1

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

𝐹𝑌(𝑦) = {

0, 𝑦 ≤ 0𝑦, 0 < 𝑦 ≤ 11, 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)

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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

𝑑𝑢𝐹𝑈(𝑢) =

𝑑

𝑑𝑢(1 − 𝑒−𝑢) = 𝑒−𝑢

Summarizing,

𝑓𝑈(𝑢) = {𝑒−𝑢, 𝑢 > 00, otherwise

This is an exponential pdf with mean 1

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

𝐹𝑌(𝑦) =

{

0, 𝑦 ≤ −𝜋 2⁄

𝑦 + 𝜋 2⁄

𝜋, −𝜋 2⁄ < 𝑦 ≤ 𝜋 2⁄

1, 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⁄

𝜋for − 𝜋 2⁄ < 𝑦 < 𝜋 2⁄ ; i. e. , for 𝑢 ∈ ℛ ,

we have

𝐹𝑈(𝑢) = 𝐹𝑌[tan−1(𝑢)] =

tan−1(𝑢) + 𝜋 2⁄

𝜋

The pdf of U, for 𝑢 ∈ ℛ, is given by

𝑓𝑈(𝑢) =𝑑

𝑑𝑢𝐹𝑈(𝑢) =

𝑑

𝑑𝑢[tan−1(𝑢) + 𝜋 2⁄

𝜋] =

1

𝜋(1 + 𝑢2).

Summarizing,

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𝑓𝑈(𝑢) = {

1

𝜋(1 + 𝑢2), − ∞ < 𝑢 < ∞

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 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

𝑑𝑢𝐹𝑈(𝑢) =

𝑑

𝑑𝑢𝐹𝑌[𝑔

−1(𝑢)]

= 𝑓𝑌[𝑔−1(𝑢)]

𝑑

𝑑𝑢𝑔−1(𝑢) (by chain rule)

Now as 𝑔 is increasing, so is 𝑔−1; thus, 𝑑

𝑑𝑢𝑔−1(𝑢) > 0. If 𝑔(𝑦) is

strictly decreasing, then

𝐹𝑈(𝑢) = 1 − 𝐹𝑌[𝑔−1(𝑢)] and

𝑑

𝑑𝑢𝑔−1(𝑢) < 0, which gives

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𝑓𝑈(𝑢) =𝑑

𝑑𝑢𝐹𝑈(𝑢) =

𝑑

𝑑𝑢{1 − 𝐹𝑌[𝑔

−1(𝑢)]}

= −𝑓𝑌[𝑔−1(𝑢)]

𝑑

𝑑𝑢𝑔−1(𝑢)

Combining both cases, we have shown that the pdf of U, where

nonzero, is given by

𝑓𝑈(𝑢) = 𝑓𝑌[𝑔−1(𝑢)] |

𝑑

𝑑𝑢𝑔−1(𝑢) |.

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

𝑓𝑌(𝑦) = {

1

𝛽𝑒−𝑦/𝛽 , 𝑦 > 0

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. 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:

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𝑑

𝑑𝑢𝑔−1(𝑢) =

𝑑

𝑑𝑢(𝑢2) = 2𝑢.

Thus, for u > 0,

𝑓𝑈(𝑢) = 𝑓𝑌[𝑔−1(𝑢)] |

𝑑

𝑑𝑢𝑔−1(𝑢) |

=1

𝛽𝑒−𝑢2

𝛽 × |2𝑢| =2𝑢

𝛽𝑒−𝑢2

𝛽 .

Summarizing,

𝑓𝑈(𝑢) = {2𝑢

𝛽𝑒−𝑢2

𝛽 , 𝑢 > 0

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

𝑓𝑌(𝑦) = {42𝑦5(1 − 𝑦), 0 < 𝑦 < 10, 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

𝑑𝑢𝑔−1(𝑢) =

𝑑

𝑑𝑢(1 − 𝑢) = −1.

Thus, for 0 < u < 1,

𝑓𝑈(𝑢) = 𝑓𝑌[𝑔−1(𝑢)] |

𝑑

𝑑𝑢𝑔−1(𝑢) |

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= 42(1 − 𝑢)5[1 − (1 − 𝑢)] × |−1| = 42𝑢(1 − 𝑢)5.

Summarizing,

𝑓𝑈(𝑢) = {42𝑢(1 − 𝑢)5, 0 < 𝑢 < 10, 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

𝑓𝑈(𝑢) = {∑𝑓𝑌[𝑔𝑖

−1(𝑢)] |𝑑

𝑑𝑢𝑔𝑖−1(𝑢) |

𝑘

𝑖=1

, 𝑢 ∈ 𝑅𝑈

0, otherwise.

That is, writing the pdf of U can be done by adding up the terms

𝑓𝑌[𝑔𝑖−1(𝑢)] |

𝑑

𝑑𝑢𝑔𝑖−1(𝑢) | corresponding to each disjoint set Ai,

i = 1, 2,…,k.

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Example 5. Suppose that Y ~ N(0, 1); that is, Y has a standard normal distribution; i.e.,

𝑓𝑌(𝑦) = {

1

√2𝜋𝑒−𝑦

2/2, −∞ < 𝑦 < ∞

0, otherwise.

Consider the transformation: 𝑈 = 𝑔(𝑌) = 𝑌2.

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) = 𝑦2 is decreasing on A1 and 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 Transformation Inverse transformation

A1 = (−∞, 0) 𝑔(𝑦) = 𝑦2 = 𝑢 𝑔1−1(𝑢) = −√𝑢 = 𝑦

A2 = (0, ∞) 𝑔(𝑦) = 𝑦2 = 𝑢 𝑔2−1(𝑢) = √𝑢 = 𝑦

And, on both sets A1 and A2,

|𝑑

𝑑𝑢𝑔𝑖−1(𝑢) | =

1

2√𝑢.

Clearly, 𝑢 = 𝑦2 > 0; thus, 𝑅𝑈 = {𝑢: 𝑢 > 0}, and the pdf of U is given by

𝑓𝑈(𝑢)

= {1

√2𝜋𝑒−(−√𝑢)

2

2 (1

2√𝑢) +

1

√2𝜋𝑒−(√𝑢)

2

2 (1

2√𝑢) , 𝑢 > 0

0, otherwise.

Thus, for u > 0, and recalling that Γ (1

2) = √𝜋, 𝑓(𝑢) collapses to

𝑓𝑈(𝑢) =2

√2𝜋𝑒−𝑢2 (

1

2√𝑢)

=1

√2𝜋𝑢12−1𝑒−

𝑢2 =

1

Γ (12) 2

12

𝑢12−1𝑒−

𝑢2 .

Summarizing, the pdf of U is

𝑓𝑈(𝑢) = {

1

Γ (12) 2

12

𝑢12−1𝑒−

𝑢2 , 𝑢 > 0

0, otherwise.

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That is, U ~ gamma(1/2, 2). Recall that the gamma(1/2, 2)

distribution is the same as a 𝜒2 distribution with 1 degree of

freedom; that is, 𝑈 ~ 𝜒2(1). □

Solution 2 (the cdf technique):

𝐹𝑈(𝑢) = 𝑃(𝑈 ≤ 𝑢) = 𝑃(𝑌2 ≤ 𝑢) = 1 − 𝑃(𝑌2 > 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

𝑑𝑢+ 𝑓𝑌(−√𝑢)

𝑑√𝑢

𝑑𝑢

=1

√2𝜋𝑒−(−√𝑢)

2

2 (1

2√𝑢) +

1

√2𝜋𝑒−(√𝑢)

2

2 (1

2√𝑢)

=1

√2𝜋𝑢12−1𝑒−

𝑢2 =

1

Γ (12)2

12

𝑢12−1𝑒−

𝑢2 , 𝑢 > 0

That is, U ~ gamma(1/2, 2). Recall that the gamma(1/2, 2)

distribution is the same as a 𝜒2 distribution with 1 degree of

freedom; that is, 𝑈 ~ 𝜒2(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 one-to-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: ℛ2 → ℛ2 be a continuous one-to-

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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”,

𝐽 = det ||

𝜕𝑔1−1(𝑢1, 𝑢2)

𝜕𝑢1

𝜕𝑔1−1(𝑢1, 𝑢2)

𝜕𝑢2𝜕𝑔2

−1(𝑢1, 𝑢2)

𝜕𝑢1

𝜕𝑔2−1(𝑢1, 𝑢2)

𝜕𝑢2

|| ≠ 0,

then 𝑓𝑈1,𝑈2(𝑢1, 𝑢2)

= {𝑓𝑌1,𝑌2[𝑔1

−1(𝑢1, 𝑢2), 𝑔2−1(𝑢1, 𝑢2)]|𝐽|, (𝑢1, 𝑢2) ∈ 𝑅𝑈1,𝑈2

0, otherwise,

where |J| denotes the absolute value of J. RECALL: The determinant of a 2 × 2 matrix, e.g.,

det |𝑎 𝑏𝑐 𝑑

| = 𝑎𝑑 − 𝑏𝑐.

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).

4. Find the Jacobian, J, of the inverse transformation. 5. Use the formula above to find 𝑓𝑈1,𝑈2(𝑢1, 𝑢2), the joint

distribution of 𝑈1and 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

𝑈2 = 𝑔2(𝑌1, 𝑌2) =𝑌1

𝑌1 + 𝑌2.

Find each of the following distributions: (a) 𝑓𝑈1,𝑈2(𝑢1, 𝑢2), the joint distribution of 𝑈1and 𝑈2,

(b) 𝑓𝑈1(𝑢1), the marginal distribution of 𝑈1, and

(c) 𝑓𝑈2(𝑢2), the marginal distribution of 𝑈2.

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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

Γ(𝛼)𝑦1𝛼−1𝑒−𝑦1 ×

1

Γ(𝛽)𝑦2𝛽−1𝑒−𝑦2

=1

Γ(𝛼)Γ(𝛽)𝑦1𝛼−1𝑦2

𝛽−1𝑒−(𝑦1+𝑦2),

for 𝑦1 > 0, 𝑦2 > 0, and 0, otherwise. Here, 𝑅𝑌1,𝑌2 =

{(𝑦1, 𝑦2): 𝑦1 > 0, 𝑦2 > 0}. By inspection, we see that 𝑢1 = 𝑦1 +

𝑦2 > 0, and 𝑢2 =𝑦1

𝑦1+ 𝑦2 must fall between 0 and 1.

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

𝑢2 =𝑦1

𝑦1 + 𝑦2

⇒𝑦1 = 𝑔1

−1(𝑢1, 𝑢2) = 𝑢1𝑢2𝑦2 = 𝑔2

−1(𝑢1, 𝑢2) = 𝑢1 − 𝑢1𝑢2

The Jacobian is given by

𝐽 = det ||

𝜕𝑔1−1(𝑢1, 𝑢2)

𝜕𝑢1

𝜕𝑔1−1(𝑢1, 𝑢2)

𝜕𝑢2𝜕𝑔2

−1(𝑢1, 𝑢2)

𝜕𝑢1

𝜕𝑔2−1(𝑢1, 𝑢2)

𝜕𝑢2

|| = det |

𝑢2 𝑢11 − 𝑢2 −𝑢1

|

= −𝑢1𝑢2 − 𝑢1(1 − 𝑢2) = −𝑢1. 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

Γ(𝛼)Γ(𝛽)(𝑢1𝑢2)

𝛼−1(𝑢1 − 𝑢1𝑢2)𝛽−1𝑒−[(𝑢1𝑢2)+(𝑢1−𝑢1𝑢2)]

× | − 𝑢1|

Note: We see that 𝑈1and 𝑈2are 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 ℎ1(𝑢1) = 𝑢1

𝛼+𝛽−1𝑒−𝑢1 and

ℎ2(𝑢2) =𝑢2𝛼−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,

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𝑓𝑈1(𝑢1) = ∫ 𝑓𝑈1,𝑈2(𝑢1, 𝑢2)1

𝑢2=0

𝑑𝑢2

= ∫𝑢2𝛼−1(1 − 𝑢2)

𝛽−1

Γ(𝛼)Γ(𝛽)𝑢1𝛼+𝛽−1𝑒−𝑢1

1

𝑢2=0

𝑑𝑢2

=1

Γ(𝛼)Γ(𝛽)𝑢1𝛼+𝛽−1𝑒−𝑢1∫ 𝑢2

𝛼−1(1 − 𝑢2)𝛽−1

1

𝑢2=0

𝑑𝑢2

(𝑢2𝛼−1(1−𝑢2)

𝛽−1 is beta(𝛼,𝛽) kernel)⇔

1

Γ(𝛼)Γ(𝛽)𝑢1𝛼+𝛽−1𝑒−𝑢1

×Γ(𝛼)Γ(𝛽)

Γ(𝛼 + 𝛽)

=1

Γ(𝛼 + 𝛽)𝑢1𝛼+𝛽−1𝑒−𝑢1

Summarizing,

𝑓𝑈1(𝑢1) = {

1

Γ(𝛼 + 𝛽)𝑢1𝛼+𝛽−1𝑒−𝑢1 , 𝑢1 > 0

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)∞

𝑢1=0

𝑑𝑢1

= ∫𝑢2𝛼−1(1 − 𝑢2)

𝛽−1

Γ(𝛼)Γ(𝛽)𝑢1𝛼+𝛽−1𝑒−𝑢1

∞

𝑢1=0

𝑑𝑢1

=𝑢2𝛼−1(1 − 𝑢2)

𝛽−1

Γ(𝛼)Γ(𝛽)∫ 𝑢1

𝛼+𝛽−1𝑒−𝑢1∞

𝑢1=0

𝑑𝑢1

=Γ(𝛼 + 𝛽)

Γ(𝛼)Γ(𝛽)𝑢2𝛼−1(1 − 𝑢2)

𝛽−1.

Summarizing,

𝑓𝑈2(𝑢2) = {

Γ(𝛼 + 𝛽)

Γ(𝛼)Γ(𝛽)𝑢2𝛼−1(1 − 𝑢2)

𝛽−1, 0 < 𝑢2 < 1

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)

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

12

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 3, Question 1) Suppose that 𝑌1 and 𝑌2

are random variables with joint pdf

𝑓𝑌1,𝑌2(𝑦1, 𝑦2) = {8𝑦1𝑦2, 0 < 𝑦1 < 𝑦2 < 10, 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-to-one 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 = (𝑈1 =

𝑔1(𝑌1, 𝑌2), 𝑈2 = 𝑔1(𝑌1, 𝑌2)) such that (𝑔1𝑖−1(∙), 𝑔2𝑖

−1(∙)) is a 1-to-1

inverse mapping of Y 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

𝑓𝑈1,𝑈2(𝑢1, 𝑢2)

= {∑𝑓𝑌1,𝑌2[𝑔1𝑖

−1(𝑢, 𝑣), 𝑔2𝑖−1(𝑢, 𝑣)]|𝐽𝑖|

𝑘

𝑖=1

, 𝑢 ∈ 𝐵 = 𝑅𝑈

0, otherwise.

Example 7. Suppose that 𝑌1~ N(0, 1), 𝑌2~ N(0, 1), and that 𝑌1 and 𝑌2 are independent. Define the transformation

𝑈1 = 𝑔1(𝑌1, 𝑌2) =𝑌1𝑌2

𝑈2 = 𝑔2(𝑌1, 𝑌2) = |𝑌2|. Find each of the following distributions:

13

(a) 𝑓𝑈1,𝑈2(𝑢1, 𝑢2), the joint distribution of 𝑈1and 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)

=1

2π𝑒−𝑦1

2/2𝑒−𝑦22/2

Here, 𝑅𝑌1,𝑌2 = {(𝑦1, 𝑦2):−∞ < 𝑦1 < ∞,−∞ < 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) = 𝑢1𝑢2

𝑦2 = 𝑔21−1(𝑢1, 𝑢2) = 𝑢2

and 𝑦1 = 𝑔12

−1(𝑢1, 𝑢2) = −𝑢1𝑢2 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:

𝑓𝑈1,𝑈2(𝑢1, 𝑢2) =∑𝑓𝑌1 ,𝑌2[𝑔1𝑖−1(𝑢, 𝑣), 𝑔2𝑖

−1(𝑢, 𝑣)]|𝐽𝑖|

2

𝑖=1

Plugging in, we have:

𝑓𝑈1,𝑈2(𝑢1, 𝑢2) =1

2π𝑒−(𝑢1𝑢2)

2/2𝑒−𝑢22/2|𝑢2|

+1

2π𝑒−(−𝑢1𝑢2)

2/2𝑒−(−𝑢2)2/2|𝑢2|

Simplifying, we have:

𝑓𝑈1 ,𝑈2(𝑢1, 𝑢2) =𝑢2π𝑒−(𝑢1

2+1)𝑢22/2, −∞ < 𝑢1 < ∞, 𝑢2 > 0

(b) To obtain the marginal distribution of 𝑈1, we integrate the joint pdf𝑓𝑈1,𝑈2(𝑢1, 𝑢2)

over 𝑢2. That is,

𝑓𝑈1(𝑢1) = ∫ 𝑓𝑈1,𝑈2(𝑢1, 𝑢2)∞

0

𝑑𝑢2

14

= 1

𝜋(𝑢12 + 1)

, −∞ < 𝑢1 < ∞

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)

⋮ 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

𝑥 =𝑢 + 𝑣

2, 𝑎𝑛𝑑 𝑦 =

𝑢 − 𝑣

2.

It is clearly one-to-one. The Jacobian is

𝐽 =𝜕(𝑥,𝑦)

𝜕(𝑢,𝑣)= |1/2 1/21/2 −1/2

| = −1

2, so |𝐽| =

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.

15

Now, we have

𝑓𝑈,𝑉(𝑢, 𝑣) = 1 ∗1

2=1

2, {

−𝑢 ≤ 𝑣 ≤ 𝑢, 0 ≤ 𝑢 ≤ 1𝑢 − 2 ≤ 𝑣 ≤ 2 − 𝑢, 1 ≤ 𝑢 ≤ 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∞

−∞

= {

∫1

2𝑑𝑣, 0 ≤ 𝑢 ≤ 1

𝑢

−𝑢

∫1

2𝑑𝑣, 1 ≤ 𝑢 ≤ 2

2−𝑢

𝑢−2

0, 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒

= {𝑢, 0 ≤ 𝑢 ≤ 12 − 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

2𝑒−

𝑥

2, 𝑥 ∈ [0,∞), find the distribution

of(𝑋 − 𝑌)/2. We note that the joint pdf of 𝑋 and 𝑌 is

𝑓𝑋,𝑌(𝑥, 𝑦) =1

4𝑒𝑥+𝑦2 , 0 ≤ 𝑥 < ∞, 0 ≤ 𝑦 < ∞.

16

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 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,

𝐽 = |2 10 1

| = 2

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

4𝑒−2𝑢+𝑣+𝑣

2 ∗ 2 =1

2𝑒−(𝑢+𝑣), (𝑢, 𝑣) ∈ ℬ.

17

The marginal pdf of 𝑈 is obtained by integrating 𝑓𝑈,𝑉(𝑢, 𝑣) with

respect to𝑣, giving

𝑓𝑈(𝑢) =

{

∫

1

2𝑒−(𝑢+𝑣)𝑑𝑣, 𝑢 < 0

∞

−2𝑢

∫1

2𝑒−(𝑢+𝑣)

∞

0

𝑑𝑣, 𝑢 > 0

= {

1

2𝑒𝑢, 𝑢 < 0

1

2𝑒−𝑢, 𝑢 > 0

=1

2𝑒−|𝑢| , − ∞ < 𝑢 < ∞

[This is sometimes called the 𝑓𝑜𝑙𝑑𝑒𝑑 (𝑜𝑟 𝑑𝑜𝑢𝑏𝑙𝑒)𝑒𝑥𝑝𝑜𝑛𝑒𝑛𝑡𝑖𝑎𝑙 distribution.]

18

Homework #3. Question 1 is the exercise on page 9 of this handout. Questions 2-9 (from our textbook): 2.9, 2.15, 2.24, 2.30, 2.31, 2.32, 2.33, 2.34