# Transformations

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) ๐น๐ (๐ฆ) โก ๐(๐ โค ๐ฆ). Let ๐ = ๐(๐) be a function of Y, and our goal is to find the distribution of U. The cdf technique is especially convenient when the cdf ๐น๐ (๐ฆ) 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. Write๐น๐ (๐ข) = ๐(๐ โค ๐ข), the cdf of U, in terms of ๐น๐ (๐ฆ), the cdf of Y . 3. Differentiate ๐น๐ (๐ข) to obtain the pdf of U, ๐๐ (๐ข). Example 1. Suppose that ๐ ~ ๐(0,1). Find the distribution of ๐ = ๐(๐) = โ ln ๐. Solution. The cdf of ๐ ~ ๐(0,1) is given by 0, ๐ฆโค0 ๐น๐ (๐ฆ) = {๐ฆ, 0 < ๐ฆ โค 1 1, ๐ฆโฅ1 The domain (*domain is the region where the pdf is non-zero) for ๐ ~ ๐(0,1) is ๐๐ = {๐ฆ: 0 < ๐ฆ < 1}; thus, because ๐ข = โ ln ๐ฆ > 0, it follows that the domain for U is ๐๐ = {๐ข: ๐ข > 0}. The cdf of U is: ๐น๐ (๐ข) = ๐(๐ โค ๐ข) = ๐(โ ln ๐ โค ๐ข) = ๐(ln ๐ > โ๐ข) = ๐(๐ > ๐ โ๐ข ) = 1 โ ๐(๐ โค ๐ โ๐ข ) = 1 โ ๐น๐ (๐ โ๐ข )

1

Because ๐น๐ (๐ฆ) = ๐ฆ for 0 < y < 1; i.e., for u > 0, we have ๐น๐ (๐ข) = 1 โ ๐น๐ (๐ โ๐ข ) = 1 โ ๐ โ๐ข Taking derivatives, we get, for u > 0, ๐๐ (๐ข) =

๐ ๐ (1 โ ๐ โ๐ข ) = ๐ โ๐ข ๐น๐ (๐ข) = ๐๐ข ๐๐ข

Summarizing, ๐๐ (๐ข) = {

๐ โ๐ข , 0,

๐ข>0 otherwise 1

This is an exponential pdf with mean ฮป = 1; that is, U ~ exponential(ฮป = 1). โก Example 2. Suppose that ๐ ~ ๐ (โ ๐โ2 , ๐โ2) . Find the distribution of the random variable defined by U = g(Y ) = tan(Y ). Solution. The cdf of ๐ ~ ๐ (โ ๐โ2 , ๐โ2) is given by 0, ๐ฆ + ๐โ2 ๐น๐ (๐ฆ) = , ๐ { 1,

๐ฆ โค โ ๐โ2 โ ๐โ2 < ๐ฆ โค ๐โ2 ๐ฆ โฅ ๐โ2

The domain for Y is ๐๐ = {๐ฆ: โ ๐โ2 < ๐ฆ < ๐โ2}. Sketching a graph of the tangent function from โ ๐โ2 to ๐โ2, we see that โโ < ๐ข < โ . Thus, ๐๐ = { ๐ข: โ โ < ๐ข < โ} โก ๐, the set of all reals. The cdf of U is: ๐น๐ (๐ข) = ๐(๐ โค ๐ข) = ๐[tan(๐) โค ๐ข] = ๐[๐ โค tanโ1(๐ข)] = ๐น๐ [tanโ1(๐ข)] Because ๐น๐ (๐ฆ) = we have

๐ฆ+๐โ2 ๐

for โ ๐โ2 < ๐ฆ < ๐โ2 ; i. e. , for ๐ข โ โ ,

๐น๐ (๐ข) = ๐น๐ [tanโ1 (๐ข)] =

tanโ1(๐ข) + ๐โ2 ๐

The pdf of U, for ๐ข โ โ, is given by ๐ ๐ tanโ1 (๐ข) + ๐โ2 1 ๐๐ (๐ข) = ๐น๐ (๐ข) = [ ]= . ๐๐ข ๐๐ข ๐ ๐(1 + ๐ข2 ) Summarizing,

2

1 , โโ0 ๐๐ (๐ฆ) = { ๐ฝ 0, otherwise. Let ๐ = ๐(๐) = โ๐, . Use the method of transformations to find the pdf of U. Solution. First, we note that the transformation ๐(๐) = โ๐ is a continuous strictly increasing function of y over ๐๐ = {๐ฆ: ๐ฆ > 0}, and, thus, ๐(๐) is one-to-one. Next, we need to find the domain of U. This is easy since y > 0 implies ๐ข = โ๐ฆ > 0 as well. Thus, ๐๐ = {๐ข: ๐ข > 0}. Now, we find the inverse transformation: ๐(๐ฆ) = ๐ข = โ๐ฆ โ ๐ฆ = ๐โ1 (๐ข) = ๐ข2 (by inverse transformation) and its derivative:

4

๐ โ1 ๐ (๐ข2 ) = 2๐ข. ๐ (๐ข) = ๐๐ข ๐๐ข Thus, for u > 0, ๐๐ (๐ข) = ๐๐ [๐โ1 (๐ข)] |

๐ โ1 ๐ (๐ข) | ๐๐ข 2

2

1 โ๐ข 2๐ข โ๐ข๐ฝ = ๐ ๐ฝ ร |2๐ข| = ๐ . ๐ฝ ๐ฝ Summarizing, 2

2๐ข โ๐ข๐ฝ ๐ข>0 ๐๐ (๐ข) = { ๐ฝ ๐ , 0, otherwise. This is a Weibull distribution. The Weibull family of distributions is common in life science (survival analysis), engineering and actuarial science applications. โก Example 4. Suppose that Y ~ beta(ฮฑ = 6; ฮฒ = 2); i.e., the pdf of Y is given by 42๐ฆ 5 (1 โ ๐ฆ), ๐๐ (๐ฆ) = { 0,

0 0 for all i โ  0, and ๐๐1 ,๐2 (๐ฆ1 , ๐ฆ2 ) is continuous on each Ai , i โ  0. Furthermore, suppose that the transformation is 1-to-1 from Ai (i = 1, 2, โฆ, k,) to B, where B is the domain of ๐ผ = (๐1 = โ1 โ1 (โ), ๐2๐ (โ)) is a 1-to-1 ๐1 (๐1 , ๐2 ), ๐2 = ๐1 (๐1 , ๐2 )) such that (๐1๐ inverse mapping of Y to U from B to Ai. Let Ji denotes the Jacobina computed from the ith inverse, i = 1, 2, โฆ, k. Then, the pdf of U is given by ๐๐1 ,๐2 (๐ข1 , ๐ข2 ) ๐

={

โ ๐๐1 ,๐2 [๐1๐ โ1 (๐ข, ๐ฃ), ๐2๐ โ1 (๐ข, ๐ฃ)]|๐ฝ๐ | ,

๐ข โ ๐ต = ๐๐

๐=1

0,

otherwise.

Example 7. Suppose that ๐1 ~ N(0, 1), ๐2 ~ N(0, 1), and that ๐1 and ๐2 are independent. Define the transformation ๐1 ๐1 = ๐1 (๐1 , ๐2 ) = ๐2 ๐2 = ๐2 (๐1 , ๐2 ) = |๐2 |. Find each of the following distributions: 12

(a) ๐๐1 ,๐2 (๐ข1 , ๐ข2 ), the joint distribution of ๐1 and ๐2 , (b) ๐๐1 (๐ข1 ), the marginal distribution of ๐1 . Solutions. (a) Since ๐1 and ๐2 are independent, the joint distribution of ๐1 and ๐2 is ๐๐1 ,๐2 (๐ฆ1 , ๐ฆ2 ) = ๐๐1 (๐ฆ1 )๐๐2 (๐ฆ2 ) 1 โ๐ฆ 2/2 โ๐ฆ2 /2 = ๐ 1 ๐ 2 2ฯ Here, ๐๐1 ,๐2 = {(๐ฆ1 , ๐ฆ2 ): โโ < ๐ฆ1 < โ, โโ < ๐ฆ2 < โ}. The transformation of Y to U is not one-to-one because the points (๐ฆ1 , ๐ฆ2 ) and (โ๐ฆ1 , โ๐ฆ2 ) are both mapped to the same (๐ข1 , ๐ข2 ) point. But if we restrict considerations to either positive or negative values of ๐ฆ2 , then the transformation is one-to-one. We note that the three sets below form a partition of ๐ด = ๐๐1 ,๐2 as defined above with ๐ด1 = {(๐ฆ1 , ๐ฆ2 ): ๐ฆ2 > 0}, ๐ด2 = {(๐ฆ1 , ๐ฆ2 ): ๐ฆ2 < 0} , and ๐ด0 = {(๐ฆ1 , ๐ฆ2 ): ๐ฆ2 = 0}. The domain of U, ๐ต = {(๐ข1 , ๐ข2 ): โโ < ๐ข1 < โ, ๐ข2 > 0} is the image of both ๐ด1 and ๐ด2 under the transformation. The inverse transformation from ๐ต ๐ก๐ ๐ด1 and ๐ต ๐ก๐ ๐ด2 are given by: ๐ฆ1 = ๐11 โ1 (๐ข1 , ๐ข2 ) = ๐ข1 ๐ข2 ๐ฆ2 = ๐21 โ1 (๐ข1 , ๐ข2 ) = ๐ข2 and ๐ฆ1 = ๐12 โ1 (๐ข1 , ๐ข2 ) = โ๐ข1 ๐ข2 ๐ฆ2 = ๐22 โ1 (๐ข1 , ๐ข2 ) = โ๐ข2 The Jacobians from the two inverses are ๐ฝ1 = ๐ฝ1 = ๐ข2 The pdf of U on its domain B is thus: 2

๐๐1 ,๐2 (๐ข1 , ๐ข2 ) = โ ๐๐1 ,๐2 [๐1๐ โ1 (๐ข, ๐ฃ), ๐2๐ โ1 (๐ข, ๐ฃ)]|๐ฝ๐ | Plugging in, we have:

๐=1

1 โ(๐ข ๐ข )2 /2 โ๐ข2 /2 ๐ 1 2 ๐ 2 |๐ข2 | 2ฯ 1 โ(โ๐ข ๐ข )2 /2 โ(โ๐ข )2 /2 1 2 2 |๐ข2 | + ๐ ๐ 2ฯ

๐๐1 ,๐2 (๐ข1 , ๐ข2 ) =

Simplifying, we have: ๐ข2 2 2 ๐๐1 ,๐2 (๐ข1 , ๐ข2 ) = ๐ โ(๐ข1 +1)๐ข2 /2 , โโ < ๐ข1 < โ, ๐ข2 > 0 ฯ (b) To obtain the marginal distribution of ๐1 , we integrate the joint pdf๐๐1 ,๐2 (๐ข1 , ๐ข2 ) over ๐ข2 . That is, โ

๐๐1 (๐ข1 ) = โซ ๐๐1 ,๐2 (๐ข1 , ๐ข2 ) ๐๐ข2 0

13

1 , โโ < ๐ข1 < โ + 1) Thus, marginally, U1 follows the standard Cauchy distribution. โก =

๐(๐ข12

REMARK: The transformation method can also be extended to handle n-variate transformations. Suppose that ๐1 , ๐2 , โฆ , ๐๐ are continuous random variables with joint pdf ๐๐ (๐ฆ) and define ๐1 = ๐1 (๐1 , ๐2 , โฆ , ๐๐ ) ๐2 = ๐2 (๐1 , ๐2 , โฆ , ๐๐ ) โฎ (๐ ๐๐ = ๐๐ 1 , ๐2 , โฆ , ๐๐ ). Example 8. Given independent random variables ๐ and๐, 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 ๐ and ๐ is๐๐,๐ (๐ฅ, ๐ฆ) = 1, 0 โค ๐ฅ โค 1, 0 โค ๐ฆ โค 1. The inverse transformation, written in terms of observed values is ๐ข+๐ฃ ๐ขโ๐ฃ ๐ฅ= , ๐๐๐ ๐ฆ = . 2 2 It is clearly one-to-one. The Jacobian is 1/2 1/2 ๐(๐ฅ,๐ฆ) 1 1 ๐ฝ = ๐(๐ข,๐ฃ) = | | = โ 2, so |๐ฝ| = 2. 1/2 โ1/2 We will use ๐ to denote the range space of (๐, ๐), and โฌ to denote that of (๐, ๐), and these are shown in the diagrams below. Firstly, note that there are 4 inequalities specifying ranges of ๐ฅ and๐ฆ, and these give 4 inequalities concerning ๐ข and๐ฃ, from which โฌcan be determined. That is, ๐ฅ โฅ 0 โ ๐ข + ๐ฃ โฅ 0, that is, ๐ฃ โฅ โ๐ข ๐ฅ โค 1 โ ๐ข + ๐ฃ โค 2, that is ๐ฃ โค 2 โ ๐ข ๐ฆ โฅ 0 โ ๐ข โ ๐ฃ โฅ 0, that is ๐ฃ โค ๐ข ๐ฆ โค 1 โ ๐ข โ ๐ฃ โค 2, that is ๐ฃ โฅ ๐ข โ 2 Drawing the four lines ๐ฃ = โ๐ข, ๐ฃ = 2 โ ๐ข, ๐ฃ = ๐ข, ๐ฃ = ๐ข โ 2 On the graph, enables us to see the region specified by the 4 inequalities.

14

Now, we have 1 1 โ๐ข โค ๐ฃ โค ๐ข, 0 โค ๐ข โค 1 = , { ๐ข โ 2 โค ๐ฃ โค 2 โ ๐ข, 1 โค ๐ข โค 2 2 2 The importance of having the range space correct is seen when we find marginal pdf of ๐. โ ๐๐ (๐ข) = โซโโ ๐๐,๐ (๐ข, ๐ฃ)๐๐ฃ ๐๐,๐ (๐ข, ๐ฃ) = 1 โ

๐ข 1

โซโ๐ข 2 ๐๐ฃ, = { โซ2โ๐ข 1 ๐๐ฃ, ๐ขโ2 2 0, ={

0โค๐ขโค1 1โค๐ขโค2 ๐๐กโ๐๐๐ค๐๐ ๐

๐ข, 0โค๐ขโค1 2 โ ๐ข, 1 โค ๐ข โค 2

= ๐ข๐ผ[0,1] (๐ข) + (2 โ ๐ข)๐ผ(1,2] (๐ข), using indicator functions. Example 9. Given ๐and ๐ are independent random variables 1

๐ฅ

each with pdf๐๐ (๐ฅ) = 2 ๐ โ2 , ๐ฅ โ [0, โ), find the distribution of(๐ โ ๐)/2. We note that the joint pdf of ๐ and ๐ is 1 ๐ฅ+๐ฆ ๐๐,๐ (๐ฅ, ๐ฆ) = ๐ 2 , 0 โค ๐ฅ < โ, 0 โค ๐ฆ < โ. 4 15

Define๐ = (๐ โ ๐)/2. Now we need to introduce a second random variable ๐ which is a function of ๐ and๐. 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 ๐ is of course, not unique. Let us define๐ = ๐. Then the transformation is, (using๐ข, ๐ฃ, ๐ฅ, ๐ฆ, since we are really dealing with the range spaces here). ๐ฅ = 2๐ข + ๐ฃ ๐ฆ=๐ฃ From it, we find the Jacobian, 2 1 ๐ฝ=| |=2 0 1 To determineโฌ, the range space of ๐ and๐, we note that ๐ฅ โฅ 0 โ 2๐ข + ๐ฃ โฅ 0 , that is ๐ฃ โฅ โ2๐ข ๐ฅ < โ โ 2๐ข + ๐ฃ < โ ๐ฆโฅ0 โ ๐ฃโฅ0 ๐ฆ

## Transformations

Transformations Dear students, Since we have covered the mgf technique extensively already, here we only review the cdf and the pdf techniques, first ...

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