- Email: [email protected]
Z built from independent random variables (X,Y) and Z may not always be dependent
Statistical Methodology 14 (2013) 62–66
Contents lists available at SciVerse ScienceDirect
Statistical Methodology journal homepage: www.elsevier.com/locate/stamet
Ratios X /Z , Y /Z built from independent random variables (X , Y ) and Z may not always be dependent Nitis Mukhopadhyay a,∗ , Mun S. Son b a
Department of Statistics, University of Connecticut, Storrs, CT 06269-4120, USA
b
Department of Mathematics & Statistics, University of Vermont, Burlington, VT 05401-1455, USA
article
info
Article history: Received 26 October 2012 Accepted 26 February 2013 Keywords: Bivariate t distribution Continuous distribution Discrete distribution Lognormal distribution Multivariate normal Normal distribution Symmetric distribution
abstract It is a commonly held ‘‘belief’’ in many quarters that the ratios U = XZ , V = YZ are necessarily dependent random variables when the random vector (X , Y ) is independent of the random variable Z simply because both U , V involve Z . Any outpouring support behind such ‘‘belief’’ often gets louder when (X , Y ) are assumed dependent. The purpose of this note is to emphasize that such ‘‘beliefs’’ may be false. Concrete examples are given when (i) X , Y are independent but U , V may be dependent or independent, (ii) X , Y are dependent but U , V may be dependent or independent. Finally, a simple general approach is given for beginners without exploiting joint and/or conditional densities. © 2013 Elsevier B.V. All rights reserved.
1. Introduction Suppose that we have three independent random variables X , Y and Z with Z > 0 w.p.1. We may wish to address the following problem: Let X ∼ N (0, 1), Y ∼ N (0, 1) and ν Z 2 ∼ χν2 and define ratios X Y U = , V = . Prove whether or not (U , V ) are dependent Z Z without exploiting their joint and conditional densities.
∗
Corresponding author. E-mail addresses: [email protected] (N. Mukhopadhyay), [email protected] (M.S. Son).
1572-3127/$ – see front matter © 2013 Elsevier B.V. All rights reserved. http://dx.doi.org/10.1016/j.stamet.2013.02.002
(1)
N. Mukhopadhyay, M.S. Son / Statistical Methodology 14 (2013) 62–66
63
As a quick response, we may be immediately told that (U , V ) has a bivariate t-distribution which is certainly correct. Then, we may also be informed by many responders that U , V from (1.1) are necessarily dependent random variables. At this point, one may genuinely start to wonder about how such a conclusion may follow right away, especially in the absence of an otherwise complicated joint density function of (U , V ) and the conditional densities. We may look up a multivariate t distribution from Johnson and Kotz [1, Chapter 37], Tong [5, Chapter 9], Mukhopadhyay [2, pp. 218–219] or another appropriate source. But, many may admit that they cannot recall from memory a general multivariate t density function or its associated conditional density functions correctly just when we may need them. Personally, we belong to that group. How can one be so sure that U , V from Eq. (1) are dependent random variables? One may often hear an emphatic response such as: It is obvious. Both U , V involve Z , and so U , V must be dependent. Unfortunately, this line of argument is not satisfactory and one may continue to ask whether a solid justification can be constructed to prove dependence of U , V from Eq. (1) without explicitly referring to their joint and conditional density functions. This short note arose from such a heightened curiosity, especially for beginners. However, before we resolve the query stated in Eq. (1), we feel tempted to exploit this opportunity to raise a broader question: Suppose that we have three independent random variables X , Y and Z with P (Z = 0) = 0 and define U = XZ and V = YZ . Again, many may be tempted to argue that since U and V both involve Z , U , V must be dependent. But, then, obviously we will be back exactly where we began. Admittedly U , V are frequently dependent, but they may not always be so in every circumstance! Moreover, one must come up with a logical proof of dependence or independence as a case may warrant. Hence, we give explicit examples where U , V are decisively independent. One will find airtight proofs of independence without explicitly writing down the joint and conditional densities of U , V . In Example 1 from Section 2, we have X , Y independent, (X , Y ) independent of Z , and yet U , V are independent. In Example 2 from Section 3, we have X , Y dependent, (X , Y ) independent of Z , and yet U , V are independent. Then, we supply a simple resolution of the opening problem from Eq. (1) in Section 4 with the help of an interesting approach involving moments. Section 5 briefly touches upon a generalization of the basic idea that we are trying to get across. It is our earnest hope that inquisitive readers may find the experience of constructing pretty examples of U and V with logically sound but simple approaches to verify their dependence or independence interesting. 2. First example: X , Y are independent but U , V are independent Let X , Y , and Z be independent random variables, X ∼ N (0, 1), Y ∼ N (0, 1), whereas P (Z = −1) = P (Z = 1) = 12 . In what follows, we prove that U = XZ , V = YZ are independent. With arbitrary real numbers a and b, we can express: P ( U ≤ a ∩ V ≤ b) =
=
1 2 1 2
P
X Z
≤a∩
Y X 1 ≤ b Z = −1 + P ≤ a ∩ ≤ b Z = 1 Z 2 Z Z
Y
P (−X ≤ a ∩ −Y ≤ b) +
1 2
P ( X ≤ a ∩ Y ≤ b) ,
since(X , Y ), Z are assumed independent.
(2)
But, the joint distribution of (X , Y ) is exactly same as that of (−X , −Y ). Thus, we can rewrite Eq. (2) as: P (U ≤ a ∩ V ≤ b) = P (X ≤ a ∩ Y ≤ b) = P (X ≤ a)P (Y ≤ b), since X , Y are assumed independent.
(3)
With analogous arguments, one can claim: P (U ≤ a) = P (X ≤ a) and
P (V ≤ b) = P (Y ≤ b),
since X , Y are assumed independent.
(4)
64
N. Mukhopadhyay, M.S. Son / Statistical Methodology 14 (2013) 62–66
A combination of Eqs. (3) and (4) obviously leads to: P (U ≤ a ∩ V ≤ b) = P (U ≤ a)P (V ≤ b),
for all real numbers a, b.
(5)
Eq. (5) is a restatement of independence of U , V . Remark 2.1. In this example, it is not crucial that both X , Y must be independent N (0.1) random variables. The conclusion from this example continues to hold if X , Y are independent, but their probability distributions (not necessarily same) are both symmetric around zero. Also, X , Y may be continuous or discrete. 3. Second example: X , Y are dependent but U , V are independent Suppose that X1 , X2 , and X3 are independent random variables, each having the following lognormal probability density:
1 f ( x) = √ exp −(log x)2 /2 , x 2π
0 < x < ∞, and 0 otherwise.
(6)
We define Y , V = where X = X1−1 , Y = X1 X2−1 Z Z Obviously, we can rewrite: U =
X
log U = − log X1 − log X3
and
and Z = X3 .
(7)
log V = log X1 − log X2 − log X3 .
(8)
Clearly, in view of Eq. (6), log Xi ’s are independent standard normal random variables. Hence, any linear function of log U , log V from Eq. (8) must be distributed as a normal random variable because every linear function of independent normal variables is distributed normally. This implies that (log U , log V ) is distributed as bivariate normal Rao [4, p. 518]. We do not exploit a bivariate normal density here. Now, we have Cov(log U , log V ) = 0 so that log U , log V must be independently distributed Rao [4, p. 520]. Since U = exp (log U ) and V = exp (log V ), clearly U , V are independent. But, how do we show that X , Y from Eq. (7) are dependent random variables? Observe that log X = − log X1 , log Y = log X1 − log X2 . Using earlier arguments, one can show that (log X , log Y ) is distributed as bivariate normal, Cov (log U , log V ) = −1(̸= 0), and hence X , Y are dependent Rao [4, p. 520]. 4. Third example: a resolution for the query from Eq. (1) With reference to Eq. (1), we note that in case U and V happen to be independent, |U | =
|X | Z
and
must also be independent. But, let us first consider Cov (|U |, |V |) assuming tacitly that ν ≥ 3. Z Mukhopadhyay [3] emphasized delicate relationships among required finiteness of covariance, zero covariance, and independence.
|V | =
|Y |
4.1. When ν exceeds 2 When ν ≥ 3, Cov (|U | , |V |) is finite, and we may express:
|Y | |X | − E E Z2 Z Z |X | |Y | 1 = E [|X | |Y |] E 2 − E E
Cov (|U | , |V |) = E
|X | |Y |
Z
Z
Z
since (X , Y ) and Z are independent
= E [|X |] E [|Y |] E
1 Z2
− E [|X |] E [|Y |] E
2
1
Z
N. Mukhopadhyay, M.S. Son / Statistical Methodology 14 (2013) 62–66
65
since X , Y are independent
1 1 = E 2 [|X |] E 2 − E 2 Z
Z
> 0,
(9)
since we have: E [|X |] > 0
and
E
1
Z2
− E2
1
Z
= Var
1
Z
> 0 as Z is non-degenerate.
But, the final conclusion from Eq. (9) is contradictory to the assumption that U and V are independent. Hence, U and V must be dependent. 4.2. When ν is either 1 or 2 When ν = 1 or 2, the derivation given in Section 4.1 will not go through because then Cov (|U | , |V |) will not be finite. But, we may continue with the basic approach described in Section 2. Let us denote the density function of Z by g (z ), z > 0, zero otherwise. With arbitrary real numbers a and b, we can express: P ( U ≤ a ∩ V ≤ b) =
∞
P
0
≤ a ∩ ≤ b Z = z g (z )dz Z Z
X
Y
∞
=
P (X ≤ az ∩ Y ≤ bz ) g (z )dz
since (X , Y ), Z are independent
P (X ≤ az )P (Y ≤ bz )g (z )dz
since X , Y independent
0
∞
= 0
= E [Φ (aZ )Φ (bZ )] ,
(10)
where Φ (.) is the customary distribution function of a standard normal random variable. With similar arguments, we will have: P (U ≤ a) = E [Φ (aZ )]
and
P (V ≤ b) = E [Φ (bZ )] .
(11)
Now, U and V will be independent if and only if for all a, b ∈ R, the following holds: P (U ≤ a ∩ V ≤ b) = P (U ≤ a)P (V ≤ b),
that is,
E [Φ (aZ )Φ (bZ )] = E [Φ (aZ )] E [Φ (bZ )] .
(12)
We will show that Eq. (12) cannot hold for all a, b ∈ R. Let us fix a = b = 1. Then, the expression on the left-hand side of Eq. (12) becomes E Φ 2 (Z ) whereas the right-hand side of Eq. (12) becomes E 2 [Φ (Z )]. Now, E Φ 2 (Z ) − E 2 [Φ (Z )] ≡ Var [Φ (Z )] which must be positive because Φ (Z ) is a nondegenerate random variable. We note that Var [Φ (Z )] is finite since Φ (Z ) is bounded w.p.1. This means that P (U ≤ 1 ∩ V ≤ 1) > P (U ≤ 1)P (V ≤ 1). So, Eq. (12) cannot hold for all a, b ∈ R. Hence, U , V are dependent random variables. Remark 4.1. The approach in Section 4.2 is slightly different from that in Section 4.1 and it proves the desired conclusion for all ν ≥ 1. Arguably, however, our approach in Eq. (9) appears less complex and hence it may be accessible to more readers under the assumption that ν ≥ 3. 5. A general approach The core approach that we have exploited before may be summarized as follows: Two vectorvalued random variables S and T are independent implies that the random variables g (S) and h(T) are independent for every (measurable) functions g (.), h(.). Here, g (.), h(.) may be allowed to be vectorvalued, but for simplicity we treat them as real-valued. So, we summarize a result with important
66
N. Mukhopadhyay, M.S. Son / Statistical Methodology 14 (2013) 62–66
implications. For brevity, we omit its proof because the result is after all well-known. The point of this section is to emphasize the usefulness of this result as a tool in the context of our inquiry. The beginners will appreciate our illustrations. Theorem 5.1. Consider two vector-valued random variables S and T. Suppose that we exhibit two specific (measurable) functions g (.), h(.) such that Cov(g (S), h(T)) is finite and non-zero. Then, S and T must be dependent. We have already used this theorem before. In Section 4.1, if we had found Cov (U , V ) with g (u) = u, h(v) = v , that would indeed be zero when ν ≥ 3. But, a zero covariance does not necessarily lead to a decisive conclusion, and hence we moved to g (u) = |u| , h(v) = |v|. Alternatively,the functions g (u) = u2 , h(v) = v 2 would work fine too, but then one would require ν ≥ 5 for Cov U 2 , V 2 to be finite. A fruitful implementation of Theorem 5.1 requires that one would use rather simple expressions for the two functions g (.) and h(.). After all, one ought to be able to evaluate Cov (g (S), h(T)) both quickly and easily. Next, some new illustrations are given where a direct approach via joint and conditional densities involving S, T is expected to be complicated and less tractable. 5.1. Some additional illustrations Example 5.1. Suppose that X1 , X2 are independent standard normal random variables, S = X1 − X2 and T = X1 X2 . Surely, Cov(S , T ) = 0. This in itself does not prove or disprove that S , T are dependent. Let us consider g (S ) = S 2 , h(T ) = T , argue that Cov(g (S ), h(T )) is finite, and write: Cov(g (S ), h(T )) = E
=E
X12 − 2X1 X2 + X22 X1 X2
X13 X2
X13 X2
=E = −2.
2X12 X22
X23 X1
since E [T ] = 0
− + 2 2 − 2E X1 X2 + E X23 X1
Thus, Theorem 5.1 shows that S , T are dependent. Example 5.2. Suppose that X1 , X2 are independent Poisson (λ) random variables with λ > 0, S = 1/2 X1 − X2 and T = X1 . Let us consider g (S ) = S , h(T ) = T 2 , argue that Cov(g (S ), h(T )) is finite, and write: Cov(g (S ), h(T )) = E [(X1 − X2 ) X1 ]
since E [(X1 − X2 )] = 0
= (λ2 + λ) − λ2 =λ > 0. Thus, Theorem 5.1 shows that S , T are dependent. References [1] N. Johnson, S. Kotz, Distributions in Statistics: Continuous Multivariate Distributions, Wiley, New York, 1972. [2] N. Mukhopadhyay, Probability and Statistical Inference, Dekker, New York, 2000. [3] N. Mukhopadhyay, When finiteness matters: counterexamples to notions of covariance, correlation, and independence, The American Statistician 64 (2010) 231–233. [4] C.R. Rao, Linear Statistical Inference and its Applications, second ed., Wiley, New York, 1973. [5] Y.L. Tong, The Multivariate Normal Distributions, Springer, New York, 1990.
Contents lists available at SciVerse ScienceDirect
Statistical Methodology journal homepage: www.elsevier.com/locate/stamet
Ratios X /Z , Y /Z built from independent random variables (X , Y ) and Z may not always be dependent Nitis Mukhopadhyay a,∗ , Mun S. Son b a
Department of Statistics, University of Connecticut, Storrs, CT 06269-4120, USA
b
Department of Mathematics & Statistics, University of Vermont, Burlington, VT 05401-1455, USA
article
info
Article history: Received 26 October 2012 Accepted 26 February 2013 Keywords: Bivariate t distribution Continuous distribution Discrete distribution Lognormal distribution Multivariate normal Normal distribution Symmetric distribution
abstract It is a commonly held ‘‘belief’’ in many quarters that the ratios U = XZ , V = YZ are necessarily dependent random variables when the random vector (X , Y ) is independent of the random variable Z simply because both U , V involve Z . Any outpouring support behind such ‘‘belief’’ often gets louder when (X , Y ) are assumed dependent. The purpose of this note is to emphasize that such ‘‘beliefs’’ may be false. Concrete examples are given when (i) X , Y are independent but U , V may be dependent or independent, (ii) X , Y are dependent but U , V may be dependent or independent. Finally, a simple general approach is given for beginners without exploiting joint and/or conditional densities. © 2013 Elsevier B.V. All rights reserved.
1. Introduction Suppose that we have three independent random variables X , Y and Z with Z > 0 w.p.1. We may wish to address the following problem: Let X ∼ N (0, 1), Y ∼ N (0, 1) and ν Z 2 ∼ χν2 and define ratios X Y U = , V = . Prove whether or not (U , V ) are dependent Z Z without exploiting their joint and conditional densities.
∗
Corresponding author. E-mail addresses: [email protected] (N. Mukhopadhyay), [email protected] (M.S. Son).
1572-3127/$ – see front matter © 2013 Elsevier B.V. All rights reserved. http://dx.doi.org/10.1016/j.stamet.2013.02.002
(1)
N. Mukhopadhyay, M.S. Son / Statistical Methodology 14 (2013) 62–66
63
As a quick response, we may be immediately told that (U , V ) has a bivariate t-distribution which is certainly correct. Then, we may also be informed by many responders that U , V from (1.1) are necessarily dependent random variables. At this point, one may genuinely start to wonder about how such a conclusion may follow right away, especially in the absence of an otherwise complicated joint density function of (U , V ) and the conditional densities. We may look up a multivariate t distribution from Johnson and Kotz [1, Chapter 37], Tong [5, Chapter 9], Mukhopadhyay [2, pp. 218–219] or another appropriate source. But, many may admit that they cannot recall from memory a general multivariate t density function or its associated conditional density functions correctly just when we may need them. Personally, we belong to that group. How can one be so sure that U , V from Eq. (1) are dependent random variables? One may often hear an emphatic response such as: It is obvious. Both U , V involve Z , and so U , V must be dependent. Unfortunately, this line of argument is not satisfactory and one may continue to ask whether a solid justification can be constructed to prove dependence of U , V from Eq. (1) without explicitly referring to their joint and conditional density functions. This short note arose from such a heightened curiosity, especially for beginners. However, before we resolve the query stated in Eq. (1), we feel tempted to exploit this opportunity to raise a broader question: Suppose that we have three independent random variables X , Y and Z with P (Z = 0) = 0 and define U = XZ and V = YZ . Again, many may be tempted to argue that since U and V both involve Z , U , V must be dependent. But, then, obviously we will be back exactly where we began. Admittedly U , V are frequently dependent, but they may not always be so in every circumstance! Moreover, one must come up with a logical proof of dependence or independence as a case may warrant. Hence, we give explicit examples where U , V are decisively independent. One will find airtight proofs of independence without explicitly writing down the joint and conditional densities of U , V . In Example 1 from Section 2, we have X , Y independent, (X , Y ) independent of Z , and yet U , V are independent. In Example 2 from Section 3, we have X , Y dependent, (X , Y ) independent of Z , and yet U , V are independent. Then, we supply a simple resolution of the opening problem from Eq. (1) in Section 4 with the help of an interesting approach involving moments. Section 5 briefly touches upon a generalization of the basic idea that we are trying to get across. It is our earnest hope that inquisitive readers may find the experience of constructing pretty examples of U and V with logically sound but simple approaches to verify their dependence or independence interesting. 2. First example: X , Y are independent but U , V are independent Let X , Y , and Z be independent random variables, X ∼ N (0, 1), Y ∼ N (0, 1), whereas P (Z = −1) = P (Z = 1) = 12 . In what follows, we prove that U = XZ , V = YZ are independent. With arbitrary real numbers a and b, we can express: P ( U ≤ a ∩ V ≤ b) =
=
1 2 1 2
P
X Z
≤a∩
Y X 1 ≤ b Z = −1 + P ≤ a ∩ ≤ b Z = 1 Z 2 Z Z
Y
P (−X ≤ a ∩ −Y ≤ b) +
1 2
P ( X ≤ a ∩ Y ≤ b) ,
since(X , Y ), Z are assumed independent.
(2)
But, the joint distribution of (X , Y ) is exactly same as that of (−X , −Y ). Thus, we can rewrite Eq. (2) as: P (U ≤ a ∩ V ≤ b) = P (X ≤ a ∩ Y ≤ b) = P (X ≤ a)P (Y ≤ b), since X , Y are assumed independent.
(3)
With analogous arguments, one can claim: P (U ≤ a) = P (X ≤ a) and
P (V ≤ b) = P (Y ≤ b),
since X , Y are assumed independent.
(4)
64
N. Mukhopadhyay, M.S. Son / Statistical Methodology 14 (2013) 62–66
A combination of Eqs. (3) and (4) obviously leads to: P (U ≤ a ∩ V ≤ b) = P (U ≤ a)P (V ≤ b),
for all real numbers a, b.
(5)
Eq. (5) is a restatement of independence of U , V . Remark 2.1. In this example, it is not crucial that both X , Y must be independent N (0.1) random variables. The conclusion from this example continues to hold if X , Y are independent, but their probability distributions (not necessarily same) are both symmetric around zero. Also, X , Y may be continuous or discrete. 3. Second example: X , Y are dependent but U , V are independent Suppose that X1 , X2 , and X3 are independent random variables, each having the following lognormal probability density:
1 f ( x) = √ exp −(log x)2 /2 , x 2π
0 < x < ∞, and 0 otherwise.
(6)
We define Y , V = where X = X1−1 , Y = X1 X2−1 Z Z Obviously, we can rewrite: U =
X
log U = − log X1 − log X3
and
and Z = X3 .
(7)
log V = log X1 − log X2 − log X3 .
(8)
Clearly, in view of Eq. (6), log Xi ’s are independent standard normal random variables. Hence, any linear function of log U , log V from Eq. (8) must be distributed as a normal random variable because every linear function of independent normal variables is distributed normally. This implies that (log U , log V ) is distributed as bivariate normal Rao [4, p. 518]. We do not exploit a bivariate normal density here. Now, we have Cov(log U , log V ) = 0 so that log U , log V must be independently distributed Rao [4, p. 520]. Since U = exp (log U ) and V = exp (log V ), clearly U , V are independent. But, how do we show that X , Y from Eq. (7) are dependent random variables? Observe that log X = − log X1 , log Y = log X1 − log X2 . Using earlier arguments, one can show that (log X , log Y ) is distributed as bivariate normal, Cov (log U , log V ) = −1(̸= 0), and hence X , Y are dependent Rao [4, p. 520]. 4. Third example: a resolution for the query from Eq. (1) With reference to Eq. (1), we note that in case U and V happen to be independent, |U | =
|X | Z
and
must also be independent. But, let us first consider Cov (|U |, |V |) assuming tacitly that ν ≥ 3. Z Mukhopadhyay [3] emphasized delicate relationships among required finiteness of covariance, zero covariance, and independence.
|V | =
|Y |
4.1. When ν exceeds 2 When ν ≥ 3, Cov (|U | , |V |) is finite, and we may express:
|Y | |X | − E E Z2 Z Z |X | |Y | 1 = E [|X | |Y |] E 2 − E E
Cov (|U | , |V |) = E
|X | |Y |
Z
Z
Z
since (X , Y ) and Z are independent
= E [|X |] E [|Y |] E
1 Z2
− E [|X |] E [|Y |] E
2
1
Z
N. Mukhopadhyay, M.S. Son / Statistical Methodology 14 (2013) 62–66
65
since X , Y are independent
1 1 = E 2 [|X |] E 2 − E 2 Z
Z
> 0,
(9)
since we have: E [|X |] > 0
and
E
1
Z2
− E2
1
Z
= Var
1
Z
> 0 as Z is non-degenerate.
But, the final conclusion from Eq. (9) is contradictory to the assumption that U and V are independent. Hence, U and V must be dependent. 4.2. When ν is either 1 or 2 When ν = 1 or 2, the derivation given in Section 4.1 will not go through because then Cov (|U | , |V |) will not be finite. But, we may continue with the basic approach described in Section 2. Let us denote the density function of Z by g (z ), z > 0, zero otherwise. With arbitrary real numbers a and b, we can express: P ( U ≤ a ∩ V ≤ b) =
∞
P
0
≤ a ∩ ≤ b Z = z g (z )dz Z Z
X
Y
∞
=
P (X ≤ az ∩ Y ≤ bz ) g (z )dz
since (X , Y ), Z are independent
P (X ≤ az )P (Y ≤ bz )g (z )dz
since X , Y independent
0
∞
= 0
= E [Φ (aZ )Φ (bZ )] ,
(10)
where Φ (.) is the customary distribution function of a standard normal random variable. With similar arguments, we will have: P (U ≤ a) = E [Φ (aZ )]
and
P (V ≤ b) = E [Φ (bZ )] .
(11)
Now, U and V will be independent if and only if for all a, b ∈ R, the following holds: P (U ≤ a ∩ V ≤ b) = P (U ≤ a)P (V ≤ b),
that is,
E [Φ (aZ )Φ (bZ )] = E [Φ (aZ )] E [Φ (bZ )] .
(12)
We will show that Eq. (12) cannot hold for all a, b ∈ R. Let us fix a = b = 1. Then, the expression on the left-hand side of Eq. (12) becomes E Φ 2 (Z ) whereas the right-hand side of Eq. (12) becomes E 2 [Φ (Z )]. Now, E Φ 2 (Z ) − E 2 [Φ (Z )] ≡ Var [Φ (Z )] which must be positive because Φ (Z ) is a nondegenerate random variable. We note that Var [Φ (Z )] is finite since Φ (Z ) is bounded w.p.1. This means that P (U ≤ 1 ∩ V ≤ 1) > P (U ≤ 1)P (V ≤ 1). So, Eq. (12) cannot hold for all a, b ∈ R. Hence, U , V are dependent random variables. Remark 4.1. The approach in Section 4.2 is slightly different from that in Section 4.1 and it proves the desired conclusion for all ν ≥ 1. Arguably, however, our approach in Eq. (9) appears less complex and hence it may be accessible to more readers under the assumption that ν ≥ 3. 5. A general approach The core approach that we have exploited before may be summarized as follows: Two vectorvalued random variables S and T are independent implies that the random variables g (S) and h(T) are independent for every (measurable) functions g (.), h(.). Here, g (.), h(.) may be allowed to be vectorvalued, but for simplicity we treat them as real-valued. So, we summarize a result with important
66
N. Mukhopadhyay, M.S. Son / Statistical Methodology 14 (2013) 62–66
implications. For brevity, we omit its proof because the result is after all well-known. The point of this section is to emphasize the usefulness of this result as a tool in the context of our inquiry. The beginners will appreciate our illustrations. Theorem 5.1. Consider two vector-valued random variables S and T. Suppose that we exhibit two specific (measurable) functions g (.), h(.) such that Cov(g (S), h(T)) is finite and non-zero. Then, S and T must be dependent. We have already used this theorem before. In Section 4.1, if we had found Cov (U , V ) with g (u) = u, h(v) = v , that would indeed be zero when ν ≥ 3. But, a zero covariance does not necessarily lead to a decisive conclusion, and hence we moved to g (u) = |u| , h(v) = |v|. Alternatively,the functions g (u) = u2 , h(v) = v 2 would work fine too, but then one would require ν ≥ 5 for Cov U 2 , V 2 to be finite. A fruitful implementation of Theorem 5.1 requires that one would use rather simple expressions for the two functions g (.) and h(.). After all, one ought to be able to evaluate Cov (g (S), h(T)) both quickly and easily. Next, some new illustrations are given where a direct approach via joint and conditional densities involving S, T is expected to be complicated and less tractable. 5.1. Some additional illustrations Example 5.1. Suppose that X1 , X2 are independent standard normal random variables, S = X1 − X2 and T = X1 X2 . Surely, Cov(S , T ) = 0. This in itself does not prove or disprove that S , T are dependent. Let us consider g (S ) = S 2 , h(T ) = T , argue that Cov(g (S ), h(T )) is finite, and write: Cov(g (S ), h(T )) = E
=E
X12 − 2X1 X2 + X22 X1 X2
X13 X2
X13 X2
=E = −2.
2X12 X22
X23 X1
since E [T ] = 0
− + 2 2 − 2E X1 X2 + E X23 X1
Thus, Theorem 5.1 shows that S , T are dependent. Example 5.2. Suppose that X1 , X2 are independent Poisson (λ) random variables with λ > 0, S = 1/2 X1 − X2 and T = X1 . Let us consider g (S ) = S , h(T ) = T 2 , argue that Cov(g (S ), h(T )) is finite, and write: Cov(g (S ), h(T )) = E [(X1 − X2 ) X1 ]
since E [(X1 − X2 )] = 0
= (λ2 + λ) − λ2 =λ > 0. Thus, Theorem 5.1 shows that S , T are dependent. References [1] N. Johnson, S. Kotz, Distributions in Statistics: Continuous Multivariate Distributions, Wiley, New York, 1972. [2] N. Mukhopadhyay, Probability and Statistical Inference, Dekker, New York, 2000. [3] N. Mukhopadhyay, When finiteness matters: counterexamples to notions of covariance, correlation, and independence, The American Statistician 64 (2010) 231–233. [4] C.R. Rao, Linear Statistical Inference and its Applications, second ed., Wiley, New York, 1973. [5] Y.L. Tong, The Multivariate Normal Distributions, Springer, New York, 1990.