Prevalencia ILTB en funci´ on

Proof of Lemma 3. Consider (i). Assume Y ^(h)9 X|I^XZ for some h ≥ 2.

Recall the partition Z = (U⁰, V⁰)⁰. From the argument following (6), we know Y ^(h+1)9 X|IXZ if and only if π^(h)_XZ,1π_ZY,i = 0, ∀i ≥ 1, and Y 9 V |I¹ XZ implies Y ^(h+1)9 X|IXZ if and only if π^(h)_XU,1π_{U Y,i}= 0, ∀i ≥ 1. Moreover, because U is univariate and Y → U|I^{1 XZ}, we deduce π^(h)_XU,1πU Y,i= 0 occurs for every i ≥ 1 if and only if π^(h)_XU,1 = 0. This is true irrespective of the dimensions of X and Y . Next, consider (ii), and recall we assume Y ^(h)9 X|IXZ. From part (i), Y

(h+1)

9 X|I^XZ if and only if π^(h)_XU,1 = 0, and from (3), we deduce

π^(h)_XU,1= π^(h_XU,2⁻¹⁾+π^(h_XX,1⁻¹⁾π_XU,1+π^(h_XY,1⁻¹⁾π_{Y U,1}+π^(h_XU,1⁻¹⁾π_{U U,1}+π^(h_XV,1⁻¹⁾π_{V U,1}. (15) For h = 2, Y ⁽²⁾9 X|IXZ implies π⁽¹⁾_XU,1 = 0 from part (i), above, hence

(16) π⁽²⁾_XU,1 = πXU,2+ πXX,1πXU,1+ πXY,1πY U,1+ πXU,1πU U,1+ πXV,1πV U,1

= πXU,2+ πXX,1× 0 + 0 × π^{Y U,1}+ 0 × π^{U U,1}+ +πXV,1πV U,1

= π_XU,2+ π_XV,1π_{V U,1}.

For h = 3, Y ⁽³⁾9 X|IXZ implies π⁽²⁾_XU,1 = π⁽¹⁾_XU,1 = 0 from part (i), above, hence (17) π⁽³⁾_XU,1 = π⁽²⁾_XU,2+ π⁽²⁾_XX,1πXU,1+ π⁽²⁾_XY,1πY U,1+ π⁽²⁾_XU,1πU U,1+ π⁽²⁾_XV,1πV U,1

= π⁽²⁾_XU,2+ π⁽²⁾_XX,1× 0 + 0 × πY U,1+ 0 × πU U,1+ π⁽²⁾_XV,1π_{V U,1}

= π⁽²⁾_XU,2+ π⁽²⁾_XV,1π_{V U,1} where

(18) π⁽²⁾_XU,2 = πXU,3+ πXX,1πXU,2+ πXY,1πY U,2+ πXU,1πU U,2+ πXV,1πV U,2

= π_XU,3+ π_XX,1π_XU,2+ 0 × πY U,2+ 0 × πU U,2+ π_XV,1π_{V U,2}

= π_XU,3+ π_XX,1π_XU,2+ π_XV,1π_{V U,2}. Thus,

π⁽³⁾_XU,1 = π⁽²⁾_XU,2+ π⁽²⁾_XV,1π_{V U,1} (19)

= π_XU,3+ π_XX,1π_XU,2+ π_XV,1π_{V U,2}+ π⁽²⁾_XV,1π_{V U,1}. Recursively we deduce

π^(h)_XU,1= πXU,h+X^h−1

i=1 π^(h_XX,1⁻ⁱ⁾πXU,i+X^h−1

i=1 π^(h_XV,1⁻ⁱ⁾πV U,i, (20)

where we include the term π^(h_XX,1⁻¹⁾π_XU,1 = 0.

Finally, consider (iii). From part (i), Y ^(h)9 X|IXZif and only if π^(h)_XU,1, and from part (ii), we have the identity

π^(h)_XU,1= πXU,h+Xh−1

i=1 π^(h_XX,1⁻ⁱ⁾πXU,i+Xh−1

i=1 π^(h_XV,1⁻ⁱ⁾πV U,i. (21) Now, let π_XU,i = π_{V U,i} = 0, i = 1...h − 1. Then

π^(h)_XU,1= π_XU,h+Xh−1

i=1 π^(h_XX,1⁻ⁱ⁾× 0 +Xh−1

i=1 π^(h_XV,1⁻ⁱ⁾× 0 = πXU,h. (22) This completes the proof.

Proof of Theorem 4. For (i), assume m_z = m_u = 1 (i.e. U = Z, Y →¹ Z|IXZ), such that m_v = 0 by convention.

Consider h = 2 and assume Y 9 X|I¹ XZ. By Lemma 3.i, we have Y 9² X|I^XZ if and only if π⁽¹⁾_XZ,1 = πXZ,1 = 0. This proves the result for h = 2..

Now, let Y ^(h)9 X|I^XZfor any h ≥ 2. From Lemma 3.ii, it follows that π^(k)XZ,1

has the representation π⁽²⁾_XZ,1 = π_XZ,2, and for 2 < k ≤ h π^(k)_XZ,1= πXZ,k+Xk−1

i=1

³

π^(k_XX,1⁻ⁱ⁾πXZ,i

´

. (23)

We will prove the result, Y ^(h+1)9 X|I^XZ if and only if πXZ,h = 0, by induction.

Assume Y ^(h)9 X|I^XZ if and only if πXZ,h−1 = 0 for some h ≥ 2. Then, notice that Y ^(h)9 X|I^XZimplies Y ^(k)9 X|I^XZ for each k = 1...h, thus, by the induction assumption we deduce Y ^(h)9 X|I^XZ if and only if πXZ,k = 0, k = 1...h − 1.

Consequently, by the induction assumption the equality in (23) reduces to π^(h)_XZ,1 = π_XZ,h+X^h−1

i=1

³π^(h_XX,1⁻ⁱ⁾π_XZ,i´

= π_XZ,h. (24)

By Lemma 3, Y ^(h+1)9 X|I^XZ if and only if π^(h)_XZ,1 = 0, and π^(h)_XZ,1 = πXZ,h, therefore Y ^(h+1)9 X|I^XZ if and only if πXZ,h= 0. Because h ≥ 2 is arbitrary, this proves claim (i) by induction.

Consider (ii), and assume m_z > 1 and 1 < m_u ≤ mz for the remainder of the proof. From (6) and (7), we know Y ⁽²⁾9 X|I^XZ if and only if πXU,1πU Y,j

= 0, j ≥ 1. By the assumption Y → U, we have π^{U Y,j} 6= 0 for some j, hence πXU,1πU Y,j = 0 if and only if πXU,1 = 0.

For (iii), let Y ⁽³⁾9 X|I^XZ and U 9 V |I^{1 XZ}. By Lemma 3, we deduce Y ⁽³⁾9 X|IXZ if and only if

π⁽²⁾_XU,1= π_XU,2+ π_XX,1π_XU,1+ π_XV,1π_{V U,1}= 0. (25)

If U 9 V |I^{1 XZ}, then πV U,i = 0, i ≥ 1, thusY ⁽³⁾9 X|I^XZ if and only if

π⁽²⁾_XU,1= πXU,2+ πXX,1πXU,1= 0. (26) Now, we know Y ⁽²⁾9 X|I^XZ if and only if πXU,1 = 0, cf. part (ii), hence Y ⁽³⁾9 X|I^XZ if and only if

0 = π⁽²⁾_XU,1= πXU,2+X¹

i=1π^(h_XX,1⁻ⁱ⁾πXU,i (27)

= πXU,2+X1

i=1π^(h_XX,1⁻ⁱ⁾× 0 = π^XU,2.

Next, consider (iv) and let Y ⁽²⁾9 X|I^XZ and V 9 (U, X)|I^{1 XY U}. By Lemma 3, we know Y ⁽³⁾9 X|IXZ if and only if

π⁽²⁾_XU,1= π_XU,2+ π_XX,1π_XU,1+ π_XV,1π_{V U,1}= 0, (28) and by Theorem 2, V 9 (U, X)|I^{1 XY U} implies πU V,i = πXV,i = 0, for each i ≥ 1. Because Y ⁽²⁾9 X|I^XZ, by part (i), πXU,1= 0, and Y ⁽³⁾9 X|I^XZif and only if

0 = π⁽²⁾_XU,1= πXU,2+ πXX,1πXU,1+ πXV,1πV U,1 (29)

= π_XU,2+ π_XX,1× 0 + 0 × πV U,1= π_XU,2.

Finally, for (v), assume π_XU,i= π_{V U,j}= 0, i = 1...2, j = 1. Because π_XU,1= 0, from part (ii) we immediately deduceY ⁽²⁾9 X|I^XZ. From Lemma 3, therefore, Y ⁽³⁾9 X|IXZ if and only if

π⁽²⁾_XU,1= π_XU,2+ π_XX,1π_XU,1+ π_XV,1π_{V U,1}= 0, (30) where πXU,i = πV U,j = 0, i = 1...2, j = 1,, hence

π⁽²⁾_XU,1 = π_XU,2+ π_XX,1π_XU,1+ π_XV,1π_{V U,1} (31)

= 0 + πXX,1× 0 + π^XV,1× 0 = 0.

Therefore, Y ⁽³⁾9 X|I^XZ follows.

Proof of Corollary 5. The only claims not immediately implied by Lemma 3 and (8) are (iii) and (iv).

For (iii), we assume Y ⁽²⁾9 X|I^XZ and πXU,2 = πXV,1 = 0. From (8) this implies π⁽²⁾_XU,1 = 0, hence Y ⁽³⁾9 X|I^XZ, cf. Lemma 3.i. Using Lemma 3.ii, the assumption πXU,2= πXV,1= 0, and πXU,1 = 0 due to Y ⁽²⁾9 X|I^XZ, cf. Lemma

3.i, we then deduce π⁽³⁾_XU,1 = πXU,3+X2

i=1π⁽³_XX,1⁻ⁱ⁾πXU,i+X2

i=1π⁽³_XV,1⁻ⁱ⁾πV U,i (32)

= π_XU,3+ π⁽²⁾_XX,1π_XU,1+ π_XX,1π_XU,2+ π⁽²⁾_XV,1π_{V U,1}+ +π_XV,1π_{V U,2}

= π_XU,3+ π⁽²⁾_XX,1× 0 + πXX,1× 0 + π⁽²⁾XV,1π_{V U,1}+ +0 × πV U,2

= π_XU,3+ π⁽²⁾_XV,1π_{V U,1}.

Using (5) and Y 9 X|I^{1 XZ} if and only if πXY,i = 0, i ≥ 1, we have

(33) π⁽²⁾_XV,1 = πXV,2+ πXX,1πXV,1+ πXY,1πY V,1+ πXU,1πU V,1+ πXV,1πV V,1

= π_XV,2+ π_XX,1× 0 + 0 × πY V,1+ 0 × πU V,1+ 0 × πV V,1

= πXV,2,

hence π⁽³⁾_XU,1= πXU,3+ πXV,2πV U,1. From Lemma 3.i we conclude Y ⁽⁴⁾9 X|I^XZ if and only if πXU,3 + πXV,2πV U,1 = 0.

For (iv), we assume Y ⁽²⁾9 X|IXZ and π_XU,2 = π_{V U,1} = 0, hence π⁽³⁾_XU,1 reduces to

(34) π⁽³⁾_XU,1 = π_XU,3+ π⁽²⁾_XX,1π_XU,1+ π_XX,1π_XU,2+ π⁽²⁾_XV,1π_{V U,1}+ π_XV,1π_{V U,2}

= πXU,3+ π⁽²⁾_XX,1× 0 + π^XX,1× 0 + π⁽²⁾XV,1× 0 + π^XV,1πV U,2

= πXU,3+ πXV,1πV U,2. The claim then follows from Lemma 3.i.

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