
NOT one-to-one on X support.
Suppose X is uniform on (-1,1).
0.4
0.0
0.2
f(x)
0.6
pdf of X
-1.5
-1.0
-0.5
0.0
0.5
1.0
1.5
X
fX (x) =
−1 < x < 1
otherwise
1/2
0
Let Y = X 4
Find the pdf of Y using the Jacobian technique.
12
Break the support of X into intervals where
the function Y = X 4 IS one-to-one.
0.4
0.2
0.0
Y
0.6
0.8
1.0
Y=X4
-1.0
-0.5
0.0
0.5
1.0
X
− 1 < x < 0 (First of 2 intervals):
√
1) x = − 4 y = −y 1/4
d
1/4
1/4
2) fY (y) =fX (−y ) · −y
dy
1 1 −3/4
= · − y
2
4
1 −3/4
= y
8
13
0.4
0.0
0.2
Y
0.6
0.8
1.0
Y=X4
-1.0
-0.5
0.0
0.5
1.0
X
0 ≤ x < 1 (Second of 2 intervals):
√
1) x = 4 y = y 1/4
d
1/4
1/4
y
2) fY (y) =fX (y ) · dy
1 1 −3/4
= · y
2 4
1 −3/4
= y
8
14
Then, we sum the pieces...
fY (y) = 18 y −3/4 + 81 y −3/4 = 41 y −3/4
or
−3/4
y
4
0
1
fY (y) =
0<y<1
otherwise
NOTE: This is exactly what we got from
the cdf technique last time.
15