Transformations
Dear students,
Since we have covered the mgf technique extensively already, here we only review the cdf
and the pdf techniques, first for univariate (one-to-one and more-to-one) and then for
bivariate (one-to-one and more-to-one) transformations.
1. The cumulative distribution function (cdf) technique
Suppose Y is a continuous random variable with cumulative distribution function (cdf)
𝐹𝑌 (𝑦) ≡ 𝑃(𝑌 ≤ 𝑦). Let 𝑈 = 𝑔(𝑌) be a function of Y, and our goal is to find the distribution
of U. The cdf technique is especially convenient when the cdf 𝐹𝑌 (𝑦) has closed form
analytical expression. This method can be used for both univariate and bivariate
transformations.
Steps of the cdf technique:
1. Identify the domain of Y and U.
2. Write𝐹𝑈 (𝑢) = 𝑃(𝑈 ≤ 𝑢), the cdf of U, in terms of 𝐹𝑌 (𝑦), the cdf of Y .
3. Differentiate 𝐹𝑈 (𝑢) to obtain the pdf of U, 𝑓𝑈 (𝑢).
Example 1. Suppose that 𝑌 ~ 𝑈(0,1). Find the distribution of 𝑈 = 𝑔(𝑌) = − ln 𝑌.
Solution. The cdf of 𝑌 ~ 𝑈(0,1) is given by
0,
𝑦≤0
𝐹𝑌 (𝑦) = {𝑦, 0 < 𝑦 ≤ 1
1,
𝑦≥1
The domain (*domain is the region where the pdf is non-zero) for 𝑌 ~ 𝑈(0,1) is 𝑅𝑌 =
{𝑦: 0 < 𝑦 < 1}; thus, because 𝑢 = − ln 𝑦 > 0, it follows that the domain for U is 𝑅𝑈 =
{𝑢: 𝑢 > 0}.
The cdf of U is:
𝐹𝑈 (𝑢) = 𝑃(𝑈 ≤ 𝑢) = 𝑃(− ln 𝑌 ≤ 𝑢)
= 𝑃(ln 𝑌 > −𝑢)
= 𝑃(𝑌 > 𝑒 −𝑢 ) = 1 − 𝑃(𝑌 ≤ 𝑒 −𝑢 ) = 1 − 𝐹𝑌 (𝑒 −𝑢 )
Because 𝐹𝑌 (𝑦) = 𝑦 for 0 < y < 1; i.e., for u > 0, we have
𝐹𝑈 (𝑢) = 1 − 𝐹𝑌 (𝑒 −𝑢 ) = 1 − 𝑒 −𝑢
Taking derivatives, we get, for u > 0,
𝑓𝑈 (𝑢) =
𝑑
𝑑
(1 − 𝑒 −𝑢 ) = 𝑒 −𝑢
𝐹𝑈 (𝑢) =
𝑑𝑢
𝑑𝑢
1
Summarizing,
𝑓𝑈 (𝑢) = {
𝑒 −𝑢 ,
0,
𝑢>0
otherwise
1
This is an exponential pdf with mean λ = 1; that is, U ~ exponential(λ = 1). □
Example 2. Suppose that 𝑌 ~ 𝑈 (− 𝜋⁄2 , 𝜋⁄2) . Find the distribution of the random variable
defined by U = g(Y ) = tan(Y ).
Solution. The cdf of 𝑌 ~ 𝑈 (− 𝜋⁄2 , 𝜋⁄2) is given by
0,
𝑦 + 𝜋⁄2
𝐹𝑌 (𝑦) =
,
𝜋
{
1,
𝑦 ≤ − 𝜋⁄2
− 𝜋⁄2 < 𝑦 ≤ 𝜋⁄2
𝑦 ≥ 𝜋⁄2
The domain for Y is 𝑅𝑌 = {𝑦: − 𝜋⁄2 < 𝑦 < 𝜋⁄2}. Sketching a graph of the tangent function
from − 𝜋⁄2 to 𝜋⁄2, we see that −∞ < 𝑢 < ∞ .
Thus, 𝑅𝑈 = { 𝑢: − ∞ < 𝑢 < ∞} ≡ 𝑅, the set of all reals. The cdf of U is:
𝐹𝑈 (𝑢) = 𝑃(𝑈 ≤ 𝑢) = 𝑃[tan(𝑌) ≤ 𝑢]
= 𝑃[𝑌 ≤ tan−1(𝑢)] = 𝐹𝑌 [tan−1(𝑢)]
Because 𝐹𝑌 (𝑦) =
𝑦+𝜋⁄2
for − 𝜋⁄2 < 𝑦 < 𝜋⁄2 ; i. e. , for 𝑢 ∈ ℛ , we have
tan−1(𝑢) + 𝜋⁄2
𝐹𝑈 (𝑢) = 𝐹𝑌 [tan−1 (𝑢)] =
𝜋
The pdf of U, for 𝑢 ∈ ℛ, is given by
𝑑
𝑑 tan−1(𝑢) + 𝜋⁄2
1
𝑓𝑈 (𝑢) =
𝐹𝑈 (𝑢) =
[
]=
.
𝑑𝑢
𝑑𝑢
𝜋
𝜋(1 + 𝑢2 )
Summarizing,
𝜋
1
, −∞<𝑢 <∞
𝑓𝑈 (𝑢) = {𝜋(1 + 𝑢2 )
0,
otherwise.
A random variable with this pdf is said to have a (standard) Cauchy distribution. One
interesting fact about a Cauchy random variable is that none of its moments are finite. Thus,
if U has a Cauchy distribution, E(U), and all higher order moments, do not exist.
Exercise: If U is standard Cauchy, show that𝐸(|𝑈|) = +∞. □
2. The probability density function (pdf) technique, univariate
Suppose that Y is a continuous random variable with cdf 𝐹𝑌 (𝑦) and domain 𝑅𝑌 , and let 𝑈 =
𝑔(𝑌), where 𝑔: 𝑅𝑌 → ℛ is a continuous, one-to-one function defined over 𝑅𝑌 . Examples of
such functions include continuous (strictly) increasing/decreasing functions. Recall from
2
calculus that if 𝑔 is one-to-one, it has an unique inverse 𝑔−1 . Also recall that if 𝑔 is
increasing (decreasing), then so is 𝑔−1 .
Derivation of the pdf technique formula using the cdf method:
Suppose that 𝑔(𝑦) is a strictly increasing function of y defined over 𝑅𝑌 . Then, it follows
that 𝑢 = 𝑔(𝑦) ⇔ 𝑔−1 (𝑢) = 𝑦 and
𝐹𝑈 (𝑢) = 𝑃(𝑈 ≤ 𝑢) = 𝑃[𝑔(𝑌) ≤ 𝑢] = 𝑃[𝑌 ≤ 𝑔−1 (𝑢)] = 𝐹𝑌 [𝑔−1 (𝑢)]
Differentiating 𝐹𝑈 (𝑢) with respect to u, we get
𝑓𝑈 (𝑢) =
𝑑
𝑑
𝑑 −1
𝐹𝑈 (𝑢) =
𝐹𝑌 [𝑔−1 (𝑢)] = 𝑓𝑌 [𝑔−1 (𝑢)]
𝑔 (𝑢) (by chain rule)
𝑑𝑢
𝑑𝑢
𝑑𝑢
𝑑
Now as 𝑔 is increasing, so is 𝑔−1 ; thus, 𝑑𝑢 𝑔−1 (𝑢) > 0. If 𝑔(𝑦) is strictly decreasing, then
𝑑
𝐹𝑈 (𝑢) = 1 − 𝐹𝑌 [𝑔−1 (𝑢)] and 𝑑𝑢 𝑔−1 (𝑢) < 0, which gives
𝑓𝑈 (𝑢) =
𝑑
𝑑
𝑑 −1
𝐹𝑈 (𝑢) =
{1 − 𝐹𝑌 [𝑔−1 (𝑢)]} = −𝑓𝑌 [𝑔−1 (𝑢)]
𝑔 (𝑢)
𝑑𝑢
𝑑𝑢
𝑑𝑢
Combining both cases, we have shown that the pdf of U, where nonzero, is given by
𝑑
𝑓𝑈 (𝑢) = 𝑓𝑌 [𝑔−1 (𝑢)] | 𝑔−1 (𝑢) |.
𝑑𝑢
It is again important to keep track of the domain for U. If 𝑅𝑌 denotes the domain of Y, then
𝑅𝑈 , the domain for U, is given by 𝑅𝑈 = {𝑢: 𝑢 = 𝑔(𝑦); 𝑦 ∈ 𝑅𝑌 }.
Steps of the pdf technique:
1. Verify that the transformation u = g(y) is continuous and one-to-one over 𝑅𝑌 .
2. Find the domains of Y and U.
3. Find the inverse transformation 𝑦 = 𝑔−1 (𝑢) and its derivative (with respect to u).
4. Use the formula above for 𝑓𝑈 (𝑢).
Example 3. Suppose that Y ~ exponential(β); i.e., the pdf of Y is
1 −𝑦/𝛽
𝑒
,
𝑦>0
𝑓𝑌 (𝑦) = { 𝛽
0,
otherwise.
Let 𝑈 = 𝑔(𝑌) = √𝑌, . Use the method of transformations to find the pdf of U.
Solution. First, we note that the transformation 𝑔(𝑌) = √𝑌 is a continuous strictly
increasing function of y over 𝑅𝑌 = {𝑦: 𝑦 > 0}, and, thus, 𝑔(𝑌) is one-to-one. Next, we need
to find the domain of U. This is easy since y > 0 implies 𝑢 = √𝑦 > 0 as well.
3
Thus, 𝑅𝑈 = {𝑢: 𝑢 > 0}. Now, we find the inverse transformation:
𝑔(𝑦) = 𝑢 = √𝑦 ⇔ 𝑦 = 𝑔−1 (𝑢) = 𝑢2 (by inverse transformation)
and its derivative:
𝑑 −1
𝑑
(𝑢2 ) = 2𝑢.
𝑔 (𝑢) =
𝑑𝑢
𝑑𝑢
Thus, for u > 0,
𝑓𝑈 (𝑢) = 𝑓𝑌 [𝑔−1 (𝑢)] |
2
𝑑 −1
𝑔 (𝑢) |
𝑑𝑢
2
1 −𝑢
2𝑢 −𝑢𝛽
= 𝑒 𝛽 × |2𝑢| =
𝑒 .
𝛽
𝛽
Summarizing,
2
2𝑢 −𝑢𝛽
𝑢>0
𝑓𝑈 (𝑢) = { 𝛽 𝑒 ,
0,
otherwise.
This is a Weibull distribution. The Weibull family of distributions is common in life
science (survival analysis), engineering and actuarial science applications. □
Example 4. Suppose that Y ~ beta(α = 6; β = 2); i.e., the pdf of Y is given by
42𝑦 5 (1 − 𝑦),
𝑓𝑌 (𝑦) = {
0,
0<𝑦<1
otherwise.
What is the distribution of U = g(Y ) = 1 −Y ?
Solution. First, we note that the transformation g(y) = 1 −Y is a continuous decreasing
function of y over 𝑅𝑌 = {𝑦: 0 < 𝑦 < 1}, and, thus, g(y) is one-to-one. Next, we need
to find the domain of U. This is easy since 0 < y < 1 clearly implies 0 < u < 1. Thus,
𝑅𝑈 = {𝑦: 0 < 𝑢 < 1}. Now, we find the inverse transformation:
𝑔(𝑦) = 𝑢 = 1 − 𝑦 ⇔ 𝑦 = 𝑔−1 (𝑢) = 1 − 𝑢 (by inverse transformation)
and its derivative:
𝑑 −1
𝑑
(1 − 𝑢) = −1.
𝑔 (𝑢) =
𝑑𝑢
𝑑𝑢
Thus, for 0 < u < 1,
4
𝑓𝑈 (𝑢) = 𝑓𝑌 [𝑔−1 (𝑢)] |
𝑑 −1
𝑔 (𝑢) |
𝑑𝑢
= 42(1 − 𝑢)5 [1 − (1 − 𝑢)] × |−1| = 42𝑢(1 − 𝑢)5 .
Summarizing,
𝑓𝑈 (𝑢) = {
42𝑢(1 − 𝑢)5 ,
0,
0<𝑢<1
otherwise.
We recognize this is a beta distribution with parameters α = 2 and β = 6. □
More-to-one transformation: What happens if u = g(y) is not a one-to-one transformation?
In this case, we can still use the method of transformations, but we have “break up" the
transformation 𝑔: 𝑅𝑌 → 𝑅𝑈 into disjoint regions where g is one-to-one.
RESULT: Suppose that Y is a continuous random variable with pdf𝑓𝑌 (𝑦) and that U = g(Y ),
not necessarily a one-to-one (but continuous) function of y over RY . However, suppose that
we can partition 𝑅𝑌 into a finite collection of sets, say, A0, A1, A2, … , Ak, where 𝑃(𝑌 ∈ 𝐴0 ) =
0, 𝑎𝑛𝑑 𝑃(𝑌 ∈ 𝐴𝑖 ) > 0 for all i ≠ 0, and 𝑓𝑌 (𝑦) is continuous on each Ai , i ≠ 0. Furthermore,
suppose that the transformation is 1-to-1 from Ai (i = 1, 2, …, k,) to B, where B is the domain
of U = g(Y ) such that 𝑔𝑖−1 (∙) is a 1-to-1 inverse mapping of Y to U = g(Y ) from B to Ai.
Then, the pdf of U is given by
𝑘
𝑑
[𝑔𝑖 −1 (𝑢)] | 𝑔𝑖 −1 (𝑢) | ,
∑
𝑓
𝑌
𝑓𝑈 (𝑢) = {
𝑑𝑢
𝑢 ∈ 𝑅𝑈
𝑖=1
0,
otherwise.
𝑑
That is, writing the pdf of U can be done by adding up the terms 𝑓𝑌 [𝑔𝑖 −1 (𝑢)] |𝑑𝑢 𝑔𝑖 −1 (𝑢) |
corresponding to each disjoint set Ai, i = 1, 2,…,k.
Example 5. Suppose that Y ~ N(0, 1); that is, Y has a standard normal distribution; i.e.,
1
𝑓𝑌 (𝑦) = {√2𝜋
0,
𝑒 −𝑦
2 /2
, −∞ < 𝑦 < ∞
otherwise.
Consider the transformation: 𝑈 = 𝑔(𝑌) = 𝑌 2 .
Solution 1 (the pdf technique):
This transformation is not one-to-one on 𝑅𝑌 = ℛ = {𝑦: −∞ < 𝑦 < ∞} , but it is one-to-one
on A1 = (−∞, 0) and A2 = (0, ∞) (separately) since 𝑔(𝑦) = 𝑦 2 is decreasing on A1 and
5
increasing on A2, and A0 = {0} where 𝑃(𝑌 ∈ 𝐴0 ) = 𝑃(𝑌 = 0) = 0. Furthermore, note that A0,
A1 and A2 partitions 𝑅𝑌 . Summarizing,
Partition
A1 = (−∞, 0)
A2 = (0, ∞)
Transformation
𝑔(𝑦) = 𝑦 2 = 𝑢
𝑔(𝑦) = 𝑦 2 = 𝑢
Inverse transformation
𝑔1 −1 (𝑢) = −√𝑢 = 𝑦
𝑔2 −1 (𝑢) = √𝑢 = 𝑦
And, on both sets A1 and A2,
𝑑
1
| 𝑔𝑖 −1 (𝑢) | =
.
𝑑𝑢
2 √𝑢
Clearly, 𝑢 = 𝑦 2 > 0; thus, 𝑅𝑈 = {𝑢: 𝑢 > 0}, and the pdf of U is given by
2
2
1 −(−√𝑢)
1
1 −(√𝑢)
1
2
(
)+
𝑒 2 (
), 𝑢 > 0
𝑓𝑈 (𝑢) = { √2𝜋 𝑒
2√𝑢
2 √𝑢
√2𝜋
0,
otherwise.
1
Thus, for u > 0, and recalling that Γ (2) = √𝜋, 𝑓(𝑢) collapses to
2 −𝑢 1
𝑓𝑈 (𝑢) =
𝑒 2(
)
2 √𝑢
√2𝜋
1
𝑢
1 1−1 −𝑢
1
−1 −
2 𝑒 2.
=
𝑢2 𝑒 2 =
𝑢
1 1
√2𝜋
Γ (2) 22
Summarizing, the pdf of U is
1
𝑢
1
−1 −
2 𝑒 2, 𝑢 > 0
𝑢
1
𝑓𝑈 (𝑢) = { Γ (1) 22
2
0,
otherwise.
That is, U ~ gamma(1/2, 2). Recall that the gamma(1/2, 2) distribution is the same as a 𝜒 2
distribution with 1 degree of freedom; that is, 𝑈 ~ 𝜒 2 (1). □
Solution 2 (the cdf technique):
𝐹𝑈 (𝑢) = 𝑃(𝑈 ≤ 𝑢) = 𝑃(𝑌 2 ≤ 𝑢) = 1 − 𝑃(𝑌 2 > 𝑢)
= 1 − 𝑃(𝑌 > √𝑢 𝑜𝑟 𝑌 < −√𝑢) = 1 − 𝑃(𝑌 > √𝑢 ) − 𝑃( 𝑌 < −√𝑢)
= 𝑃(𝑌 ≤ √𝑢 ) − 𝑃( 𝑌 ≤ −√𝑢) = 𝐹𝑌 (√𝑢) − 𝐹𝑌 (−√𝑢)
Taking derivative with respect to 𝑢 at both sides, we have:
𝑓𝑈 (𝑢) = 𝑓𝑌 (√𝑢)
𝑑 √𝑢
𝑑 √𝑢
+ 𝑓𝑌 (−√𝑢)
𝑑𝑢
𝑑𝑢
6
=
1
√2𝜋
=
𝑒
−
(−√𝑢)
2
1
√2𝜋
1
2
1
1
(
)+
𝑒
2 √𝑢
√2𝜋
𝑢
𝑢2−1 𝑒 − 2 =
−
(√𝑢)
2
2
(
1
2√𝑢
)
1
𝑢
1
−1 −
2 𝑒 2, 𝑢 > 0
𝑢
1 1
Γ (2) 22
That is, U ~ gamma(1/2, 2). Recall that the gamma(1/2, 2) distribution is the same as a 𝜒 2
distribution with 1 degree of freedom; that is, 𝑈 ~ 𝜒 2 (1). □
3. The probability density function (pdf) technique, bivariate
Here we discuss transformations involving two random variable 𝑌1 , 𝑌2 . The bivariate
transformation is
𝑈1 = 𝑔1 (𝑌1 , 𝑌2 )
𝑈2 = 𝑔2 (𝑌1 , 𝑌2 )
Assuming that 𝑌1 and 𝑌2 are jointly continuous random variables, we will discuss the oneto-one transformation first. Starting with the joint distribution of 𝒀 = (𝑌1 , 𝑌2 ), our goal is to
derive the joint distribution of 𝑼 = (𝑈1 , 𝑈2 ).
Suppose that 𝒀 = (𝑌1 , 𝑌2 ) is a continuous random vector with joint pdf𝑓𝑌1 ,𝑌2 (𝑦1 , 𝑦2 ). Let
𝑔: ℛ 2 → ℛ 2 be a continuous one-to-one vector-valued mapping from 𝑅𝑌1 ,𝑌2 to 𝑅𝑈1 ,𝑈2 , where
𝑈1 = 𝑔1 (𝑌1 , 𝑌2 ) and 𝑈2 = 𝑔2 (𝑌1 , 𝑌2 ), and where 𝑅𝑌1 ,𝑌2 and 𝑅𝑈1 ,𝑈2 denote the two-dimensional
domain of 𝒀 = (𝑌1 , 𝑌2 ) and 𝑼 = (𝑈1 , 𝑈2 ), respectively. If 𝑔1 −1 (𝑢1 , 𝑢2 ) and 𝑔2 −1 (𝑢1 , 𝑢2 ) have
continuous partial derivatives with respect to both 𝑢1 and 𝑢2 , and the Jacobian, J, where,
with “det” denoting “determinant”,
𝜕𝑔1 −1 (𝑢1 , 𝑢2 ) 𝜕𝑔1 −1 (𝑢1 , 𝑢2 )
𝜕𝑢1
𝜕𝑢2
| ≠ 0,
𝐽 = det ||
|
−1 (𝑢
−1 (𝑢
)
𝜕𝑔2
𝜕𝑔2
1 , 𝑢2
1 , 𝑢2 )
𝜕𝑢1
𝜕𝑢2
then
𝑓 [𝑔 −1 (𝑢1 , 𝑢2 ), 𝑔2 −1 (𝑢1 , 𝑢2 )]|𝐽|, (𝑢1 , 𝑢2 ) ∈ 𝑅𝑈1 ,𝑈2
𝑓𝑈1 ,𝑈2 (𝑢1 , 𝑢2 ) = { 𝑌1 ,𝑌2 1
0,
otherwise,
where |J| denotes the absolute value of J.
RECALL: The determinant of a 2 × 2 matrix, e.g.,
𝑎 𝑏
det |
| = 𝑎𝑑 − 𝑏𝑐.
𝑐 𝑑
Steps of the pdf technique:
1. Find 𝑓𝑌1 ,𝑌2 (𝑦1 , 𝑦2 ), the joint distribution of 𝑌1 and 𝑌2 . This may be given in the problem. If
Y1 and Y2 are independent, then 𝑓𝑌1 ,𝑌2 (𝑦1 , 𝑦2 ) = 𝑓𝑌1 (𝑦1 )𝑓𝑌2 (𝑦2 ).
2. Find 𝑅𝑈1 ,𝑈2 , the domain of 𝑼 = (𝑈1 , 𝑈2 ).
3. Find the inverse transformations 𝑦1 = 𝑔1 −1 (𝑢1 , 𝑢2 ) and 𝑦2 = 𝑔2 −1 (𝑢1 , 𝑢2 ).
7
4. Find the Jacobian, J, of the inverse transformation.
5. Use the formula above to find 𝑓𝑈1 ,𝑈2 (𝑢1 , 𝑢2 ), the joint distribution of 𝑈1 and 𝑈2 .
NOTE: If desired, marginal distributions 𝑓𝑈1 (𝑢1 ) and 𝑓𝑈2 (𝑢2 ). can be found by integrating
the joint distribution 𝑓𝑈1 ,𝑈2 (𝑢1 , 𝑢2 ).
Example 6. Suppose that 𝑌1 ~ gamma(α, 1), 𝑌2 ~ gamma(β, 1), and that𝑌1 and 𝑌2 are
independent. Define the transformation
𝑈1 = 𝑔1 (𝑌1 , 𝑌2 ) = 𝑌1 + 𝑌2
𝑌1
𝑈2 = 𝑔2 (𝑌1 , 𝑌2 ) =
.
𝑌1 + 𝑌2
Find each of the following distributions:
(a) 𝑓𝑈1 ,𝑈2 (𝑢1 , 𝑢2 ), the joint distribution of 𝑈1 and 𝑈2 ,
(b) 𝑓𝑈1 (𝑢1 ), the marginal distribution of 𝑈1 , and
(c) 𝑓𝑈2 (𝑢2 ), the marginal distribution of 𝑈2 .
Solutions. (a) Since 𝑌1 and 𝑌2 are independent, the joint distribution of 𝑌1 and 𝑌2 is
𝑓𝑌1 ,𝑌2 (𝑦1 , 𝑦2 ) = 𝑓𝑌1 (𝑦1 )𝑓𝑌2 (𝑦2 )
1
1
𝛽−1
=
𝑦1𝛼−1 𝑒 −𝑦1 ×
𝑦2 𝑒 −𝑦2
Γ(𝛼)
Γ(𝛽)
1
𝛽−1
=
𝑦 𝛼−1 𝑦2 𝑒 −(𝑦1 +𝑦2 ) ,
Γ(𝛼)Γ(𝛽) 1
for 𝑦1 > 0, 𝑦2 > 0, and 0, otherwise. Here, 𝑅𝑌1 ,𝑌2 = {(𝑦1 , 𝑦2 ): 𝑦1 > 0, 𝑦2 > 0}. By inspection,
𝑦
we see that 𝑢1 = 𝑦1 + 𝑦2 > 0, and 𝑢2 = 𝑦 +1𝑦 must fall between 0 and 1.
1
2
Thus, the domain of 𝑼 = (𝑈1 , 𝑈2 ) is given by
𝑅𝑈1 ,𝑈2 = {(𝑢1 , 𝑢2 ): 𝑢1 > 0, 0 < 𝑢2 < 1}.
The next step is to derive the inverse transformation. It follows that
𝑢1 = 𝑦1 + 𝑦2
𝑦1 = 𝑔1 −1 (𝑢1 , 𝑢2 ) = 𝑢1 𝑢2
𝑦1
⇒
−1
𝑢2 =
𝑦1 + 𝑦2 𝑦2 = 𝑔2 (𝑢1 , 𝑢2 ) = 𝑢1 − 𝑢1 𝑢2
The Jacobian is given by
𝜕𝑔1 −1 (𝑢1 , 𝑢2 ) 𝜕𝑔1 −1 (𝑢1 , 𝑢2 )
𝑢1
𝜕𝑢1
𝜕𝑢2
| = det | 𝑢2
𝐽 = det ||
| = −𝑢1 𝑢2 − 𝑢1 (1 − 𝑢2 ) = −𝑢1 .
|
−1 (𝑢
−1
1
−
𝑢
−𝑢
2
1
𝜕𝑔2
1 , 𝑢2 ) 𝜕𝑔2 (𝑢1 , 𝑢2 )
𝜕𝑢1
𝜕𝑢2
We now write the joint distribution for 𝑼 = (𝑈1 , 𝑈2 ). For 𝑢1 > 0 and 0 < 𝑢2 < 1, we have
that
𝑓𝑈1 ,𝑈2 (𝑢1 , 𝑢2 ) = 𝑓𝑌1 ,𝑌2 [𝑔1 −1 (𝑢1 , 𝑢2 ), 𝑔2 −1 (𝑢1 , 𝑢2 )]|𝐽|
1
=
(𝑢 𝑢 )𝛼−1 (𝑢1 − 𝑢1 𝑢2 )𝛽−1 𝑒 −[(𝑢1 𝑢2 )+(𝑢1 −𝑢1 𝑢2 )] × | − 𝑢1 |
Γ(𝛼)Γ(𝛽) 1 2
Note: We see that 𝑈1 and 𝑈2 are independent since the domain 𝑅𝑈1 ,𝑈2 = {(𝑢1 , 𝑢2 ): 𝑢1 >
0, 0 < 𝑢2 < 1} does not constrain 𝑢1 by 𝑢2 or vice versa and since the nonzero part
of 𝑓𝑈1 ,𝑈2 (𝑢1 , 𝑢2 ) can be factored into the two expressions ℎ1 (𝑢1 ) and ℎ2 (𝑢2 ), where
8
ℎ1 (𝑢1 ) = 𝑢1 𝛼+𝛽−1 𝑒 −𝑢1
and
𝑢2 𝛼−1 (1 − 𝑢2 )𝛽−1
.
Γ(𝛼)Γ(𝛽)
(b) To obtain the marginal distribution of 𝑈1 , we integrate the joint pdf𝑓𝑈1 ,𝑈2 (𝑢1 , 𝑢2 )
over 𝑢2 . That is, for 𝑢1 > 0,
ℎ2 (𝑢2 ) =
1
𝑓𝑈1 (𝑢1 ) = ∫
𝑓𝑈1 ,𝑈2 (𝑢1 , 𝑢2 ) 𝑑𝑢2
𝑢2 =0
1
𝑢2 𝛼−1 (1 − 𝑢2 )𝛽−1 𝛼+𝛽−1 −𝑢
=∫
𝑢1
𝑒 1 𝑑𝑢2
Γ(𝛼)Γ(𝛽)
𝑢2 =0
1
1
=
𝑢1 𝛼+𝛽−1 𝑒 −𝑢1 ∫ 𝑢2 𝛼−1 (1 − 𝑢2 )𝛽−1 𝑑𝑢2
Γ(𝛼)Γ(𝛽)
𝑢2 =0
𝛼−1
𝛽−1
(1−𝑢2 )
(𝑢2
is beta(𝛼,𝛽) kernel)
1
Γ(𝛼)Γ(𝛽)
⇔
𝑢1 𝛼+𝛽−1 𝑒 −𝑢1 ×
Γ(𝛼)Γ(𝛽)
Γ(𝛼 + 𝛽)
1
=
𝑢 𝛼+𝛽−1 𝑒 −𝑢1
Γ(𝛼 + 𝛽) 1
Summarizing,
1
𝑢 𝛼+𝛽−1 𝑒 −𝑢1 , 𝑢1 > 0
𝑓𝑈1 (𝑢1 ) = {Γ(𝛼 + 𝛽) 1
0,
otherwise.
We recognize this as a gamma(𝛼 + 𝛽, 1) pdf; thus, marginally, 𝑈1 ~gamma(𝛼 + 𝛽, 1).
(c) To obtain the marginal distribution of 𝑈2 , we integrate the joint pdf 𝑓𝑈1 ,𝑈2 (𝑢1 , 𝑢2 ) over
𝑢2 . That is, for 0 < 𝑢2 < 1,
∞
𝑓𝑈2 (𝑢2 ) = ∫
𝑓𝑈1 ,𝑈2 (𝑢1 , 𝑢2 ) 𝑑𝑢1
𝑢1 =0
∞
𝑢2 𝛼−1 (1 − 𝑢2 )𝛽−1 𝛼+𝛽−1 −𝑢
𝑢1
𝑒 1 𝑑𝑢1
Γ(𝛼)Γ(𝛽)
𝑢1 =0
𝑢2 𝛼−1 (1 − 𝑢2 )𝛽−1 ∞
=
∫ 𝑢1 𝛼+𝛽−1 𝑒 −𝑢1 𝑑𝑢1
Γ(𝛼)Γ(𝛽)
𝑢1 =0
Γ(𝛼 + 𝛽) 𝛼−1
(1 − 𝑢2 )𝛽−1 .
=
𝑢
Γ(𝛼)Γ(𝛽) 2
=∫
Summarizing,
Γ(𝛼 + 𝛽) 𝛼−1
(1 − 𝑢2 )𝛽−1 , 0 < 𝑢2 < 1
𝑢
𝑓𝑈2 (𝑢2 ) = {Γ(𝛼)Γ(𝛽) 2
0,
otherwise.
Thus, marginally, U2 ~ beta(𝛼, 𝛽). □
REMARK: Suppose that 𝒀 = (𝑌1 , 𝑌2 ) is a continuous random vector with joint pdf
𝑓𝑌1 ,𝑌2 (𝑦1 , 𝑦2 ), and suppose that we would like to find the distribution of a single random
variable
𝑈1 = 𝑔1 (𝑌1 , 𝑌2 )
9
Even though there is no 𝑈2 present here, the bivariate transformation technique can still be
useful. In this case, we can devise an “extra variable” 𝑈2 = 𝑔2 (𝑌1 , 𝑌2 ), perform the bivariate
transformation to obtain 𝑓𝑈1 ,𝑈2 (𝑢1 , 𝑢2 ), and then find the marginal distribution of 𝑈1 by
integrating 𝑓𝑈1 ,𝑈2 (𝑢1 , 𝑢2 ) out over the dummy variable 𝑢2 . While the choice of 𝑈2 is
arbitrary, there are certainly bad choices. Stick with something easy; usually 𝑈2 =
𝑔2 (𝑌1 , 𝑌2 ) = 𝑌2 does the trick.
Exercise: (Homework 2, Question 1) Suppose that 𝑌1 and 𝑌2 are random variables with
joint pdf
8𝑦 𝑦 , 0 < 𝑦1 < 𝑦2 < 1
𝑓𝑌1 ,𝑌2 (𝑦1 , 𝑦2 ) = { 1 2
0,
otherwise.
Find the pdf of 𝑈1 = 𝑌1 /𝑌2 .
More-to-one transformation: What happens if the transformation of Y to U is not a one-toone transformation? In this case, similar to the univariate transformation, we can still use
the pdf technique, but we have to “break up" the transformation 𝒈: 𝑅𝒀 → 𝑅𝑼 into disjoint
regions where g is one-to-one.
RESULT: Suppose that Y is a continuous bivariate random variable with pdf𝑓𝑌1 ,𝑌2 (𝑦1 , 𝑦2 ) and
that 𝑈1 = 𝑔1 (𝑌1 , 𝑌2 ), 𝑈2 = 𝑔2 (𝑌1 , 𝑌2 ), where 𝑼 = (𝑈1 , 𝑈2 ) is not necessarily a one-to-one
(but continuous) function of y over RY = 𝑅𝑌1 ,𝑌2 . Furthermore, suppose that we can partition
𝑅𝒀 into a finite collection of sets, say, A0, A1, A2, … , Ak, where 𝑃(𝒀 ∈ 𝐴0 ) = 0, 𝑎𝑛𝑑 𝑃(𝒀 ∈
𝐴𝑖 ) > 0 for all i ≠ 0, and 𝑓𝑌1 ,𝑌2 (𝑦1 , 𝑦2 ) is continuous on each Ai , i ≠ 0. Furthermore, suppose
that the transformation is 1-to-1 from Ai (i = 1, 2, …, k,) to B, where B is the domain of 𝑼 =
−1
−1
(∙), 𝑔2𝑖
(∙)) is a 1-to-1 inverse mapping of Y
(𝑈1 = 𝑔1 (𝑌1 , 𝑌2 ), 𝑈2 = 𝑔1 (𝑌1 , 𝑌2 )) such that (𝑔1𝑖
to U from B to Ai.
Let Ji denotes the Jacobina computed from the ith inverse, i = 1, 2, …, k. Then, the pdf of U is
given by
𝑘
𝑓𝑈1 ,𝑈2 (𝑢1 , 𝑢2 ) = {
∑ 𝑓𝑌1 ,𝑌2 [𝑔1𝑖 −1 (𝑢, 𝑣), 𝑔2𝑖 −1 (𝑢, 𝑣)]|𝐽𝑖 | ,
𝑢 ∈ 𝐵 = 𝑅𝑈
𝑖=1
0,
otherwise.
Example 7. Suppose that 𝑌1 ~ N(0, 1), 𝑌2 ~ N(0, 1), and that 𝑌1 and 𝑌2 are independent.
Define the transformation
𝑌1
𝑈1 = 𝑔1 (𝑌1 , 𝑌2 ) =
𝑌2
𝑈2 = 𝑔2 (𝑌1 , 𝑌2 ) = |𝑌2 |.
Find each of the following distributions:
(a) 𝑓𝑈1 ,𝑈2 (𝑢1 , 𝑢2 ), the joint distribution of 𝑈1 and 𝑈2 ,
(b) 𝑓𝑈1 (𝑢1 ), the marginal distribution of 𝑈1 .
Solutions. (a) Since 𝑌1 and 𝑌2 are independent, the joint distribution of 𝑌1 and 𝑌2 is
𝑓𝑌1 ,𝑌2 (𝑦1 , 𝑦2 ) = 𝑓𝑌1 (𝑦1 )𝑓𝑌2 (𝑦2 )
10
1 −𝑦 2/2 −𝑦2 /2
𝑒 1 𝑒 2
2π
= {(𝑦1 , 𝑦2 ): −∞ < 𝑦1 < ∞, −∞ < 𝑦2 < ∞}.
=
Here, 𝑅𝑌1 ,𝑌2
The transformation of Y to U is not one-to-one because the points (𝑦1 , 𝑦2 ) and (−𝑦1 , −𝑦2 )
are both mapped to the same (𝑢1 , 𝑢2 ) point. But if we restrict considerations to either
positive or negative values of 𝑦2 , then the transformation is one-to-one. We note that the
three sets below form a partition of 𝐴 = 𝑅𝑌1 ,𝑌2 as defined above with 𝐴1 = {(𝑦1 , 𝑦2 ): 𝑦2 > 0},
𝐴2 = {(𝑦1 , 𝑦2 ): 𝑦2 < 0} , and 𝐴0 = {(𝑦1 , 𝑦2 ): 𝑦2 = 0}.
The domain of U, 𝐵 = {(𝑢1 , 𝑢2 ): −∞ < 𝑢1 < ∞, 𝑢2 > 0} is the image of both 𝐴1 and 𝐴2 under
the transformation. The inverse transformation from 𝐵 𝑡𝑜 𝐴1 and 𝐵 𝑡𝑜 𝐴2 are given by:
𝑦1 = 𝑔11 −1 (𝑢1 , 𝑢2 ) = 𝑢1 𝑢2
𝑦2 = 𝑔21 −1 (𝑢1 , 𝑢2 ) = 𝑢2
and
𝑦1 = 𝑔12 −1 (𝑢1 , 𝑢2 ) = −𝑢1 𝑢2
𝑦2 = 𝑔22 −1 (𝑢1 , 𝑢2 ) = −𝑢2
The Jacobians from the two inverses are 𝐽1 = 𝐽1 = 𝑢2
The pdf of U on its domain B is thus:
2
𝑓𝑈1 ,𝑈2 (𝑢1 , 𝑢2 ) = ∑ 𝑓𝑌1 ,𝑌2 [𝑔1𝑖 −1 (𝑢, 𝑣), 𝑔2𝑖 −1 (𝑢, 𝑣)]|𝐽𝑖 |
𝑖=1
Plugging in, we have:
𝑓𝑈1 ,𝑈2 (𝑢1 , 𝑢2 ) =
Simplifying, we have:
1 −(𝑢 𝑢 )2 /2 −𝑢2/2
1 −(−𝑢 𝑢 )2 /2 −(−𝑢 )2 /2
1 2
2
|𝑢2 |
𝑒 1 2
𝑒 2 |𝑢2 | +
𝑒
𝑒
2π
2π
𝑢2 −(𝑢2 +1)𝑢2 /2
2 , −∞ < 𝑢 < ∞, 𝑢 > 0
𝑒 1
1
2
π
(b) To obtain the marginal distribution of 𝑈1 , we integrate the joint pdf𝑓𝑈1 ,𝑈2 (𝑢1 , 𝑢2 )
over 𝑢2 . That is,
𝑓𝑈1 ,𝑈2 (𝑢1 , 𝑢2 ) =
∞
𝑓𝑈1 (𝑢1 ) = ∫ 𝑓𝑈1 ,𝑈2 (𝑢1 , 𝑢2 ) 𝑑𝑢2
0
1
, −∞ < 𝑢1 < ∞
𝜋(𝑢12 + 1)
Thus, marginally, U1 follows the standard Cauchy distribution. □
=
REMARK: The transformation method can also be extended to handle n-variate
transformations. Suppose that 𝑌1 , 𝑌2 , … , 𝑌𝑛 are continuous random variables with joint pdf
𝑓𝒀 (𝑦) and define
𝑈1 = 𝑔1 (𝑌1 , 𝑌2 , … , 𝑌𝑛 )
𝑈2 = 𝑔2 (𝑌1 , 𝑌2 , … , 𝑌𝑛 )
⋮
11
𝑈𝑛 = 𝑔𝑛 (𝑌1 , 𝑌2 , … , 𝑌𝑛 ).
Example 8. Given independent random variables 𝑋 and𝑌, each with uniform distributions
on (0, 1), find the joint pdf of U and V defined by U=X+Y, V=X-Y, and the marginal pdf of U.
The joint pdf of 𝑋 and 𝑌 is𝑓𝑋,𝑌 (𝑥, 𝑦) = 1, 0 ≤ 𝑥 ≤ 1, 0 ≤ 𝑦 ≤ 1.
The inverse transformation, written in terms of observed values is
𝑢+𝑣
𝑢−𝑣
𝑥=
, 𝑎𝑛𝑑 𝑦 =
.
2
2
It is clearly one-to-one. The Jacobian is
1/2 1/2
𝜕(𝑥,𝑦)
1
1
𝐽 = 𝜕(𝑢,𝑣) = |
| = − 2, so |𝐽| = 2.
1/2 −1/2
We will use 𝒜 to denote the range space of (𝑋, 𝑌), and ℬ to denote that of (𝑈, 𝑉), and these
are shown in the diagrams below. Firstly, note that there are 4 inequalities specifying
ranges of 𝑥 and𝑦, and these give 4 inequalities concerning 𝑢 and𝑣, from which ℬcan be
determined. That is,
𝑥 ≥ 0 ⇒ 𝑢 + 𝑣 ≥ 0, that is, 𝑣 ≥ −𝑢
𝑥 ≤ 1 ⇒ 𝑢 + 𝑣 ≤ 2, that is 𝑣 ≤ 2 − 𝑢
𝑦 ≥ 0 ⇒ 𝑢 − 𝑣 ≥ 0, that is 𝑣 ≤ 𝑢
𝑦 ≤ 1 ⇒ 𝑢 − 𝑣 ≤ 2, that is 𝑣 ≥ 𝑢 − 2
Drawing the four lines
𝑣 = −𝑢, 𝑣 = 2 − 𝑢, 𝑣 = 𝑢, 𝑣 = 𝑢 − 2
On the graph, enables us to see the region specified by the 4 inequalities.
12
Now, we have
1 1
−𝑢 ≤ 𝑣 ≤ 𝑢, 0 ≤ 𝑢 ≤ 1
= ,
{
𝑢 − 2 ≤ 𝑣 ≤ 2 − 𝑢, 1 ≤ 𝑢 ≤ 2
2 2
The importance of having the range space correct is seen when we find marginal pdf of 𝑈.
∞
𝑓𝑈 (𝑢) = ∫−∞ 𝑓𝑈,𝑉 (𝑢, 𝑣)𝑑𝑣
𝑓𝑈,𝑉 (𝑢, 𝑣) = 1 ∗
𝑢 1
∫−𝑢 2 𝑑𝑣,
= { ∫2−𝑢 1 𝑑𝑣,
𝑢−2 2
0,
={
0≤𝑢≤1
1≤𝑢≤2
𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
𝑢,
0≤𝑢≤1
2 − 𝑢, 1 ≤ 𝑢 ≤ 2
= 𝑢𝐼[0,1] (𝑢) + (2 − 𝑢)𝐼(1,2] (𝑢), using indicator functions.
Example 9. Given 𝑋and 𝑌 are independent random variables each with pdf𝑓𝑋 (𝑥) =
1 −𝑥
𝑒 2 , 𝑥 ∈ [0, ∞), find the distribution of(𝑋 − 𝑌)/2.
2
We note that the joint pdf of 𝑋 and 𝑌 is
1 𝑥+𝑦
𝑓𝑋,𝑌 (𝑥, 𝑦) = 𝑒 2 , 0 ≤ 𝑥 < ∞, 0 ≤ 𝑦 < ∞.
4
Define𝑈 = (𝑋 − 𝑌)/2. Now we need to introduce a second random variable 𝑉 which is a
function of 𝑋 and𝑌. We wish to do this in such a way that the resulting bivariate
13
transformation is one-to-one and our actual task of finding the pdf of U is as easy as
possible. Our choice for 𝑉 is of course, not unique. Let us define𝑉 = 𝑌. Then the
transformation is, (using𝑢, 𝑣, 𝑥, 𝑦, since we are really dealing with the range spaces here).
𝑥 = 2𝑢 + 𝑣
𝑦=𝑣
From it, we find the Jacobian,
2 1
𝐽=|
|=2
0 1
To determineℬ, the range space of 𝑈 and𝑉, we note that
𝑥 ≥ 0 ⇒ 2𝑢 + 𝑣 ≥ 0 , that is 𝑣 ≥ −2𝑢
𝑥 < ∞ ⇒ 2𝑢 + 𝑣 < ∞
𝑦≥0 ⇒ 𝑣≥0
𝑦<∞ ⇒ 𝑢<∞
So ℬ is as indicated in the diagram below.
Now, we have
𝑓𝑈,𝑉 (𝑢, 𝑣) =
1 −2𝑢+𝑣+𝑣
1
2
𝑒
∗ 2 = 𝑒 −(𝑢+𝑣) , (𝑢, 𝑣) ∈ ℬ.
4
2
The marginal pdf of 𝑈 is obtained by integrating 𝑓𝑈,𝑉 (𝑢, 𝑣) with respect to𝑣, giving
∞
1 −(𝑢+𝑣)
∫
𝑒
𝑑𝑣, 𝑢 < 0
−2𝑢 2
𝑓𝑈 (𝑢)
=
∞
1 −(𝑢+𝑣)
∫
𝑒
𝑑𝑣, 𝑢 > 0
{ 0 2
14
1 𝑢
𝑒 ,
𝑢<0
= {2
1 −𝑢
𝑒 ,
𝑢>0
2
1
= 𝑒 −|𝑢| , − ∞ < 𝑢 < ∞
2
[This is sometimes called the 𝑓𝑜𝑙𝑑𝑒𝑑 (𝑜𝑟 𝑑𝑜𝑢𝑏𝑙𝑒)𝑒𝑥𝑝𝑜𝑛𝑒𝑛𝑡𝑖𝑎𝑙 distribution.]
Homework #2.
Question 1 is the exercise on page 9 of this handout.
Questions 2-8 (from our textbook): 2.3, 2.9, 2.11, 2.13, 2.18, 2.34, 2.38
15