Transformations

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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‘Œ (๐‘ฆ) โ‰ก ๐‘ƒ(๐‘Œ โ‰ค ๐‘ฆ). Let ๐‘ˆ = ๐‘”(๐‘Œ) 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‘Œ (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‘ˆ (๐‘ข) = ๐‘ƒ(๐‘ˆ โ‰ค ๐‘ข), the cdf of U, in terms of ๐น๐‘Œ (๐‘ฆ), the cdf of Y . 3. Differentiate ๐น๐‘ˆ (๐‘ข) to obtain the pdf of U, ๐‘“๐‘ˆ (๐‘ข). Example 1. Suppose that ๐‘Œ ~ ๐‘ˆ(0,1). Find the distribution of ๐‘ˆ = ๐‘”(๐‘Œ) = โˆ’ ln ๐‘Œ. Solution. The cdf of ๐‘Œ ~ ๐‘ˆ(0,1) is given by 0, ๐‘ฆโ‰ค0 ๐น๐‘Œ (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‘Œ ~ ๐‘ˆ(0,1) is ๐‘…๐‘Œ = {๐‘ฆ: 0 < ๐‘ฆ < 1}; thus, because ๐‘ข = โˆ’ ln ๐‘ฆ > 0, it follows that the domain for U is ๐‘…๐‘ˆ = {๐‘ข: ๐‘ข > 0}. The cdf of U is: ๐น๐‘ˆ (๐‘ข) = ๐‘ƒ(๐‘ˆ โ‰ค ๐‘ข) = ๐‘ƒ(โˆ’ ln ๐‘Œ โ‰ค ๐‘ข) = ๐‘ƒ(ln ๐‘Œ > โˆ’๐‘ข) = ๐‘ƒ(๐‘Œ > ๐‘’ โˆ’๐‘ข ) = 1 โˆ’ ๐‘ƒ(๐‘Œ โ‰ค ๐‘’ โˆ’๐‘ข ) = 1 โˆ’ ๐น๐‘Œ (๐‘’ โˆ’๐‘ข )

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Because ๐น๐‘Œ (๐‘ฆ) = ๐‘ฆ for 0 < y < 1; i.e., for u > 0, we have ๐น๐‘ˆ (๐‘ข) = 1 โˆ’ ๐น๐‘Œ (๐‘’ โˆ’๐‘ข ) = 1 โˆ’ ๐‘’ โˆ’๐‘ข Taking derivatives, we get, for u > 0, ๐‘“๐‘ˆ (๐‘ข) =

๐‘‘ ๐‘‘ (1 โˆ’ ๐‘’ โˆ’๐‘ข ) = ๐‘’ โˆ’๐‘ข ๐น๐‘ˆ (๐‘ข) = ๐‘‘๐‘ข ๐‘‘๐‘ข

Summarizing, ๐‘“๐‘ˆ (๐‘ข) = {

๐‘’ โˆ’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‘Œ ~ ๐‘ˆ (โˆ’ ๐œ‹โ„2 , ๐œ‹โ„2) . Find the distribution of the random variable defined by U = g(Y ) = tan(Y ). Solution. The cdf of ๐‘Œ ~ ๐‘ˆ (โˆ’ ๐œ‹โ„2 , ๐œ‹โ„2) is given by 0, ๐‘ฆ + ๐œ‹โ„2 ๐น๐‘Œ (๐‘ฆ) = , ๐œ‹ { 1,

๐‘ฆ โ‰ค โˆ’ ๐œ‹โ„2 โˆ’ ๐œ‹โ„2 < ๐‘ฆ โ‰ค ๐œ‹โ„2 ๐‘ฆ โ‰ฅ ๐œ‹โ„2

The domain for Y is ๐‘…๐‘Œ = {๐‘ฆ: โˆ’ ๐œ‹โ„2 < ๐‘ฆ < ๐œ‹โ„2}. Sketching a graph of the tangent function from โˆ’ ๐œ‹โ„2 to ๐œ‹โ„2, we see that โˆ’โˆž < ๐‘ข < โˆž . Thus, ๐‘…๐‘ˆ = { ๐‘ข: โˆ’ โˆž < ๐‘ข < โˆž} โ‰ก ๐‘…, the set of all reals. The cdf of U is: ๐น๐‘ˆ (๐‘ข) = ๐‘ƒ(๐‘ˆ โ‰ค ๐‘ข) = ๐‘ƒ[tan(๐‘Œ) โ‰ค ๐‘ข] = ๐‘ƒ[๐‘Œ โ‰ค tanโˆ’1(๐‘ข)] = ๐น๐‘Œ [tanโˆ’1(๐‘ข)] Because ๐น๐‘Œ (๐‘ฆ) = we have

๐‘ฆ+๐œ‹โ„2 ๐œ‹

for โˆ’ ๐œ‹โ„2 < ๐‘ฆ < ๐œ‹โ„2 ; i. e. , for ๐‘ข โˆˆ โ„› ,

๐น๐‘ˆ (๐‘ข) = ๐น๐‘Œ [tanโˆ’1 (๐‘ข)] =

tanโˆ’1(๐‘ข) + ๐œ‹โ„2 ๐œ‹

The pdf of U, for ๐‘ข โˆˆ โ„›, is given by ๐‘‘ ๐‘‘ tanโˆ’1 (๐‘ข) + ๐œ‹โ„2 1 ๐‘“๐‘ˆ (๐‘ข) = ๐น๐‘ˆ (๐‘ข) = [ ]= . ๐‘‘๐‘ข ๐‘‘๐‘ข ๐œ‹ ๐œ‹(1 + ๐‘ข2 ) Summarizing,

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1 , โˆ’โˆž0 ๐‘“๐‘Œ (๐‘ฆ) = { ๐›ฝ 0, otherwise. Let ๐‘ˆ = ๐‘”(๐‘Œ) = โˆš๐‘Œ, . Use the method of transformations to find the pdf of U. Solution. First, we note that the transformation ๐‘”(๐‘Œ) = โˆš๐‘Œ is a continuous strictly increasing function of y over ๐‘…๐‘Œ = {๐‘ฆ: ๐‘ฆ > 0}, and, thus, ๐‘”(๐‘Œ) is one-to-one. Next, we need to find the domain of U. This is easy since y > 0 implies ๐‘ข = โˆš๐‘ฆ > 0 as well. Thus, ๐‘…๐‘ˆ = {๐‘ข: ๐‘ข > 0}. Now, we find the inverse transformation: ๐‘”(๐‘ฆ) = ๐‘ข = โˆš๐‘ฆ โ‡” ๐‘ฆ = ๐‘”โˆ’1 (๐‘ข) = ๐‘ข2 (by inverse transformation) and its derivative:

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๐‘‘ โˆ’1 ๐‘‘ (๐‘ข2 ) = 2๐‘ข. ๐‘” (๐‘ข) = ๐‘‘๐‘ข ๐‘‘๐‘ข Thus, for u > 0, ๐‘“๐‘ˆ (๐‘ข) = ๐‘“๐‘Œ [๐‘”โˆ’1 (๐‘ข)] |

๐‘‘ โˆ’1 ๐‘” (๐‘ข) | ๐‘‘๐‘ข 2

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1 โˆ’๐‘ข 2๐‘ข โˆ’๐‘ข๐›ฝ = ๐‘’ ๐›ฝ ร— |2๐‘ข| = ๐‘’ . ๐›ฝ ๐›ฝ Summarizing, 2

2๐‘ข โˆ’๐‘ข๐›ฝ ๐‘ข>0 ๐‘“๐‘ˆ (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 โˆ’ ๐‘ฆ), ๐‘“๐‘Œ (๐‘ฆ) = { 0,

0 0 for all i โ‰  0, and ๐‘“๐‘Œ1 ,๐‘Œ2 (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‘ˆ1 = โˆ’1 โˆ’1 (โˆ™), ๐‘”2๐‘– (โˆ™)) is a 1-to-1 ๐‘”1 (๐‘Œ1 , ๐‘Œ2 ), ๐‘ˆ2 = ๐‘”1 (๐‘Œ1 , ๐‘Œ2 )) 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‘ˆ1 ,๐‘ˆ2 (๐‘ข1 , ๐‘ข2 ) ๐‘˜

={

โˆ‘ ๐‘“๐‘Œ1 ,๐‘Œ2 [๐‘”1๐‘– โˆ’1 (๐‘ข, ๐‘ฃ), ๐‘”2๐‘– โˆ’1 (๐‘ข, ๐‘ฃ)]|๐ฝ๐‘– | ,

๐‘ข โˆˆ ๐ต = ๐‘…๐‘ˆ

๐‘–=1

0,

otherwise.

Example 7. Suppose that ๐‘Œ1 ~ N(0, 1), ๐‘Œ2 ~ N(0, 1), and that ๐‘Œ1 and ๐‘Œ2 are independent. Define the transformation ๐‘Œ1 ๐‘ˆ1 = ๐‘”1 (๐‘Œ1 , ๐‘Œ2 ) = ๐‘Œ2 ๐‘ˆ2 = ๐‘”2 (๐‘Œ1 , ๐‘Œ2 ) = |๐‘Œ2 |. Find each of the following distributions: 12

(a) ๐‘“๐‘ˆ1 ,๐‘ˆ2 (๐‘ข1 , ๐‘ข2 ), the joint distribution of ๐‘ˆ1 and ๐‘ˆ2 , (b) ๐‘“๐‘ˆ1 (๐‘ข1 ), the marginal distribution of ๐‘ˆ1 . Solutions. (a) Since ๐‘Œ1 and ๐‘Œ2 are independent, the joint distribution of ๐‘Œ1 and ๐‘Œ2 is ๐‘“๐‘Œ1 ,๐‘Œ2 (๐‘ฆ1 , ๐‘ฆ2 ) = ๐‘“๐‘Œ1 (๐‘ฆ1 )๐‘“๐‘Œ2 (๐‘ฆ2 ) 1 โˆ’๐‘ฆ 2/2 โˆ’๐‘ฆ2 /2 = ๐‘’ 1 ๐‘’ 2 2ฯ€ Here, ๐‘…๐‘Œ1 ,๐‘Œ2 = {(๐‘ฆ1 , ๐‘ฆ2 ): โˆ’โˆž < ๐‘ฆ1 < โˆž, โˆ’โˆž < ๐‘ฆ2 < โˆž}. The transformation of Y to U is not one-to-one because the points (๐‘ฆ1 , ๐‘ฆ2 ) and (โˆ’๐‘ฆ1 , โˆ’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‘Œ1 ,๐‘Œ2 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 ): โˆ’โˆž < ๐‘ข1 < โˆž, ๐‘ข2 > 0} is the image of both ๐ด1 and ๐ด2 under the transformation. The inverse transformation from ๐ต ๐‘ก๐‘œ ๐ด1 and ๐ต ๐‘ก๐‘œ ๐ด2 are given by: ๐‘ฆ1 = ๐‘”11 โˆ’1 (๐‘ข1 , ๐‘ข2 ) = ๐‘ข1 ๐‘ข2 ๐‘ฆ2 = ๐‘”21 โˆ’1 (๐‘ข1 , ๐‘ข2 ) = ๐‘ข2 and ๐‘ฆ1 = ๐‘”12 โˆ’1 (๐‘ข1 , ๐‘ข2 ) = โˆ’๐‘ข1 ๐‘ข2 ๐‘ฆ2 = ๐‘”22 โˆ’1 (๐‘ข1 , ๐‘ข2 ) = โˆ’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‘ˆ1 ,๐‘ˆ2 (๐‘ข1 , ๐‘ข2 ) = โˆ‘ ๐‘“๐‘Œ1 ,๐‘Œ2 [๐‘”1๐‘– โˆ’1 (๐‘ข, ๐‘ฃ), ๐‘”2๐‘– โˆ’1 (๐‘ข, ๐‘ฃ)]|๐ฝ๐‘– | Plugging in, we have:

๐‘–=1

1 โˆ’(๐‘ข ๐‘ข )2 /2 โˆ’๐‘ข2 /2 ๐‘’ 1 2 ๐‘’ 2 |๐‘ข2 | 2ฯ€ 1 โˆ’(โˆ’๐‘ข ๐‘ข )2 /2 โˆ’(โˆ’๐‘ข )2 /2 1 2 2 |๐‘ข2 | + ๐‘’ ๐‘’ 2ฯ€

๐‘“๐‘ˆ1 ,๐‘ˆ2 (๐‘ข1 , ๐‘ข2 ) =

Simplifying, we have: ๐‘ข2 2 2 ๐‘“๐‘ˆ1 ,๐‘ˆ2 (๐‘ข1 , ๐‘ข2 ) = ๐‘’ โˆ’(๐‘ข1 +1)๐‘ข2 /2 , โˆ’โˆž < ๐‘ข1 < โˆž, ๐‘ข2 > 0 ฯ€ (b) To obtain the marginal distribution of ๐‘ˆ1 , we integrate the joint pdf๐‘“๐‘ˆ1 ,๐‘ˆ2 (๐‘ข1 , ๐‘ข2 ) over ๐‘ข2 . That is, โˆž

๐‘“๐‘ˆ1 (๐‘ข1 ) = โˆซ ๐‘“๐‘ˆ1 ,๐‘ˆ2 (๐‘ข1 , ๐‘ข2 ) ๐‘‘๐‘ข2 0

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1 , โˆ’โˆž < ๐‘ข1 < โˆž + 1) Thus, marginally, U1 follows the standard Cauchy distribution. โ–ก =

๐œ‹(๐‘ข12

REMARK: The transformation method can also be extended to handle n-variate transformations. Suppose that ๐‘Œ1 , ๐‘Œ2 , โ€ฆ , ๐‘Œ๐‘› are continuous random variables with joint pdf ๐‘“๐’€ (๐‘ฆ) and define ๐‘ˆ1 = ๐‘”1 (๐‘Œ1 , ๐‘Œ2 , โ€ฆ , ๐‘Œ๐‘› ) ๐‘ˆ2 = ๐‘”2 (๐‘Œ1 , ๐‘Œ2 , โ€ฆ , ๐‘Œ๐‘› ) โ‹ฎ (๐‘Œ ๐‘ˆ๐‘› = ๐‘”๐‘› 1 , ๐‘Œ2 , โ€ฆ , ๐‘Œ๐‘› ). Example 8. Given independent random variables ๐‘‹ and๐‘Œ, 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‘Œ is๐‘“๐‘‹,๐‘Œ (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‘ขโˆ’๐‘ฃ ๐‘ฅ= , ๐‘Ž๐‘›๐‘‘ ๐‘ฆ = . 2 2 It is clearly one-to-one. The Jacobian is 1/2 1/2 ๐œ•(๐‘ฅ,๐‘ฆ) 1 1 ๐ฝ = ๐œ•(๐‘ข,๐‘ฃ) = | | = โˆ’ 2, so |๐ฝ| = 2. 1/2 โˆ’1/2 We will use ๐’œ to denote the range space of (๐‘‹, ๐‘Œ), and โ„ฌ to denote that of (๐‘ˆ, 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 โˆ’๐‘ข ๐‘ฅ โ‰ค 1 โ‡’ ๐‘ข + ๐‘ฃ โ‰ค 2, that is ๐‘ฃ โ‰ค 2 โˆ’ ๐‘ข ๐‘ฆ โ‰ฅ 0 โ‡’ ๐‘ข โˆ’ ๐‘ฃ โ‰ฅ 0, that is ๐‘ฃ โ‰ค ๐‘ข ๐‘ฆ โ‰ค 1 โ‡’ ๐‘ข โˆ’ ๐‘ฃ โ‰ค 2, that is ๐‘ฃ โ‰ฅ ๐‘ข โˆ’ 2 Drawing the four lines ๐‘ฃ = โˆ’๐‘ข, ๐‘ฃ = 2 โˆ’ ๐‘ข, ๐‘ฃ = ๐‘ข, ๐‘ฃ = ๐‘ข โˆ’ 2 On the graph, enables us to see the region specified by the 4 inequalities.

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Now, we have 1 1 โˆ’๐‘ข โ‰ค ๐‘ฃ โ‰ค ๐‘ข, 0 โ‰ค ๐‘ข โ‰ค 1 = , { ๐‘ข โˆ’ 2 โ‰ค ๐‘ฃ โ‰ค 2 โˆ’ 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‘ˆ. โˆž ๐‘“๐‘ˆ (๐‘ข) = โˆซโˆ’โˆž ๐‘“๐‘ˆ,๐‘‰ (๐‘ข, ๐‘ฃ)๐‘‘๐‘ฃ ๐‘“๐‘ˆ,๐‘‰ (๐‘ข, ๐‘ฃ) = 1 โˆ—

๐‘ข 1

โˆซโˆ’๐‘ข 2 ๐‘‘๐‘ฃ, = { โˆซ2โˆ’๐‘ข 1 ๐‘‘๐‘ฃ, ๐‘ขโˆ’2 2 0, ={

0โ‰ค๐‘ขโ‰ค1 1โ‰ค๐‘ขโ‰ค2 ๐‘œ๐‘กโ„Ž๐‘’๐‘Ÿ๐‘ค๐‘–๐‘ ๐‘’

๐‘ข, 0โ‰ค๐‘ขโ‰ค1 2 โˆ’ ๐‘ข, 1 โ‰ค ๐‘ข โ‰ค 2

= ๐‘ข๐ผ[0,1] (๐‘ข) + (2 โˆ’ ๐‘ข)๐ผ(1,2] (๐‘ข), using indicator functions. Example 9. Given ๐‘‹and ๐‘Œ are independent random variables 1

๐‘ฅ

each with pdf๐‘“๐‘‹ (๐‘ฅ) = 2 ๐‘’ โˆ’2 , ๐‘ฅ โˆˆ [0, โˆž), find the distribution of(๐‘‹ โˆ’ ๐‘Œ)/2. We note that the joint pdf of ๐‘‹ and ๐‘Œ is 1 ๐‘ฅ+๐‘ฆ ๐‘“๐‘‹,๐‘Œ (๐‘ฅ, ๐‘ฆ) = ๐‘’ 2 , 0 โ‰ค ๐‘ฅ < โˆž, 0 โ‰ค ๐‘ฆ < โˆž. 4 15

Define๐‘ˆ = (๐‘‹ โˆ’ ๐‘Œ)/2. Now we need to introduce a second random variable ๐‘‰ which is a function of ๐‘‹ and๐‘Œ. 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‘Œ. 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‘ˆ and๐‘‰, we note that ๐‘ฅ โ‰ฅ 0 โ‡’ 2๐‘ข + ๐‘ฃ โ‰ฅ 0 , that is ๐‘ฃ โ‰ฅ โˆ’2๐‘ข ๐‘ฅ < โˆž โ‡’ 2๐‘ข + ๐‘ฃ < โˆž ๐‘ฆโ‰ฅ0 โ‡’ ๐‘ฃโ‰ฅ0 ๐‘ฆ
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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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