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
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• (1) Distribution function (cdf )
technique
Suppose X has pdf fX (x).
We want to find the pdf of Y = g(X).
Procedure:
1) Determine the cdf of Y , which is
FY (y) = P (Y ≤ y).
2) Then, fy (y) can be found through differentiation as
d F (y)
fY (y) = dy
Y
where the derivative exists.
2
– Example 1 (cdf 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 distribution
function (cdf) technique.
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To determine FY (y), start by thinking of
P (Y ≤ y) for some fixed y in the graphic
below.
0.8
1.2
Y=X2
0.0
0.4
Y
y
y1
0.0
0.5
2
1.0
1.5
X
For the r.v. Y to be less than a fixed y,
√
X must be between 0 and y.
We can find P (0 < X <
4
√
y) using fX (x).
1) FY (y) =P (Y ≤ y)
=P (X 2 ≤ y)
√
√
=P (− y < X < y)
√
=P (0 < X < y) (support of X is (0,1)
Z √y
√ y
=
4x3dx = x4
0
0
= y2

0
Thus, FY (y) = y 2

1
y<0
0≤y<1
y≥1
d F (y)
2) fY (y) = dy
Y
2y
0<y<1
=
0
otherwise
With this pdf for Y , we can calculate the
probability that Y is in any particular interval using integration.
5
INTUITION: The idea is that if we can determine the values of X that lead to any
particular values of Y , then we can obtain the probabilities related to Y using
the probabilities of X (remember, this is
a mapping from X-space to Y -space).
6
– Example 2 (cdf technique):
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
(NOT one-to-one on X support).
Find the pdf of Y using the distribution
function (cdf) technique.
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To determine FY (y), start by thinking of
P (Y ≤ y) for some fixed y in the graphic
below.
0.4
y
0.0
0.2
Y
0.6
0.8
1.0
Y=X4
− y1
-1.0
4
y1
-0.5
0.0
0.5
4
1.0
X
For the r.v. Y to be less than a fixed y,
√
√
4
X must be between − y and 4 y.
√
4
We can find P (− y < X <
fX (x).
8
√
4
y) using
1) FY (y) =P (Y ≤ y)
=P (X 4 ≤ y)
√
√
4
=P (− y < X < 4 y)
√
Z √
4y
4y
1
1 = √ dx = x √
2 −4y
−4y 2
1 √
√
√
4
4
= ( y + y) = 4 y
2
y≤0
0
√
4y
0<y<1
Thus, FY (y) =

1
y≥1
d F (y)
2) fY (y) = dy
Y
1
−3/4
y
= 4
0
0<y<1
otherwise
With this pdf for Y , we can calculate the
probability that Y is in any particular interval using integration.
9
What does fY (y) look like?
3
0
1
2
f(y)
4
5
6
f(y) for y in (0,1)
0.0
0.2
0.4
0.6
0.8
1.0
Y
Check:
Z 1
Z 1
1 −3/4
y
dy
fY (y)dy =
0
0 4
1
√
= 4y =1
0
10
– Example 3 (cdf technique):
Suppose X is a r.v. with
fX (x) = x8 for 0 ≤ x < 4.
Let Y = 2X + 4. Find fY (y).
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(cont.)
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