Regression of Random Vectors for Multivariate t-Student Distribution

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(1)Zeszyty Uniwersytet Ekonomiczny w Krakowie Naukowe Metody analizy danych. 876 Kraków 2011. Paweł Najman. Department of Mathematics. Regression of Random Vectors for Multivariate t-Student Distribution 1. Regression of a Two-dimensional Random Vector in Relation to a Three-dimensional Random Vector 1.1. Introduction The volumes which were the subject of economic research are in general random variables. In practice, there is often a need for simultaneous analysis of the “set”. In this case we are dealing not with a one-dimensional random variable, but with a random vector, also known as a multidimensional random variable. The starting point for our discussion are the works of J. Tatar (2000, 2002, 2008a), who in 2008 (2008b) examined the regression function for a discrete distribution. Najman and Tatar in 2010 did the research for multivariate normal distribution. We are here looking for an efficient curvilinear regression function (Jakubowski & Sztencel 2001) two-dimensional vector provided three-dimensional vector for t-Student distribution – in other words, regression of the first kind (Gerstenkorn & Śródka 1973). Multivariate t-Student distribution is a generalisation of a classical t-Student distribution for a one-dimensional random variable. The method for constructing the distribution, in the case of a p-dimensional variable, rests on the observation that if variables V and U are independent and have appropriate distributions: multivariate normal parameters (0, Σ) and chi-square with n degrees of freedom (n ≥ 3);.

(2) Paweł Najman. 106 where Σ is the p × p – dimensional matrix and V the density:. n U = X – m, then variable X has. n+ p 1 2 Γ b 2 l Σ –1 –_ n + pi 2 X ~ f _ xi = , a1 + 1n _ x – miT Σ –1 _ x –mik p 2 n Γ a 2 k_ nπi. which we call multivariate t-Student distribution with parameters (Σ, m, n). Let p = 5 and the vector X = (X1, X2, X3, X4, X5) have multivariate t-Student distribution with parameters (Σ, m, n), where m = [m1, m 2, m 3, m 4, m 5] and Σ = [σij]5 × 5 and Σ –1 = A = [aij]5 × 5, thus X ~ T(Σ, m, n). In considering random variable Yi = (Xi – m i). For i = 1, 2, …, 5 we have EYi = (E(Xi) – mi) = 0 and VarYi = Var(Xi – E(Xi)) = VarXi. Furthermore:. cov _Yi, Y j i = E _Yi Y j i – EYi EY j = E _ Xi – mii_ X j – m j i =. = ` E _ Xi X j i – mi EX j – m j EXi + mi m j j = E _ Xi X j i – EXi EX j = σij .. Without the loss of generality, we can consider the vector of random variables Y = (Y1, Y2, Y3, Y4, Y5) with an expected value equal to the zero vector and covariance matrix Σ. In order to calculate the corresponding integrals (necessary at some stage of the process), we apply the transformation yi = xi – mi, i = 1, …, 5. This gives us Y ~ T (Σ, (0, 0, 0, 0, 0), n), while the density function is given by the formula:. Y ~ f _ yi =. 1 2 (n + 5) 5 Γb n + 2 lA n. Γ a 2n k_ nπi5. 2. 2. _ n + yT Ayi. –_ n + 5i 2. , where y = (y1, y2, y3, y4, y5).. 1.2. Regression Function E((X1, X2)  (X3, X4, X5)) We are considering a probability distribution which is not degenerate, and since matrix Σ is positive, definite and symmetric (Feller 1981), matrix A satisfies the following conditions: a) it is symmetric; b) A > 0; c) aii > 0 for i = 1, …, 5; 2 d) a11 a22 – a12 > 0. Let us consider two cases based on the parity number of degrees of freedom n. Case 1: n = 2k – 1, k ≥ 2. First of all we determine the marginal probability density distribution f_Y3 , Y4 , Y5i _ y3 , y4 , y5i of random vector (Y3, Y4, Y5). For this purpose, we calculate the following two-dimensional integral:.

(3) Regression of Random Vectors…. 107. f_Y3 , Y4 , Y5i _ y3, y4 , y5i =. By introducing an additional constant: d1–1 =. 1 2 (n + 5) 5 Γb n + 2 lA n. Γ a 2n k_ nπi5. we obtain:. d1 I1 = d1. #. R. f _ y i dy1 dy2 .. #. R2. 2. 2. ,. 5. –_ k + 2i. 5. # _n + yT Ayi–_k + 2i dy1 = # fn + / / yi y j aij p. f _ y i dy1 =. R. R. i =1 j =1. dy1.. Let Δ1 be the discriminant of the following quadratic trinomial (of variable y1). 5. n + yAyT = n + /. 5. 5. 5. 5. / yi y j aij = y12 $ a11 + y1 $ 2 / yi a1i + fn + / / yi y j aij p ,. i =1 j =1. i=2. i=2 j=2. and a1, b1, c1, its coefficients. This is a form of quadratic trinomial positively defined, hence Δ1 < 0. We shall prove now the following lemma: Lemma 1 If X = ax2 + bx + c, m ≥ 2, a > 0, Δ < 0, then Proof Since. thus. #. R. #. R. dx = _2m – 3i 2a # dx , X m _ m– 1i_ –Δi R X m – 1. dx = _2m – 5i 2a # dx , f, X m – 1 _ m – 2i_ –Δi R X m – 2. #. and. R. dx = _2m – 3i !! 2a m – 1 2π . b l Xm –Δ _ m – 1i ! –Δ. _2m – 3i 2a dx = 2ax + b + # dx + c, X m _ m – 1i_ –Δi X m – 1 _ m – 1i_ –Δi X m – 1 2 2ax + b # dx X = –Δ arctg –Δ + c,. #. #. R. dx = _ 1 i 2a # dx X 2 _ 1 i_ –Δi R X. #. R. dx = 2π . X –Δ.

(4) Paweł Najman. 108 Therefore. #. R. dx = _2m – 3i_2m – 5i $ f $ _ 1 i 2a m – 1 # dx = _2m – 3i !! 2a m - 1 2π . b l b l X Xm –Δ _ m – 1i_ m – 2i $ f $ _ 1 i –Δ _ m – 1i ! –Δ R. This completes the proof of the lemma 1. Using lemma 1 and inserting a = a1, b = b1, c = c1, Δ = Δ1, m = k + 2, we get _2k + 1i !! 1 I1 = 2π 2a k + 1 . d1 _ k + 1i ! _ 1i _ –Δ1ik + 1, 5. Thus, since. _2k + 1i !! d2–1 = 2≠ 2a k + 1, d1 _ k + 1i ! _ 1i. then. d2 I2 = d2. # I1 dy2 = # _ –Δ1i–_k + 1, 5i dy2.. R. R. Let Δ2 be the discriminant of the following quadratic trinomial (of variable y2): 5. –Δ1 = 4 >a11 f n + /. 5. i=2 j=2. =. y 22. $ 4 _ a11 a22 –. 2 i a12. 5. 2. 5. / yi y j aij p – e / yi a1i o H = i=2. 5. + y2 $ 8 / yi _ a11 a2i – a1i a12i + 4 f a11 n + / i=2. ,. 5. / yi y j _a1i a1j – a11 aijip. i=3 j=3. whereas a2, b2, c2, its coefficients. This is a form of quadratic trinomial positively defined, hence Δ2 < 0. We next prove the following lemma: Lemma 2 If X = ax + bx + c, m ≥ 2, a > 0, Δ < 0, then 2. Proof Since. thus. #. X. dx. m+1 2. =. #. R. dx. Xm + 1 2. _ m – 1i ! _8ai 1 . = m _2m – 1i !! a _ –Δi. _ m – 1i 8a 2ax + b m –1 2 + _2m – 1i_ –Δi _ m – 0, 5i_ –Δi X # dx3 2 = 2ax1 2+ b + c 0, 5X _ –Δi X. #. R. #. R. X. dx = _ m – 1i 8a # dx , X m + 1 2 _2m – 1i_ –Δi R X m – 1 2 dx. m –1 2. =. m. #. _ m – 2 i 8a # dx , f, _2m – 3i_ –Δi R X m – 3/2. dx + c, Xm – 1 2.

(5) Regression of Random Vectors…. 109. #. R. and. dx = 8 a . –Δ X3 2. #. R. Therefore. dx = _ 1i 8a # dx X 2, 5 _ 3i_ –Di R X3 2. #. R. dx. Xm + 1. =. 2. _ m – 1i_ m – 2i $ f $ _ 1 i 8a m – 1 b l _2m – 1i_2m – 3i $ f $ _ 3i –Δ. #. R. dx = _ m – 1i ! _8ai 1 . m X3 2 _2m – 1i !! a _ –Δi m. It completes the proof of the lemma 2. Using lemma 2 and inserting a = a2, b = b2, c = c2, Δ = Δ2, m = k + 1, X = –Δ1, we get _8a2i k! I2 = 1 d2 _2k + 1i !! a2. k +1. Let us denote R. 1 k +1 . _ –Δ 2i. # y2 f_ yi dy1 dy2 by I3 and calculate it. Let us also observe that. 2. d2 I3 = d2. # y2 f_ yi dy1 dy2 = # y2 e d2 # R. R2. =. R. # y2 d2 I1 dy2 = # y2. R. R. f _ y i dy1o dy2 =. 1 k + 1, 5 dy2 . _ –Δ1i. First, however, here is yet another useful lemma: Lemma 3. If X = ax2 + bx + c, m ≥ 2, a > 0, Δ < 0, then. #. R. Proof. Indeed: if. #. then. #. R. xdx = –b _ m – 1i ! _8ai 1 . m X m + 1 2 2a _2m – 1i !! a _ –Δi. xdx = – 1 Xm + 1 2 _2m – 1i aX m – 1. xdx = – b 2a Xm + 1 2. #. R. dx. Xm + 1. Lemma 2 2. =. 2. – 2ba. m. #. dx + c, Xm + 1 2. _ m – 1i ! _8ai 1 , – 2ba m _2m – 1i !! a _ –Δ i m. this is precisely the thesis of lemma 3. Using lemma 3 and inserting a = a2, b = b2, c = c2, Δ = Δ2, m = k + 1, X = –Δ1, we get.

(6) Paweł Najman. 110 _8a2i –b k! I3 = 1 2a2 d2 2 _2k + 1i !! a2. k +1. therefore. R. # y2 fa_ y1, y2i _ y3, y4, y5ik dy1 dy2 = #. 2. R. _8a2i I –b k! = I3 = 1 2a2 d a2 2 2 _2k + 1i !! 2. k +1. =. 2. #. R2. 1 k +1 , _ –Δ 2i. y2 f _ y i. dy1 dy2 =. f _ y i dy1 dy2. # y2 f_ yi dy1 dy2. R2. #. R2. f _ y i dy1 dy2. =. _2k + 1i !! a2 –b 1 k +1 = 2a2 = k + 1 $ d2 k + 1 _ –Δ 2i k! 2 _ –Δ 2i _8a2i. 5. – / yi _ a2i a11 – a12 a1ii i=3. 2 a11 a22 – a12. .. Analogously, we determine the expected value of the variable Y1 with respect to the same conditional density: 5. # y1 fa_ y1, y2i _ y3, y4, y5ik dy1 dy2 =. – / yi _ a1i a22 – a12 a2ii i=3. R2. 2 a22 a11 – a12. .. Case 2: n = 2k, k ≥ 2.. We also determine the marginal probability density distribution f_Y3 , Y4 , Y5i _ y3, y4 , y5i of random vector (Y3, Y4, Y5). For this purpose, we calculate the following two‑dimensional integral: f_Y3 , Y4 , Y5i _ y3, y4 , y5i =. By introducing an additional constant d1–1 =. we obtain. d1 I1 = d1. #. R. f _ y i dy1 =. #. R2. f _ y i dy1 dy2 .. 1 2 (n + 5) 5 Γb n + 2 lA n. Γ a 2n k_ nπi5. 2. 2. ,. 5. –_ k + 5 2i. 5. # _n + yT Ayi–_k + 5 2i dy1 = # fn + / / yi y j aij p R. R. i =1 j =1. dy1 .. Let Δ1 once again be the discriminant of the following quadratic trinomial (of variable y1):. 5. n + yAyT = n + /. 5. 5. 5. 5. / yi y j aij = y12 $ a11 + y1 $ 2 / yi a1i + fn + / / yi y j aij p. i =1 j =1. i=2. i=2 j=2. and a1, b1, c1, its coefficients. This is a form of positively defined quadratic trinomial, hence Δ1 < 0..

(7) Regression of Random Vectors…. 111. Using lemma 2 (from case 1) and inserting a = a1, b = b1, c = c1, Δ = Δ1, m = k + 2, we get _ k + 1i ! _8a1i I1 = 1 d1 _2k + 3i !! a1. k+2. Let us define. 1 k+2. _ –Δ1i. k+2 _ k + 1i ! _8a1i d3–1 = 1 , d1 _2k + 3i !! a1. and obtain. d3 I2 = d3. # I1 dy2 = # _ –Δ1i–_k + 2i dy2 .. R. R. Let Δ2 be the discriminant of the following quadratic trinomial (of variable y2): 5. –Δ1 = 4 >a11 f n + /. 5. 2. 5. / yi y j aij p – e / yi a1i o H =. i=2 j=2. i=2. 5. 2 i + y2 $ 8 / yi _ a11 a2i – a1i a12i + = y 22 $ 4 _ a11 a22 – a12. 5. + 4 f a11 n + /. i=2. 5. / yi y j _a1i a1j – a11 aijip. i=3 j=3. and a2, b2, c2, its coefficients. Again, this is a form of positively defined quadratic trinomial, hence Δ2 < 0. Using lemma 1 (from case 1) and inserting a = a 2, b = b 2, c = c 2, Δ = Δ2, m = k + 1, X = –Δ1, we get. As in case 1, we denote. _2k + 1i !! 1 I2 = 2π 2a k + 1 . d3 _ k + 1i ! _ 2i _ –Δ 2ik + 1, 5. # y2 f _ yi dy1 dy2 by I3 and calculate it. We then observe that. R2. d3 I3 = d3. # y2 f _ yi dy1 dy2 = # y2 e d3 # R. R2. =. # y2 d3 I1 dy2 = # y2. R. R. R. f _ y i dy1o dy2 =. 1 k + 2 dy2 . _ –Δ1i. First, however, we shall show the useful lemma: Lemma 4. If X = ax2 + bx + c, m ≥ 2, a > 0, Δ < 0, then. #. R. xdx = –b _2m – 3i !! 2a m – 1 2π . b l X m 2a _ m – 1i ! –Δ –Δ.

(8) Paweł Najman. 112 Proof. Indeed: if. xdx = – 1 – b Xm _ m – 1i 2aX m – 1 2a. #. then. #. R. xdx = – b # dx 2a X m Xm R. Lemma 1. =. #. dx + c, Xm. _2m – 3i !! 2a m – 1 2π – 2ba , b l –Δ _ m – 1i ! –Δ. this is precisely the thesis of lemma 4. Using lemma 4 and inserting a = a2, b = b2, c = c2, Δ = Δ2, m = k + 2, X = –Δ1, we have –b _2k + 1i !! 1 I3 = 2π 2a2 2a k + 1 , d3 2 _ k + 1i ! _ 2i _ –Δ 2ik + 1, 5. thus finally. # y2 f a_ y1, y2i _ y3, y4 , y5ik dy1 dy2 = #. R2. R2. #. R2. y2 f _ y i. f _ y i dy1 dy2. dy1 dy2 =. # y2 f _ yi dy1 dy2. R2. #. R2. f _ y i dy1 dy2. =. d _ k + 1i ! I –b _2k + 1i !! 1 1 –Δ k + 1, 5 = = I3 = 2π 2a2 $ 3 2a k + 1 d3 2 _ k + 1i ! _ 2i _ –Δ 2ik + 1, 5 2π _2k + 1i !! _2a2ik + 1 _ 2i 2 5. –b = 2a 2 = 2. – / yi _ a2i a11 – a12 a1ii i=3. 2 a11 a22 – a12. .. Analogously, we determine the expected value of variable Y1 with respect to the same conditional density: 5. R. # y1 f a_ y1, y2i _ y3 , y4, y5ik dy1 dy2 =. – / yi _ a1i a22 – a12 a2ii i=3. 2. 2 a22 a11 – a12. .. Now it is time to return to vector (X1, X2, X3, X4, X5) by applying the substitution previously used. Substituting the relevant variables in both cases (1 and 2) in the integrals we get: 5. R. # y2 f a_ y1, y2i _ y3, y4 , y5ik dy1 dy2 =. 2. =. – / yi _ a2i a11 – a12 a1ii i=3. 2 a11 a22 – a12. # _ x2 – m2i f a_ x1, x2i _ x3 , x4 , x5ik dx1 dx2 =. R2. =.

(9) Regression of Random Vectors…. =. and. 113. # x2 f a_ x1, x2i _ x3 , x4, x5ik dx1 dx2 – m2 #. R2. =. R2. f a_ x1, x2i _ x3, x4 , x5ik dx1 dx2 =. # x2 f a_ x1, x2i _ x3 , x4 , x5ik dx1 dx2 – m2. R2. # y1 f a_ y1, y2i _ y3 , y4, y5ik dy1 dy2 = # x1 f a_ x1, x2i _ x3, x4, x5ik dx1 dx2 – m1.. R2. R2. Summing up all the previous considerations, we find that for any number of degrees of freedom (n ≥ 3) we obtain. E a_ X1, X2i _ X3, X4, X5ik = J N 5 5 K – / _ Xi – mii_ a1i a22 – a12 a2ii – / _ Xi – mii_ a2i a11 – a12 a1ii O K O = Km1 + i = 3 , m2 + i = 3 2 2 O, a22 a11 – a12 a11 a22 – a12 L P. and thus we have an effective form regression of a two-dimensional random vector in relation to a three-dimensional random vector.. 2. Regression of Random Vector – General Case 2.1. Introduction The reasoning presented in Section 2.1 extends out to the case of random vectors of any size, while the assumption that they are random vectors with multivariate t-Student distributions remains valid. Let p, j, m, n  N ∧ 1 ≤ j < j + 1 ≤ p ∧ n ≥ 3 and X = (X1, …, Xp) be p-dimensional random vector with multivariate t-Student distribution with parameters (Σ, M, n), where M = [m1, …, mp] and Σ = [σij]p × p and Σ–1 = A = [aij]p × p; hence X ~ T(Σ, M, n). As in the specific case we consider a random variable Yi = (Xi – mi) for i = 1, 2, …, p. In order to calculate the corresponding integrals (which is necessary at some stage of the process), we apply the transformation yi = xi – mi, i = 1, …, p. We have Y ~ T(Σ, (0, …, 0), n), and the density function is given by the formula:. n+ p Γ b 2 l A 1 2 n(n + p) 2 –_ n + pi 2 _ n + yT Ayi Y ~ f _ yi = , where y = (y1, …, yp). p 2 n Γ a 2 k_ nπi. 2.2. Regression Function E a_Y1, Y2, f, Y j i _Y j + 1, Y j + 2, f, Y pik. We first subdivide the random vector Y = (Y1, …, Yp) – also vector X – into two vectors, j-dimensional and (p-j)-dimensional, respectively. Without the loss.

(10) Paweł Najman. 114. of generality we assume that the j-dimensional vector includes j initial random variables from vector Y, and other random variables give us the (p-j)-dimensional vector. Therefore vector Y will be divided into vectors (Y1, …, Yj) and (Yj + 1, …, Yp). p p Let W _ y1, f, y pi = n + /. / yi y j aij be the quadratic trinomial of variable y1. i =1 j =1. (if s > 1) or variable y 2 (if s = 1). Further, for a fixed s  {1, …, j}, volumes s s s s Δ1_ i, a1_ i, b1_ i, c1_ i are, respectively, discriminant and factors of quadratic trinomial W. The next discriminants and factors we determine recursively: a) volumes Δ_i s i, a_i s i, b_i s i, c_i s i are respectively discriminant and factors of quadyi , gdy i < s , for i = 2, …, j – 1; ratic trinomial D_i s–i1 of variable * yi + 1, gdy i ≥ s. b) volumes D_js i, a_js i, b_js i, c_js i are respectively discriminant and factors of quadratic trinomial D_js–i 1 of variable ys. We use the summary record: s Δ1_ i = Δ1_ i _ y2 , f, y pi, f, Δ_i i = * s. s. Δ_i s i _ yi + 1, f, y pi, if i < s Δ. s s , Δ_j i = Δ_j i _ y j + 1, f, y pi y , yi + 2, f, y pi, if i ≥ s. _si i _ s. a1_ i = a1_ i _ y2, f, y pi, f, a_i i = *. a_i i _ yi + 1, f, y pi, if i < s. b1_ i = b1_ i _ y2, f, y pi, f, b_i i = *. b_i i _ yi + 1, f, y pi, if i < s. s. s. s. s. s. s. s. , a_j i = a_j i _ y j + 1, f, y pi s a_i i _ ys, yi + 2, f, y pi, if i ≥ s s. b. s. s. , b_j i = b_j i _ y j + 1, f, y pi y , yi + 2 , f, y pi, if i ≥ s. _si i _ s. s. s. for i = 2, …, j – 1. Note that, since matrix Σ is positively defined and symmetric, the probability distribution is not degenerated (Feller 1981) then matrix is also positively defined and symmetric, thus the quadratic form W is positive definite due to each of the variables y1, …, ys – 1, ys, ys + 1, …, yj. Hence, D_i s i < 0 / a_i s i > 0 for any i, s  {1, …, j}. We are seeking an effective formula to determine the expected value. However, since. E a_Y1, Y2, f, Y j i _Y j + 1, Y j + 2 , f, Ypik .. E a_Y1, Y2 , f, Y j i _Y j + 1, Y j + 2, f, Ypik =. = b E aY1 _Y j + 1, Y j + 2, f, Ypik, E aY2 _Y j + 1, Y j + 2 , f, Ypik, f, E aY j _Y j + 1, Y j + 2, f, Ypikl ,. it suffices to set E aYs _Y j + 1, Y j + 2 , f, Y pik for s = 1, …, j..

(11) Regression of Random Vectors…. 115. Each of the above expected values is calculated as follows, in turn: E aYs _Y j + 1, Y j + 2 , f, Ypik =. =. #. Rj. #. Rj. ys f _ y i. f _ y i dy1 dy2 fdy j. # ys f _ yi dy1 dy2 fdy j. Rj. dy1 dy2 fdy j =. #. Rj. J j, s =: I . j f _ y i dy1 dy2 fdy j. The procedure for calculating the corresponding integrals will use the thesis of the lemma proved in section 1.2. Successively we compute the integral Jj, s with respect to variables dy1, …, dys – 1, dys + 1, …, dyj, s = 2, …, j – 1 (if s = 1, then we start from dy2 and end with dyj; if s = j, then we start from dy1 and end with dyj – 1), and then with respect to dys. In the same order we calculate the integral Ij (more precisely: it is calculated at j different ways, depending on s, s  {1, …, j}). We run the reasoning for the two integrals (for a fixed s) almost identically, except for calculating the integral with respect to dys: in this case, if the integral Ij is calculated using, respectively, lemma 1 or lemma 2 (depending on the parity of s and j), then lemma 4 or lemma 3 would be appropriate to calculate the integral Jj, s. It should also be noted that the theses of lemmas 1 and 4, as well as lemmas 2 and 3, differ only in terms of factor b – 2ba l . Therefore J j, s = f –. Hence. b_js i. 2a_js i. p I j, s = 1, f, j.. E a_Y1, Y2 , f, Y j i _Y j + 1, Y j + 2, f, Ypik = f –. b_j1i. ,– _ 1i. 2a j. b_j2i. , f, – _ 2i. 2a j. b_js i. , f, – _si. 2a j. b_j j i. 2a_j. ji. p.. Determining the regression function j-dimensional random vector meant that (p-j)-dimensional random vector is reduced, and therefore j Tri square sequence of variables y1, …, ys – 1, ys + 1, …, yj, ys, needed to be “organised” and the couple coefficients a_i s i, b_i s i, i = 1, f, j of these j Tri square to be determined. We proceed now to determine the final form of regression function E aY j _Y j + 1, Y j + 2, f, Ypik .. Detailed calculations for the step case n + p = 2k, k ≥ 2, i.e. if n + p is an even number no less than four. Let s  {1, …, j} ∧ n + p = 2k, k ≥ 2, then. Ij =. #. Rj. f _ y i dy1 dy2 fdy j = d. # `W_ y1, y2, f, y pij–k dy1 dy2 fdy j ,. Rj.

(12) Paweł Najman. 116 where. n+ p Γ b 2 l A 1 2 n(n + p) 2 _ k – 1i ! A 1 2 n k = d= , Γ a 2n k_ nπip 2 Γ a 2n k_ nπip 2. we use successively: ––lemma 1 for m = k, Δ = Δ1_ s i, a = a1_ s i, b = b1_ s i, c = c1_ s i, s s s s ––lemma 2 for m = k – 1, Δ = Δ_2 i, a = a_2 i, b = b_2 i, c = c_2 i, s s s s ––lemma 1 for m = k – 1, Δ = Δ_3 i, a = a_3 i, b = b_3 i, c = c_3 i, s s s s ––lemma 2 for m = k – 2, Δ = Δ_4 i, a = a_4 i, b = b_4 i, c = c_4 i f etc. In general: a) if j is even, then: j–4 s s s s – lemma 1 for m = k – 2 , Δ = Δ_j –i 3, a = a_j –i 3, b = b_j –i 3, c = c_j –i 3 j–2 s s s s and for m = k – 2 , Δ = Δ_j –i 1, a = a_j –i 1, b = b_j –i 1, c = c_j –i 1, j–2 s s s s – lemma 2 for m = k – 2 , Δ = Δ_j –i 2 , a = a_j –i 2, b = b_j –i 2, c = c_j –i 2 j and for m = k – 2 , Δ = Δ_j s i, a = a_js i, b = b_js i, c = c_js i. b) if j is odd, then: j–3 s s s s – lemma 1 for m = k – 2 , Δ = Δ_j –i 2, a = a_j –i 2 , b = b_j –i 2, c = c_j –i 2 j –1 s s s s and for m = k – 2 , Δ = Δ_j –i 1, a = a_j –i 1, b = b_j –i 1, c = c_j –i 1, j –1 – lemma 2 for m = k – 2 , Δ = Δ_j s i, a = a_js i, b = b_js i, c = c_js i. We obtain:. Ij = >. _2k – 3i !! _ k – 1i !. _si k – 1 1 j. ` 2a. 2π H>. _si _ k – 2 i ! ` 8a 2 j. k –1. _2k – 3i !!. a. _si 2. H>. _2k – 5i !! _ k – 2i !. `2a_3s ij. k–2. 2π H $. Z R j –1 V k– ]R W VSb k – j + 1 l ! _si j –1 a8a j – 1k 2 W k– ] S _2k – j i !! WS 2 _si 2 a k 2π WS ]S W$ j – 1 2a j – 2 _si _2k – j i !! _si k – 2 S W a ] b l k ! – S W – j 1 ` j WT ]S _ k – 3i ! 8a 4 2 X V $> $f$ [T H X R j s _2k – 5i !! k – +1 W j a_4 i ]R S V s _ i 2 j k – +1 ] S_2k – j + 1i !! W WS b k – 2 l ! a8a j – 2k a2a_js–i 3k 2 2π WS ]S W$ j _si !! 2 – + 1 k j _ i S W aj – 2 ]] SS b k – + 1l ! W W 2 X \T XT.

(13) Regression of Random Vectors…. Z R V j j +1 k– ] S_2k – j – 2i !! W 1 k–2 _si 2 a k a π $ 2 2 ] S j Wf _ s i p , if j is odd j +1 ] SS b k – l! WW –Δ j 2 ] T RX j V [ k– W VS b k – j – 1l ! _si 2 j j ] R_2k – j – 1i !! a k 8a j k– W 1 k–2 WS 2 _si 2 ]$ S a2a j – 1k 2π WS Wf _ s i p , if j is even j ] S b a_js i W –Δ j WWS_2k – j – 1i !! ] SS k – 2 l ! X \ T XT. and J j, s = I j f –. 117. b_js i. 2a_js i. p, s = 1, f, j. Hence, finally:. E a_Y1, Y2 , f, Y j i _Y j + 1, Y j + 2 , f, Ypik = f –. b_j1i. ,– _ 1i. 2a j. b_j2i. ,f– _ 2i. 2a j. b_js i. , f, – _si. 2a j. b_j j i. 2a_j. ji. p.. Substituting the relevant variables in both cases in the integrals gives us (for s = 1, …, j):. # ys f a_ y1, f, y ji _ y j + 1, f, ymik dy1fdy j = f –. Rj. =. b_js i. 2a_j. si. p=. # _ xs – msi f a_ x1, f, x ji _ x j + 1, f, xmik dx1fdx j =. Rj. =. # xs f a_ x1, f, x ji _ x j + 1, f, xmik dx1fdx j – ms .. Rj. To sum up, regardless of the number of degrees of freedom n ≥ 3, we still have the following feature of the regression j-dimensional random vector terms of m‑dimensional random vector: b_j1i. E a_ X1, f, X j i _ X j + 1, f, X pik = f m1 –. , m2 – _ 1i. 2a j. b_j2i. , f, ms – _ 2i. 2a j. b_js i. , f, m j – _si. 2a j. b_j j i. 2a_j j i. p.. 3. Example As an illustration of the above, we determine E a_ X1, X2i _ X3 , X4 , X5ik, where X = (X1, X2, X3, X4, X5) has a multivariate t-Student distribution, i.e. X ~ T(Σ, m, n), described by density:. X ~ f _ xi =. 1 5 Γb n + 2 lA. Γ a 2n k_ nπi5. 2. 2. –_ n + 5i 2 , a1 + 1n _ x – miT A _ x – mik. where x = (x1, x2, x3, x4, x5), being m = [2, 5, –3, 11, 7], n > 3 and.

(14) Paweł Najman. 118. This means that:. R 9 –3 6 –3 3V S W S–3 5 0 3 –1W S W Σ –1 = A = 7aij A5 = S 6 0 21 7 6 W. i, j = 1 S–3 3 7 7 –1W S W S 3 –1 6 –1 31W T X R163 5 19 2 1 V S 288 – 96 – 48 3 24 WW S 5 7 267 9 3 W S– 80 – 25 – 200 W S 96 800 9 3 1 W. Σ = S– 19 – 3 – 20 80 8 5 S 48 W 7 3 2 W 29 S 2 – – 25 5 25 W 25 S 3 3 – 1 2 1 W SS 1 – 24 200 20 25 25 W T X. Using the model of section 1, we get the “hyperplane” regression:. E a_ X1, X2i _ X3 , X4, X5ik = b – 5 X3 + 1 X4 – 13 X5 , – 21 X3 – 21 X4 + 9l . 6 6. Using the universal method for the general case from section 2, we obtain:. hence,. D _ y1, f, y5i = 1 + 1n `5y 22 + y2 _ –6y1 + 6y4 – 2y5i + 9y12 + 21y32 + 7y 24 + 31y52 + 12y1 y3 – 6y1 y4 + 6y1 y5 + 14y3 y4 + 12y3 y5 – 2y4 y5i 1 D1_ i _ y1, y3 , f, y5i = 12 _ –6y1 + 6y4 – 2y5i2 – n. 1 2 2 2 2 + 20 n a1 + n _9y1 + 21y3 + 7y 4 + 31y5 + 12y1 y3 – 6y1 y4 + 6y1 y5 + 14y3 y4 + 12y3 y5 – 2y4 y5ik = 1 = 12 a –144y12 + y1 _ –240y3 + 48y4 – 96y5i + c_2 ik , n. therefore. and. E aY1 _Y3 , Y4 , Y5ik = – E a X1 _ X3 , X4 , X5ik = m1 +. Analogously, we determine:. hence,. b_2 i 1. 2a. _ 1i 2. =. 5y3 – y4 + 2y5 6. –5y3 + y4 – 2y5 = – 5 X3 + 1 X4 – 13 X5 . 6 6 6. D _ y1 , f, y5i = 1 + 1n `9y12 + y1 _ –6y2 + 12y3 – 6y4 + 6y5i + 5y 22 + 21y32 + 7y 24 + 31y52 + 6y2 y4 – 2y2 y5 + 14y3 y4 + 12y3 y5 – 2y4 y5i.

(15) Regression of Random Vectors…. 2 D1_ i _ y2, f, y5i = 12 _ –6y2 + 12y3 – 6y4 + 6y5i2 – n. 1 2 2 2 2 + 36 n a1 + n _5y 2 + 21y3 + 7y 4 + 31y5 + 6y2 y4 – 2y2 y5 + 14y3 y4 + 12y3 y5 – 2y4 y5ik = 2 = 12 a –144y 22 + y2 _ –144y3 – 144y4i + c_2 ik , n. Finally:. b_2 i 2. –y – y = 32 4 , 2 2a_2 i y +y E a X2 _ X3 , X4, X5ik = m2 + 3 2 4 = – 21 X3 – 21 X4 + 9. E aY2 _Y3 , Y4, Y5ik = –. 119. E a_ X1, X2i _ X3 , X4, X5ik = b – 5 X3 + 1 X4 – 13 X5 , – 21 X3 – 21 X4 + 9l . 6 6. We received a two-dimensional vector of random variables that are a linear combination of one-dimensional random variables. The above example shows that even when there are a large number of random variables, determining the regression function using the given algorithm is neither difficult nor complicated numerically. Bibliography Feller W. (1981), An Introduction to Probability Theory, vol. II, 3rd ed., PWN, Warsaw. Gerstenkorn T. and Śródka T. (1973), Combinatorics and Probability, 2nd ed., PWN, Warsaw. Jakubowski J. and Sztencel R. (2001), An Introduction to Probability Theory, 2nd ed., Script, Warsaw. Najman P. and Tatar J. (2010), Regression of Random Vectors for Multivariate Normal Distribution [in:] Badania ekonometryczne w teorii i praktyce (2010), ed. A.S. Barczak, Prace Naukowe Uniwersytetu Ekonomicznego w Katowicach, Katowice. Tatar J. (2000), A New Characterization of Multidimensional Probability Distributions, Report of the Statutory, Agreement nr 92/KM/1/99/S; AE, Krakow. Tatar J. (2002), A New Measures of Dependence of Random Vectors, Committee on Statistical and Demographic Sciences PAN, Krakow, 22 May. Tatar J. (2008a), Measures of Dependence of Random Vectors with Different Dimensions, Scientific Papers of UEK, nr 780, Krakow. Tatar J. (2008b), New Characteristics of Multivariate Conditional Distributions, XLIV SEMPP – XXVI Prof. Zbigniew Pawłowski Seminar, Osieczany, May.. Regresja wektorów losowych dla wielowymiarowego rozkładu t-Studenta Przedmiotem prezentowanej pracy jest jakże istotne dla współczesnych finansów pojęcie „warunkowej wartości oczekiwanej”. Rozważania prowadzone są dla pięciowymiaro-.

(16) 120. Paweł Najman. wego wektora losowego o wielowymiarowym rozkładzie t-Studenta. W pierwszej części pracy dokonujemy podziału jego współrzędnych w stosunku 2:3 (tzn. na podwektory odpowiednio: dwu- oraz trójwymiarowy), poszukując efektywnej postaci funkcji regresji wektora dwuwymiarowego pod warunkiem wektora trójwymiarowego. W tym celu w pierwszej kolejności wyznacza się (obliczając odpowiednią całkę podwójną) gęstość prawdopodobieństwa rozkładu brzegowego wektora trójwymiarowego, następnie zaś gęstość prawdopodobieństwa rozkładu warunkowego wektora dwuwymiarowego pod warunkiem wektora trójwymiarowego. W drugiej części pracy rozumowanie zaprezentowane w pierwszej części pracy uogólniono na przypadek wektorów losowych o dowolnych wymiarach, przy nadal obowiązującym założeniu, że są to wielowymiarowe wektory losowe o rozkładzie t-Studenta. Uzyskane wyniki (ogólna postać funkcji regresji) zilustrowano przykładem liczbowym, otrzymując konkretną „hiperpłaszczyznę” regresji.. Random Vector Regression for Multivariate t-Student Distribution The paper examines the concept of „conditional expected value,” which is of great importance in modern finance. The considerations are carried out on a five dimensional random vector with multivariate t-Student distribution. In the first part we construct a distribution of its coordinates in a 2:3 ratio (i.e., the vectors are two-and three-dimensional, respectively) in order to find an effective two-dimensional vector regression function in relation to the three-dimensional vector. To that end, the probability density distribution of the boundary three-dimensional vector is determined (by calculating the appropriate double integral), and then the conditional probability density distribution of two-dimensional vector was used to produce the three-dimensional vector. The second part of the paper discusses the reasoning presented in the first part and then generalises it for a random vector of any size that will remain applicable provided that it is a multi-dimensional random vectors with t-Student distribution. The results (the general form of the regression function) are illustrated with a specific quantitative example that maintains a „hyperplane” regression..

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