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(1)國立高雄大學統計學研究所碩士論文. A study of some different concepts of symmetry on the real line 實數上有關對稱性概念的探討. 研究生：鄧慧怡撰指導教授：黃文璋. 中華民國九十九年七月.

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(4) A study of some different concepts of symmetry on the real line. by Hui-Yi Teng Advisor Wen-Jang Huang. Institute of Statistics National University of Kaohsiung Kaohsiung, Taiwan 811, R.O.C. July 2010.

(5) Contents. Z` zZ`. ii iii. 1 Introduction. 1. 2 Preliminary Results. 2. 3 R-Symmetry on R. 3. 4 I-Symmetry. 6. 5 Doubly-Symmetry in R. 7. 6 Main Results. 7. 7 Some Interesting Examples of I-Symmetry 10 7.1 Characterization of I-Symmetry . . . . . . . . . . . . . . . . . . . . . . . . 10 7.2 I-Symmetry Arising From Trigonometric Formulas . . . . . . . . . . . . . . 15 References. 16. F. 17. i.

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(7) A study of some different concepts of symmetry on the real line Advisor: Wen-Jang Huang Department of Applied Mathematics National University of Kaohsiung. Student: Hui-Yi Teng Institute of Statistics National University of Kaohsiung. ABSTRACT Recently, different concepts of symmetry on R+ such as R-symmetry, log-symmetry, and doubly symmetry are studied. Analogue concept and their properties of these symmetries on R will be studied in this work. Based on skewing representation and previous studies, characterizations of doubly symmetry on R will be given. Among others, some interesting examples of the so-called I-symmetry, that is the analogue of log-symmetry on R, will also be presented. Key words and phrases: R-symmetry; log-symmetry; I-symmetry; doubly symmetry; skewing representation; characterization.. iii.

(8) 1. Introduction. A random variable (r.v.) X is said to be symmetric about a constant µ, if X − µ and µ − X d. have the same distribution, denote it by X − µ = µ − X. If µ = 0, we simply say X is symmetric. Recently, different concepts of symmetry on R+ are introduced and investigated. Mudholkar and Wang (2007) gave the definition of R-symmetric distribution on R+ . According to their definition, a positive r.v. X with probability density function (p.d.f.) fX is said to be R-symmetric about the R-center θ, where θ > 0, if fX (θx) = fX (θ/x) , x > 0. Earlier in 1965, Seshadri studied another non-ordinary symmetry. He characterized those nonnegative r.v.’s X’s d d on R+ such that X = 1/X. Jones (2008) refered to this as log-symmetry since log X = − log X. When X is defined on R+ , Jones (2008) also studied R-symmetry and log-symmetry about δ, d. δ > 0. The latter is, X/δ = δ/X, which is equivalent to ordinary symmetry about log δ of the d. r.v. log X. Among others, Jones (2008) pointed out that when X has a p.d.f. f , X/δ = δ/X is equivalent to x2 fX (δx) = fX (δ/x) , x > 0. Jones and Arnold (2008) studied the r.v.’s defined on R+ which are both R-symmetry and log-symmetry, the so-called doubly symmetry. An example of doubly symmetric distribution is lognormal. They also characterized the class of absolutely continuous r.v.’s defined on R+ that are doubly symmetric, which turns out to be a proper subset of absolutely continuous and moment-equivalent to the lognormal distribution. In this work, we will investigate natural analogue of the concepts of R-symmetry, logsymmetry, and doubly symmetry on R. More precisely, we call the analogue of log-symmetry on R as I-symmetry. Here ‘I’ stands for ‘inverse’. Throughout this work, unless it is stated, every r.v. is assumed to follow an absolutely continuous distribution. Also for an r.v., say X, let fX denote the p.d.f. of X. First, we give the definitions of those symmetries mentioned above. Definition 1. An r.v. X defined on R is said to be R-symmetric about the R-center θ, where θ > 0, if µ ¶ θ fX (θx) = fX , x ∈ R\{0}, (1) x or equivalently, if µ fX (x) = fX. θ2 x. ¶ , x ∈ R\{0}.. It can be shown easily from (1), if X is R-symmetric on R, then fX (0) = 0. Definition 2. An r.v. X defined on R is said to be I-symmetric about δ, where δ > 0, if X d δ = , δ X. 1.

(9) or equivalently, if x2 fX (δ) = fX. µ ¶ δ , x ∈ R\{0}. x. (2). Definition 3. An r.v. X defined on R is said to be doubly symmetric about (θ, δ), where θ, δ > 0, if X is both R-symmetric about θ and I-symmetric about δ. In Section 2, based on mixture distribution, we investigate the relationship between doubly symmetry on R+ and doubly symmetry on R. In Sections 3, 4, and 5, we give some propositions of R-symmetry, I-symmetry, and doubly symmetry, respectively. Next, in Section 6, we discuss the connection between mixture distribution representation and skewing representation. By the skewing representation, we investigate and characterize doubly symmetry. Finally, in Section 7, we give some interesting examples of I-symmetry.. 2. Preliminary Results. Let X be an r.v. defined on R. Obviously fX can have the following mixture representation: ( af1 (x), x > 0, fX (x) = (3) (1 − a)f2 (−x), x ≤ 0, where. Z. a = P (X > 0) = 0. ∞. fX (x)dx,. (4). and a ∈ [0, 1]. Then both f1 (x) = fX (x)/a and f2 (x) = fX (−x)/(1 − a) are p.d.f.’s on R+ . Note that f1 is defined to be 0 if a = 0, and f2 is defined to be 0 if a = 1. It can be seen if a = 0, then X is defined on R− ; if a = 1, then X is defined on R+ . Based on the above representation, we have the following simple lemma. Lemma 1. Let X be an r.v. defined on R with 0 < a < 1, where a is defined in (4). Then fX is doubly symmetric about (θ, δ) if and only if both f1 and f2 are doubly symmetric about (θ, δ), where f1 and f2 are given in (3). By Lemma 1, we have the following two immediate consequences. Corollary 1: Let X be an r.v. defined on R with 0 < a < 1, where a is defined in (4). Then fX is R-symmetric about θ if and only if both the f1 and f2 given in (3) are R-symmetric about θ. Corollary 2: Let X be an r.v. defined on R with 0 < a < 1, where a is defined in (4). Then fX is I-symmetric about δ if and only if both f1 and f2 given in (3) are log-symmetric about δ. 2.

(10) Remark 1: Suppose X is symmetric about 0. Clearly, the constant a in (3) is equal to 0.5 and f1 (x) = f2 (x) = f|X| (x), x > 0. Then according to Corollary 2, X is I-symmetric about δ if and only if |X| is log-symmetric about δ. Also according to Lemma 1, X is doubly symmetric about (θ, δ) if and only if |X| is doubly symmetric about (θ, δ). Furthermore, according to Corollary 1, X is R-symmetric about θ if and only if |X| is R-symmetric about θ.. 3. R-Symmetry on R. In this section, we give some simple properties of R-symmetry. The first proposition gives the mode of an R-symmetric distribution. Proposition 1. Let the r.v. X defined on R be R-symmetric about θ. Then maxx∈R fX (x) = max{fX (θ), fX (−θ)}. Proof: Obviously we only need to prove the case 0 < a < 1, where a is defined in (4). According to Mudholkar and Wang (2007), for an r.v. which is R-symmetric about θ on R+ , then θ is the mode. From Corollary 1, as X is R-symmetric about θ on R, fX (x) = af1 (x), a ∈ [0, 1], x > 0, where f1 is R-symmetric about θ on R+ , then maxx∈R+ fX (x) = maxx∈R+ af1 (x) = af1 (θ) = fX (θ). Note that as mentioned it before, being R-symmetric, fX (0) = 0. Similarly, we have maxx∈R− fX (x) = maxx∈R− (1 − a)f2 (x) = (1 − a)f2 (−θ) = fX (−θ), where f2 is R-symmetric about θ on R+ . Hence maxx∈R fX (x) = max{fX (θ), fX (−θ)}. This completes the proof. Proposition 2. Let the r.v. X defined on R be R-symmetric about θ. Then for every constant a > 0, aX is R-symmetric about aθ. Proof: Upon changing of variable, it yields ³x´ 1 , x ∈ R. faX (x) = fX a a. (5). Therefore, µ faX (aθx) = fX. aθx a. ¶. 1 = fX a. µ. aθ ax. ¶. 1 = faX a. µ. aθ x. ¶ , x ∈ R\{0},. where the first and last equalities are by (5), and the second equality is by (1). This completes the proof. For independent r.v.’s X and Y defined on R+ which are R-symmetric about θ1 and θ2 , respectively, Mudholkar and Wang (2007) proved that XY is R-symmetric about θ1 θ2 . The next proposition shows that this property also holds for R-symmetry on R. Proposition 3. Let the independent r.v.’s X and Y defined on R be R-symmetric about θ1 and θ2 , respectively. Then XY is R-symmetric about θ1 θ2 .. 3.

(11) Proof: Let X1 = X/θ1 , Y1 = Y /θ2 , and W = X1 Y1 . Then both X1 and Y1 are R-symmetric about 1 and µ ¶ µ ¶ Z ∞ Z ∞ w 1 1 1 fW (w) = fX1 fY1 (y)dy = fX1 (wt)fY1 dt |y| y |t| t −∞ Z−∞ ∞ 1 = fX1 (wt)fY1 (t)dt, w ∈ R, (6) −∞ |t| where the change of variable y = 1/t, and fY1 (t) = fY1 (1/t) are used. On the other hand, since fX1 (wt) = fX1 (1/(wt)), from (6) we obtain µ ¶ Z ∞ µ ¶ Z ∞ 1 1 1 1 fX1 fX1 (wy)fY1 (y)dy, w ∈ R\{0}. fW = fY1 (y)dy = w wy −∞ |y| −∞ |y| This proves W is R-symmetric about 1. Consequently, XY = θ1 θ2 W is R-symmetric about θ1 θ2 . This completes the proof. Obviously, Proposition 3 can be easily extended to the situation of n r.v.’s. However, if X is R-symmetric about θ, 1/X may not be R-symmetric for any center c. Consequently, under the conditions of Proposition 3, X/Y may not be R-symmetric about any center c. We give an example in the following. Example 1. Let X and Y be i.i.d. with the distribution of the root-reciprocal of IG(1, λ) (IG stands for inverse Gaussian). That is Ã r µ ¶2 ! λ 1 2λ fX (x) = fY (x) = , x > 0. exp − −x π 2 x Then as pointed out by Mudholkar and Wang (2007), both X and Y are R-symmetric in R about 1. It can be found that the p.d.f.’s of U = XY , V = 1/X, and W = X/Y are given by Ã µ ¶2 ! µ ¶ Z ∞ 2λ λ 1 λ ³ u x ´2 fU (u) = exp − −x exp − − dx, u > 0, πx 2 x 2 x u 0 Ã √ µ ¶2 ! 2λ λ 1 fV (v) = √ 2 exp − −v , v > 0, 2 v πv and. Ã µ ¶2 ! µ ¶2 ! λ 1 λ 1 x exp − −x exp − − xw dx 2 x 2 xw 0 µ µ ¶µ ¶¶ Z λe2λ ∞ λ 1 1 = exp − +w +y dy, w > 0, πw 0 2 w y. 2λ fW (w) = π. Z. ∞. Ã. respectively.. 4.

(12) Now. Ã Ã µ ¶ Z ∞ µ ¶2 ! µ ¶2 ! 1 2λ λ 1 λ 1 fU = exp − −x exp − − ux dx u πx 2 x 2 ux 0 Ã Ã µ ¶ ! µ ¶2 ! Z ∞ 2λ λ u t 2 λ 1 = exp − − exp − −t dt = fU (u), u 6= 0. πt 2 t u 2 t 0. This proves U is R-symmetric about 1. Next for v 6= 0, fV (v) = fV (θ2 /v). 2λ √ exp(− 2λ π (v π 2 2λ 2λ √ exp(− ( θ π v π. − v1 )2 ) v12 −. v 2 v2 ) ) θ4 θ2. µ µ ¶µ 2 ¶¶ 2λ v θ2 θ4 1 2 − , = 4 exp − θ − 2 v π θ θ2 v 2. (7). which can be shown easily is a strictly monotone decreasing function of v. Also fV (v)/fV (θ2 /v) = 1 when v = θ. Hence for every θ > 0, fV (v)/fV (θ2 /v) 6= 1, if v 6= θ. This proves V is not Rsymmetric about any center θ. Finally, ³ ´´ ¡ ¢ ³1 R∞ λ 1 exp − + w + y dy 2 2 w y 0 θ fW (w) ³ ³ ´ ³ ´´ = , w 6= 0. (8) 1 fW (θ2 /w) w2 R ∞ exp − λ θ2 + w + y dy 0. 2. w. θ2. y. Clearly when θ2 = 1, fW (w)/fW (c/w) 6= 1 for w 6= 1. For θ2 6= 1, since (θ2 + 1/θ2 )/2 > 1, and g(z) = e−z , z > 0, is a strictly decreasing function of z, we have ´´ ³ ³ R∞ 1 + y dy exp −λ y 0 ³ ³ ´´ > 1. ¢ ¡ R∞ λ 1 2+ 1 exp − θ + y dy 2 y 0 θ2 Hence if θ2 > 1, then θ2 /w2 > 1, and it yields fW (1)/fW (θ2 /1) > 1. Now consider the case θ2 < 1. Assume there exists a θ, 0 < θ < 1, such that fW (w) = fW (θ2 /w), for w ∈ R+ . Then from (8) we obtain µ ¶µ ¶¶ µ ¶¶ µ µ ¶µ Z ∞ Z 1 w2 ∞ w λ 1 λ θ2 1 exp − +w +y dy = 2 exp − + 2 +y dy. 2 w y θ 0 2 w θ y 0 By replacing w by w/θ2 , it yields µ µ ¶µ ¶¶ µ ¶µ ¶¶ µ Z ∞ Z λ 1 1 w4 ∞ λ θ4 w 1 exp − +w +y dy = 8 exp − + 2 +y dy. 2 w y θ 0 2 w θ y 0 Repeating this procedure, in the kth time replacing w by w/θ2k , we obtain µ ¶µ ¶¶ µ ¶µ ¶¶ µ µ Z ∞ Z ∞ 1 1 w 1 λ 1 λ θ2n 2n +w +y + 2n +y exp − exp − dy = w dy 2 w y 2 w θ y θ2n2 0 0 holds for every n ≥ 1. Let µ ¶µ ¶¶ µ 1 w 1 λ θ2n gn (y) = 2n2 exp − + 2n +y , n ≥ 1. 2 w θ y θ 5.

(13) Since for every y ≥ 0, lim gn (y) = 0,. n→∞. also it can be shown easily that for every y ≥ 0, gn (y) is decreasing in n for n large enough, by Lebesgue’s monotone convergence theorem, we arrive at the following contradiction µ µ ¶µ ¶¶ Z ∞ Z ∞ λ 1 1 2n exp − +w +y dy = lim w gn (y)dy = 0, 0 < w < 1. n→∞ 2 w y 0 0 This completes the proof that W is not R-symmetric about any center θ.. 4. I-Symmetry. We now give some simple properties of I-symmetry. Proposition 4. Let the r.v. X defined on R be I-symmetric about δ. Then P (−δ < X ≤ δ) = 1/2. Proof: From Corollary 2, for X being I-symmetric about δ, then f1 and f2 are log-symmetric about δ, where f1 and f2 are given in (3). According to Jones (2008), for an r.v. defined on R+ which is log-symmetric about δ, then δ is the median. Hence Z δ P (0 < X ≤ δ|X > 0) = f1 (x)dx = 1/2, 0. and. Z P (0 < −X ≤ δ|X ≤ 0) = 0. δ. f2 (x)dx = 1/2.. This completes the proof. The proofs of the following three propositions are similar to those of the situation of logsymmetry on R+ , hence are omitted. Proposition 5. Let the r.v. X defined on R be I-symmetric about δ. Then for every constant a > 0, aX is I-symmetric about aδ. Proposition 6. Let the independent r.v.’s X and Y defined on R be I-symmetric about δ1 and δ2 , respectively. Then XY is I-symmetric about δ1 δ2 . Again the above proposition also holds for n r.v.’s as the situation of log-symmetric on R+ (see Jones (2008)). On the other hand, although X is R-symmetric may not imply 1/X is R-symmetric, it can be seen easily that this is true for I-symmetric. We state this as follows. Proposition 7. Let the independent r.v.’s X and Y defined on R be I-symmetric about δ1 and δ2 , respectively. Then X/Y is I-symmetric about δ1 /δ2 . In particular, 1/X is I-symmetric about 1/δ1 . 6.

(14) 5. Doubly-Symmetry in R. In this section we give some simple properties of doubly symmetry. Proposition 8. Let the r.v. X defined on R be doubly symmetric about (θ, δ). Then for any constant a > 0, aX is doubly symmetric about (aθ, aδ). Proposition 9. Let the independent r.v.’s X and Y defined on R be doubly symmetric about (θ2 , δ2 ) and (θ2 , δ2 ), respectively. Then XY is doubly symmetric about (θ1 θ2 , δ1 δ2 ). The proofs of the above two propositions are trivial hence are omitted. Although the ratio of two independent R-symmetric r.v.’s may not be R-symmetric about any center, yet as pointed out in the following result, this is true if both X and Y are doubly symmetric. Proposition 10. Let the independent r.v.’s X and Y defined on R be doubly symmetric about (θ1 , δ1 ) and (θ2 , δ2 ), respectively. Then X/Y is doubly symmetric about (θ1 θ2 /δ22 , δ1 /δ2 ). In particular, 1/X is doubly symmetric about (θ1 /δ12 , 1/δ1 ). d. Proof: Based on Proposition 7, we only need to prove the “R-symmetric part”. Since Y /δ2 = d. δ2 /Y if and only if 1/Y = Y /δ22 , also by Proposition 2, Y /δ22 is R-symmetric about θ2 /δ22 , we obtain 1/Y is also R-symmetric about θ2 /δ22 . The proof now follows by Proposition 3.. 6. Main Results. For a p.d.f. fX , except (3), it can also be represented as fX (x) = 2f (x)G(x), x ∈ R, where fX (x) + fX (−x) f (x) = = 2. (9). (. 1 2 (af1 (x) + (1 − a)f2 (x)), 1 2 (af1 (−x) + (1 − a)f2 (−x)),. x > 0, x ≤ 0,. (10). is a symmetric p.d.f., and fX (x) = G(x) = fX (x) + fX (−x). (. af1 (x)/(af1 (x) + (1 − a)f2 (x)), x > 0, (1 − a)f2 (−x)/(af1 (−x) + (1 − a)f2 (−x)), x ≤ 0,. (11). is a skewing function. In this section, we will characterize doubly symmetry through skewing representation. When fX is represented as in (9), we are interested in knowing that is it possible that f is not doubly symmetric, yet fX is doubly symmetric? The next lemma will answer this question.. 7.

(15) Lemma 2. Let the r.v. X defined on R be doubly symmetric about (θ, δ). Then f is also doubly symmetric about (θ, δ), where f is the symmetric p.d.f. given in (9). Consequently, |X| is doubly symmetric about (θ, δ) on R+ . Proof: That X is doubly symmetric about (θ, δ) implies f (θx) =. fX (θx) + fX (−θx) fX (θ/x) + fX (−θ/x) = =f 2 2. µ ¶ θ , x ∈ R\{0}, x. and x2 fX (δx) + x2 fX (−δx) fX (δ/x) + fX (−δ/x) x f (δx) = = =f 2 2 2. µ ¶ δ , x ∈ R\{0}. x. This proves the first assertion. The second assertion follows immediately by noting |X| has the p.d.f. 2f (x), x > 0. This completes the proof. Jones and Arnold (2008) characterized the doubly symmetry on R+ . By using their characterization and the skewing representation of a distribution as in (9), we can characterize the doubly symmetry on R. Theorem 1. Let the r.v. X defined on R be doubly symmetric about (θ, δ). Let k = δ/θ. Also let fX be represented as in (9). Then f has the form f (x) ∝. ∞ X. θ−2i k 2i(i+1) x2i |x|ω(θ−2 k 4(i−1) x2 )I(θk −2i < |x| ≤ θk 2−2i ), x ∈ R\{0},. (12). i=−∞. where ω is a nonnegative function on (k −4 , 1] and chosen to satisfy µ ¶ 1 ψ(u) = ψ , k −4 < u ≤ 1, k4 u. (13). where ψ(u) ≡ uω(u),. (14). and G is chosen to satisfy µ ¶ µ ¶ θ δ G(θx) = G , and G(δx) = G x ∈ R\{0}. x x. (15). Proof: First from Lemma 2, we obtain X1 = |X| is doubly symmetric. Now by Jones and Arnold (2008), fX1 (x) ∝. ∞ X. θ−2i k 2i(i+1) x2i+1 ω(θ−2 k 4(i−1) x2 )I(θk −2i < x ≤ θk 2−2i ), x > 0,. i=−∞. where the nonnegative function ω defined on (k −4 , 1] satisfying (13) and (14). Note that f (x) = fX1 (|x|)/2, x ∈ R, hence (12) is obtained immediately. 8.

(16) Next due to the doubly symmetric property of X, we have (1) and (2). Then by the representation of (9), (1) and (2) in turn imply µ ¶ µ ¶ θ θ 2f (θx)G(θx) = 2f G , (16) x x and. µ ¶ µ ¶ δ δ 2x f (δx)G(δx) = 2f G , x x 2. (17). respectively. Again, from Lemma 2, we have f (θx) = f (θ/x) and x2 f (δx) = f (δ/x), these together with (16) and (17), imply (15) immediately. This completes the proof. Theorem 2. Let the p.d.f. of the r.v. X be written as in (9). Then X is doubly symmetric about (θ, δ) if and only if (i) f is doubly symmetric about (θ, δ), (ii) G satisfies (15). Proof: The “if” part is obvious. By Lemma 2 and Theorem 1, the “only if” follows immediately. This completes the proof. We give an illustration of Theorem 2. Example 2. Let the p.d.f. of the r.v. X be written as in (8), where µ ¶ (log |x| − µ)2 1 exp − , f (x) = √ 2σ 2 2 2πσ|x| and 1 ε G(x) = + sgn(x) cos 2 2. µ. 2π(log |x| − µ) σ2. ¶ , |ε| ≤ 1.. Note that X1 = |X| has LogN ormal(µ, σ 2 ) distribution, which is doubly symmetric about 2 2 (eµ−σ , eµ ). From Remark 1, f is doubly symmetric about (θ, δ) = (eµ−σ , eµ ). Also, µ ¶ 2π(log |x| − σ 2 ) 1 ε µ−σ 2 G(e x) = + sgn(x) cos 2 2 σ2 µ ¶ 1 ε 2π(log |x|) = + sgn(x) cos − 2π 2 2 σ2 µ ¶ 1 ε 2π(− log |x|) = + sgn(x) cos + 2π 2 2 σ2 µ ¶ 1 ε 2π(− log |x|) = + sgn(x) cos − 2π 2 2 σ2 Ã ! µ ¶ 2 1 ε 2π(− log |x| − σ 2 ) eµ−σ = + sgn(x) cos =G , 2 2 σ2 x 9.

(17) and. µ ¶ 1 ε 2π(log |x|) G(e x) = + sgn(x) cos 2 2 σ2 µ ¶ µ µ¶ 1 ε 2π(− log |x|) e = + sgn(x) cos =G . 2 2 2 σ x µ. That is the conditions for G in (15) are satisfied. Therefore, X is doubly symmetric about 2 (eµ−σ , eµ ).. 7 7.1. Some Interesting Examples of I-Symmetry Characterization of I-Symmetry. First we give a characterization by Seshadri (1965) of the p.d.f. of an r.v. defined on R+ which is log-symmetric about 1. d. Lemma 3. Let X be an r.v. defined on R+ . Then X = 1/X if and only if 1 g(log x), x > 0, x where g is a symmetric p.d.f.. (18). fX (x) =. The next lemma is an extension of the above lemma, which concerns r.v.’s defined on R. d. Lemma 4. Let X be an r.v. defined on R. Then X = 1/X if and only if fX (x) =. 1 g(log |x|)G(x), x ∈ R\{0}, |x|. (19). where g is a symmetric p.d.f. and G is a skewing function which satisfies G(x) = G(1/x), x ∈ R\{0}.. (20). Proof: First we prove the “if” part. Suppose (19) holds. Let Z = 1/X. Then µ ¶ µ ¯ ¯¶ µ ¶ ¯1¯ 1 1 |z| ¯ ¯ G 1 = 1 g(log |z|)G(z) = fX (z), z ∈ R\{0}, fZ (z) = fX = g log ¯z ¯ z z2 z2 z |z| where the third equality is by the symmetry of g and (20). This proves the “if” part. d. Next, assume X = 1/X. According to Corollary 2, both f1 and f2 are log-symmetric about 1, where f1 and f2 are given in (3). From Lemma 3, f1 (x) = g1 (log x)/x and f2 (x) = g2 (log x)/x, where g1 and g2 are symmetric p.d.f.’s. By (9), (10), and (11), ( 1 x > 0, 2 (ag1 (log x)/x + (1 − a)g2 (log x)/x), f (x) = 1 2 (ag1 (log(−x))/(−x) + (1 − a)g2 (log(−x)))/(−x), x < 0, 1 (ag1 (log |x|) + (1 − a)g2 (log |x|)) 2|x| 1 = g(log |x|), x ∈ R\{0}, 2|x| =. 10. (21).

(18) and. ( G(x) =. ag1 (log x)/(ag1 (log x) + (1 − a)g2 (log x)), x > 0, (1 − a)g2 (log(−x))/(ag1 (log(−x)) + (1 − a)g2 (log(−x))), x < 0,. (22). where g(x) = ag1 (x) + (1 − a)g2 (x), x ∈ R, which is a mixture p.d.f. of g1 and g2 . Obviously g is also symmetric since both g1 and g2 are symmetric. Substituting (21) into (9), (19) follows immediately. The rest to be proved is G satisfies (20). Now from (22), µ ¶ ( 1 ag1 (log 1/x)/(ag1 (log 1/x) + (1 − a)g2 (log 1/x)), x > 0, G = x (1 − a)g2 (log(−x))/(ag1 (log(−1/x)) + (1 − a)g2 (log(−1/x))), x < 0. ( ag1 (log x)/(ag1 (log x) + (1 − a)g2 (log x)), x > 0, = (1 − a)g2 (log(−x))/(ag1 (log(−x)) + (1 − a)g2 (log(−x))), x < 0. = G(x), x ∈ R\{0}, where the second equality follows by the symmetry of g1 and g2 . This completes the proof. Remark 2. If the G in (20) is G(x) = 0, x ≤ 0 and G(x) = 1, x > 0, then X > 0 and fX is given as in (18). Consider an r.v. X which is I-symmetric about δ. Then by Proposition 5 and Lemma 4, the consequence given below follows immediately. d. Theorem 3. Let X be an r.v. defined on R. Then X/δ = δ/X, δ > 0, if and only if µ ¶ ³ ´ 1 |x| x fX (x) = g log G , x ∈ R\{0}, |x| δ δ. (23). where g is a symmetric p.d.f. and G is a skewing function which satisfies (20). We now give some examples. Example 3. The function defined below is a skewing function satisfying (20), ( 1 2 (1 + h(x)), 0 ≤ |x| ≤ 1, G(x) = 1 1 2 (1 + h( x )), |x| > 1,. (24). where |h(x)| ≤ 1, |x| ≤ 1, is an odd function. In particular when h(x) = cxn , where |c| ≤ 1 and n is 0 or odd number, then ( 1 n 0 ≤ |x| ≤ 1, 2 (1 + cx ), G(x) = (25) 1 −n ), |x| > 1. (1 + cx 2 Moreover G(x) = 1/2, x ∈ R, is a skewing function satisfying (20). The following are examples to illustrate Theorem 3. 11.

(19) Example 4. Let X be C(0, 1) distributed with p.d.f. fX (x) =. 1 , x ∈ R. π(1 + x2 ). d. Obviously X = 1/X. By choosing g(x) =. 2ex , x ∈ R, π(1 + e2x ). (26). and G(x) = 1/2, x ∈ R, then fX (x) =. 1 1 = g(log |x|)G(x), x ∈ R\{0}. 2 π(1 + x ) |x|. On the other hand, if fX (x) =. 2 G(x), x ∈ R, π(1 + x2 ) d. where G is given in (24), then it is still true that X = 1/X. Example 5. Let g(x) = 21 e−|x| , x ∈ R, the p.d.f. of a Laplace distribution, and ( 0, x < 0, G(x) = 1, x ≥ 0, a skewing function which satisfies (20). Then ( 1 1 0 < x < 1, 2, fX (x) = g(log x)G(x) = 1 |x| , x ≥ 1. 2x2 d. This is the p.d.f. of U/V , where U and V are i.i.d. U(0, 1) r.v.’s. Note that if X = U/V , where U and V are i.i.d. r.v.’s, then X is I-symmetric about 1. Although the ratio of two i.i.d. r.v.’s has a distribution of I-symmetric about 1, conversely, for an r.v. Z which is log-symmetric about 1, we will show below that there may not exist two d i.i.d. r.v.’s X and Y such that Z = X/Y . As log Z is symmetric, if there is a symmetric r.v. which is not distributed as the difference of two i.i.d. r.v.’s, then this offers an example that a log-symmetric r.v. cannot be distributed as the ratio of two i.i.d. r.v.’s. Throughout the rest of this section, for an r.v. Z, let ψZ (t), t ∈ R, denote the characteristic function (ch.f.) of Z. First we give a lemma. Lemma 5. Let the r.v. Z defined on R+ be log-symmetric about 1. Also let Z1 = log Z. If d there exist two i.i.d. r.v.’s X and Y such that Z = X/Y , then ψZ1 (t) ≥ 0 for t ∈ R. d. d. Proof: That Z = X/Y implies Z1 = log X − log Y . Obviously, log X and log Y are also i.i.d. Consequently, ψZ1 (t) = ψlog X (t)ψlog Y (−t) = ψlog X (t)ψlog X (−t) = |ψlog X (t)|2 ≥ 0. This completes the proof. 12.

(20) For a symmetric r.v. Z1 , let Z = eZ1 . Then Z is log-symmetric about 1. According to Lemma 5, if ψZ1 (t) ≤ 0 for some t ∈ R, then there do not exist two i.i.d. r.v.’s X and Y such d. that Z = X/Y . The following example was given by Seshadri (1965). Example 6. Let the p.d.f. of the r.v. Z1 mentioned above be 1 2 fZ1 (z) = √ z 2 e−z /2 , z ∈ R. 2π Then. r ψZ1 (t) =. 2 2 (1 − t2 )e−t /2 , t ∈ R. π d. That there do not exist two i.i.d. r.v.’s X and Y such that Z = X/Y follows by noting ψZ1 (t) < 0 when |t| > 1. Next we give a sufficient condition for the distribution of Z defined on R+ which can be represented as X/Y , where X and Y are i.i.d. r.v.’s. Theorem 4. Let the r.v. Z defined on R+ be log-symmetric about 1, and let Z1 = log Z. If p d ψZ1 is also a ch.f., then there exist two i.i.d. r.v.’s X and Y such that Z = X/Y . p Proof: That Z is log symmetric about 1 implies ψZ1 is real and even. Hence ψZ1 is also even. p Let i.i.d. r.v.’s X1 and Y1 have ch.f. ψZ1 . Then p p p ψX1 −Y1 (t) = ψX1 ψY1 (−t) = ψZ1 (t) ψZ1 (−t) = ( ψZ1 (t))2 = ψZ1 (t), t ∈ R. d. d. Consequently, Z1 = X1 − Y1 , which in turn implies Z = X/Y , where X = eX1 , and Y = eY1 . This completes the proof. Theorem 4 has an immediate consequence. Corollary 3. Let the r.v. Z defined on R+ be log-symmetric about 1. Also let Z1 = log Z. If d ψZ1 is infinitely divisible, then there exist two i.i.d. r.v.’s X and Y such that Z = X/Y . p Proof: That ψZ1 is infinitely divisible implies ψZ1 is also a ch.f. By Theorem 1, there exist d. two i.i.d. r.v.’s X and Y such that Z = X/Y . This completes the proof. The following is the Pólya type criterion for ch.f.’s, which can be found in Chung (2001), p.191. Theorem 5. Let the function ψ on R satisfy ψ(0) = 1, ψ(t) ≥ 0, ψ(t) = ψ(−t), t ∈ R, and ψ is decreasing and continuous convex on R+ . Then ψ is a ch.f. 13. (27).

(21) If a ch.f. ψ satisfies the sufficient condition of Theorem 5, then it is said to be a Pólya type ch.f. Theorems 4 and 5 yield the following consequence. Corollary 4. Let the r.v. Z defined on R+ be log-symmetric about 1. Also let Z1 = log Z. If ψZ1 is a Pólya type ch.f. satisfying 1 ψZ1 (t) > 0, ψZ00 1 (t)ψZ1 (t) − (ψZ0 1 (t))2 ≥ 0, t > 0, 2. (28). d. then there exist two i.i.d. r.v.’s X and Y such that Z = X/Y . p p Proof: Firstly, we show that ψZ1 is also a Pólya type ch.f. Obviously, ψZ1 satisfies (27). p Also, since ψZ1 is decreasing and continuous on R+ , so is ψZ1 . Now p ψ 00 (t)ψZ1 (t) − (ψZ0 1 (t))2 /2 p ( ψZ1 (t))00 = Z1 ≥0 (29) 2ψZ1 (t) ψZ1 (t) p p by (28). Consequently, ψZ1 is convex on R+ . Therefore, ψZ1 is also a Pólya type ch.f. By d. Theorem 4, there exist two i.i.d. r.v.’s X and Y such that Z = X/Y . This completes the proof. It is known that both C(0, 1) and N (0, 1) distributions are infinitely divisible. We now present some examples to illustrate Corollary 3. Example 7. Let the p.d.f. of the r.v. Z be fZ (z) =. 1 , z > 0. πz(1 + (log z)2 ). Then Z is log symmetric about 1, and Z1 is C(0, 1) distributed, where Z1 = log Z. Since C(0, 1) distribution is infinitely divisible, according to Corollary 3, there exist two i.i.d. r.v.’s X and Y d such that Z = X/Y . As can be seen below, if the common p.d.f. of X and Y is fX (x) =. 1/2 , x > 0, πx(1/4 + (log x)2 ). then this can be served as an example. Let W = X/Y . Then Z ∞ fW (w) = yfX (wy)fX (y)dy 0 Z ∞ 1/2 1/2 = y dy 2 πwy(1/4 + (log wy) ) πy(1/4 + (log y)2 ) 0 1 = fZ (w), w > 0. = πw(1 + (log w)2 ) Example 8. Let the r.v. Z have LogN ormal(0, 1) distribution. Also let Z1 = log Z. Then Z is log-symmetric about 1, and Z1 is N (0, 1) distributed, which is infinitely divisible. Hence there d. exist two i.i.d. r.v.’s X and Y such that Z = X/Y . The r.v.’s X and Y with LogN ormal(0, 1/2) being their common distribution is an example. 14.

(22) Example 9. Let the r.v. Z1 have the following Pólya type ch.f. ψZ1 (t) =. 1 , t ∈ R. 1 + |t|. Also let Z = eZ1 . Then Z is log-symmetric about 1. Since 1 3 ψZ00 1 (t)ψZ1 (t) − (ψZ0 1 (t))2 = ≥ 0, t ∈ R+ , 2 2(1 + t)4 d. according to Corollary 4, there exist two i.i.d. r.v.’s X and Y such that Z = X/Y .. 7.2. I-Symmetry Arising From Trigonometric Formulas. Let Z = X/Y . Although Z is I-symmetry about 1 if X and Y are i.i.d., the converse is not true. That the joint p.d.f. of X, Y satisfies fX,Y (x, y) = fX,Y (y, x), x, y ∈ R,. (30). is sufficient to imply Z is I-symmetric about 1. See also the following example by Jones (1999). Example 10. Let (X, Y ) have the polar representation X = R cos Θ and Y = R sin Θ,. (31). where Θ is U(0, 2π) distributed, and R is a positive r.v. independent with Θ. Then fX,Y (x, y) =. p 1 p fR ( x2 + y 2 ), x, y ∈ R, 2π x2 + y 2. (32). which satisfies (30). Hence tan Θ (= Y /X) is I-symmetric about 1. In fact, tan Θ is C(0, 1) distributed. Example 9 shows that there exists an I-symmetric distribution about 1 arising from trigonometric functions. Jones (1999) also pointed out if the Θ given in (31) is U(a, b) distributed, where b − a = mπ, m is a positive integer, then tan Θ has a C(0, 1) distribution. It follows immediately that for S being an r.v. independent of Θ, where Θ is U(−π/2, π/2) distributed, then tan(nΘ+S) is also C(0, 1) distributed, where n is a positive integer. Furthermore Jones (1999) gave some multiple angle and angle sum formulas for tangent functions, which remain C(0, 1) distributed. For example, the double angle formula for tangent function yields (tan Θ − 1/ tan Θ)/2 is C(0, 1) distributed. Also the multiple angle and angle sum formulas for sine and cosine functions yield some functions of X and Y have the same distribution as X and some functions of X and Y √ d have a C(0, 1) distribution. For example, 2XY / X 2 + Y 2 = X and 2XY /(Y 2 − X 2 ) is C(0, 1) distributed (see Jones (1999)). Inspired by Jones (1999), we present some related results in the following. Let ³ π π´ 2 , (33) fU (u) = G(tan u), u ∈ − , π 2 2 15.

(23) where G is a skewing function satisfying (20). Let T = tan U . Then it can be shown easily that T is I-symmetric about 1 with p.d.f. 2 fT (t) = G(t), t ∈ R. (34) π(1 + t2 ) The following theorem points out that some of the results presented by Jones (1999) still hold for the r.v.’s U and T given above. Theorem 6. Let T have the p.d.f. given in (34). Then d. (i) T = 1/T ; (ii) tan(2U ) is C(0, 1) distributed; (iii) (T − 1/T )/2 is C(0, 1) distributed; (iv) tan(2U + S) is C(0, 1) distributed, where S is an r.v. independent with U ; (v) 2XY /(Y 2 − X 2 ) is C(0, 1) distributed, where X = R cos U , Y = R sin U , and R is a positive r.v. independent with U ; (vi) let V = sin(4U ), then 1 fV (v) = √ , |v| < 1. π 1 − v2 The proof of the above theorem is standard hence is omitted. Remark 3. If G(x) = 1/2, x ∈ R, that is U is U(−π/2, π/2) distributed, then 2U in (ii) and (iv) can be replaced by U , and 4U in (vi) can be replaced by U .. References [1] K.L. Chung (2001). A Course in Probability Theory, 3rd ed. Academic Press, New York. [2] M.C. Jones (1999). Distributional relationships arising from simple trigonometric formulas. Amer. Statist. 53, 99-102. [3] M.C. Jones (2008). On reciprocal symmetry. J. Statist. Planning and Inference 138, 30393043. [4] M.C. Jones and B.C. Arnold (2008). Distributions that are both log-symmetric and Rsymmetric. Electron. J. Stat. 2, 1300-1308. [5] G.S. Mudholkar, H. Wang, (2007). IG-symmetry and R-symmetry: Interrelations and applications to the inverse Gaussian theory. J. Statist. Planning and Inference 137, 3655-3671. [6] V. Seshadri (1965). On random variables which have the same distribution as their reciprocals. Canad. Math. Bull. 8, 819-824.. 16.

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