On the relation between the true and sample correlations under Bayesian modelling of gene expression datasets

Proof. (Theorem 3.1)

Due to symmetry considerations, it is enough to prove that

To this end, assume by contradiction that there exists some c ∈ (−1, 1) such that P{ρx1x2=c}=1. Since G(⋅) is a probability distribution over Θ and it is known that almost surely Fθ(⋅) is associated with finite first two moments, the probability (with respect to G(⋅)) that the correlation matrix of (X₁, X₂, Y) is positive semi-definite, equals to one. To obtain a contradiction, it is shown that with positive probability, the characteristic polynomial of this correlation matrix is associated with negative roots. In details, since P{ρx1x2=c}=1, then almost surely the characteristic polynomial of the correlation matrix of (X₁, X₂, Y) is given by

Now, set ρ₁ = 0 and obtain the following equation:

Clearly, if c∈(−1,1)∖{0} , for ρ2=2−c22∈(12,1) there exists one root of P(λ) which is given by

i.e., there exists a negative solution to the equation

Since Cardano formula [see e.g. Spiegel (1968)] implies that the solutions of the equation P(λ; ρ₁, ρ₂, c) = 0 are continuous in (ρ₁, ρ₂) at the point p:=(0,2−c22), there exists δ > 0 which is associated with a ball Bδ(p)⊆(−1,1)2 such that for any (ρ₁, ρ₂) ∈ B_δ(p) there exists a negative root for P(λ; ρ₁, ρ₂, c). In addition, the fact that ϕ(⋅) is strictly increasing continuous function on (−1,1) and ϕ(ρ1),ϕ(ρ2)∼i.i.dN(0,σq2) implies that (ρ₁, ρ₂) are continuously distributed over (−1, 1)². Therefore, with positive probability, P(λ; ρ₁, ρ₂, c) is associated with negative root implying a contradiction. To complete the proof, if c=0 , then

If we let ρi;i=1,2 be both close enough to 1, then this characteristic polynomial has negative root. Therefore the same arguments which were used for the case where c∈(−1,1)∖{0} may be carried out once again to justify existence of negative root with positive probability.

   □

Lemma 5.1. (Multivariate Delta Method)

Let Ψ be m-dimensional positive definite matrix with a sequence of m-dimensional random vectors {Xn}n=1∞ such that

If f:ℝ^m → ℝ^d is associated with ∇f(⋅) which is continuous in some neighbourhood of μ, then

where ∇f(β~)=[∂fj∂βi]ij(β~) is the partial derivative matrix of f(⋅) at the point β~ ∈ ℝ^m.

Proof.

See chapter 7 of Ferguson (1996).    □

Proof. (Theorem 3.2)

Consider the following notations:

Using these notations, the sample correlations can be written as:

Now, before going forward let us mention that the following analysis is done with respect to the probability space conditioned on θ for which there exist finite four moments of Fθ(⋅). Since it is given that Fθ(⋅) is associated with such finite moments with probability one, the results of the upcoming analysis hold with probability one. Now, with respect to this setup, an application of the multivariate central limit theorem (CLT) justifies the convergence

where the covariance matrix Σ¹ is given by Σij1=C(Zi,Zj),1≤i,j≤3k+2 and Z is as follows:

Define a function η:R3k+2→R2k+1 by

and notice that

1. η[(mx1⋮mxkmymx12⋮mxk2my2mx1y⋮mxky)]=(sx12⋮sxk2sy2sx1y⋮sxky2)
2. ∇η(z)=(−2diag(z1,…,zk+1)BIk+1×k+1Ok+1×k0k×k+1Ik×k)

   where the matrix B is given by

   B:=(−zk+1⋅Ik×k−z1,…,−zk) .

Thus, the multivariate delta method can be used in order to deduce that

where Σ² is given by

Notice that for the vector of inputs written above ∇η is given by

which means that Σ² equals to the down-right 2k+1 × 2k+1 block of Σ¹. Considering this result, define another function γ:R2k+1→Rk as follows

which satifies

1. γ[(sx12⋮sxk2sy2sx1y⋮sxky2)]=(r1⋮rk)
2. ∇γT(v)=[A|B|C]

   where

         A:=−diag(vk+1+12v13vk+1,…,vk+1+k2vk3vk+1)

     B:=−(vk+1+12v1vk+13,…,vk+1+k2vkvk+13)T

   C:=diag(1v1vk+1,…,1v1vk+1) .

Therefore, the multivariate delta method can be used once again to obtain the limit

where Σ³ is given by

Define ϕ∗:(−1,1)k→Rk as follows

and notice that ϕ(⋅) is differentiable in its domain and hence its derivative matrix is given by

If so, we shall do one more execution of the multivariate delta method to derive the convergence

where Σ⁴ is given by

Since ϕ′(⋅) is positive for any possible input and non-correlation is equivalent to independence under Gaussian law, then for any i ≠ j asymptotic independence of ϕ(r_i) and ϕ(r_j) is equivalent to [Σ3]ij=0. To see how the needed result stems from this understanding, for simplicity and w.l.o.g, consider the case where i = 1 and j = 2. In this case r₁ and r₂ are asymptotically independent iff the following equation holds

To conclude, recall that Σ2 is down-right 2k+1 × 2k+1 block of Σ1 and deduce the needed result. □

Proof. (Theorem 3.3)

For simplicity and w.l.o.g. it is enough to show that the event of having rn1 and rn2 which are not asymptotically independent occurs with positive probability. To do so, consider (i,j)=(1,2) , and notice that due to the previous theorem, it is enough to prove that Equation (2) does not hold with positive probability. Now, (X₁, X₂, Y) is a Gaussian and hence, as was shown by Isserlis (1918), each of the covariances appeared in Equation (2) can be expressed as follows

where σx12:=V(X1), σx22:=V(X2), σx1x2:=C(X1,X2), σx1y:=C(X1,Y) and σx2y:=C(X2,Y). By insertion of these expressions into Equation (2), a sufficient and necessary condition for asymptotic independence of rn1 and rn2 is that the following equation holds with probability one

where ρx1x2:=σx1,x2/σx12σx22. The next step is to show that with positive probability, (ρ₁, ρ₂) is such that this equation has no real solution. By showing this, since ρx1x2 is real with probability one, we in fact demonstrate that this equation doesn't hold with positive probability. To see this, since ϕ(x) = 0 iff x = 0, then Assumption 2.1 implies that P{ρi≠0,∀i=1,2}=1. Therefore, Equation (3) is almost surely a quadratic equation w.r.t. ρx1x2 that, depending on the values of ρ₁ and ρ₂, might not have a real solution. Indeed, if (ρ₁, ρ₂) = (0.5, 0.9) ∈ (−1, 1)², then the discriminant of the quadratic equation equals to −0.00855 < 0.

Now, the fact that the discriminant of the quadratic equation is continuous in ρ₁ and ρ₂ at the point (0.5, 0.9) implies that there exists some δ > 0 such that the discriminant is negative for any (ρ₁, ρ₂) ∈ B_δ(0.5, 0.9) ⊂ (−1, 1)². By Assumption 2.1, G(⋅) is a prior such that ϕ(ρ1),…,ϕ(ρk)∼i.i.dN(0,σq2) and hence^[5]

To proceed, since ϕ(⋅) is strictly increasing and continuous, there exists a strictly monotone and continuous inverse ϕ−1(⋅). Therefore,

i.e. with positive probability there is no real solution and hence the needed result is provided.    □

Computation of Figure (1)

Input:n, k, u

Output: Histogram and normal QQ-plot of a random realization of ϕ(rn1),…,ϕ(rnk)

1. For j = 1, …, n:

2. Draw (X1j,…,Xkj)∼Nk(0,I) and set Yj=∑i=1uXij.

3. End for.

4. For i = 1, …, k:

5. Compute the empirical correlation between the vectors (Xi1,…,Xin) and (Y1,…,Yn) and denote it by rni.

6. End for.

7. Return Histogram and normal QQ-plot of ϕ(rn1),…,ϕ(rnk).

Computation of Figure (2)

Input:n, k, u, B

Output: Estimates of the expectation and standard deviation of the proportion of genes that are selected correctly as computed by the straightforward and fast approximated approaches.

1. For t = 1, …, B:

2. For j = 1, …, n:

3. Draw (X1j,…,Xkj)∼Nk(0,I) and set Yj=∑i=1uXij.

4. End for.

5. For i = 1, …, k:

6. Compute the empirical correlation between the vectors (X_i1, …, X_in) and (Y₁, …, Y_n) and denote it by rni(t).

7. Compute dt=1u∑i=1uIi where I_i indicates whether |rni(t)| is one of the u highest values of the vector (|rn1(t)|,…,|rnk(t)|).

8. End for.

9. Compute the empirical variance of rn1(t),…,rnk(t) and denote it by W.

10. Set σ^q=W−1n−3 .

11. Draw ϕ(ρ1),…,ϕ(ρk)∼i.i.dN(0,σ^q2) .

12. Compute the set of indices which are associated with the u highest values of the vector (|ϕ(ρ1)|,…,|ϕ(ρ1)|), denote it by S₁ and Draw z1(t),…,zk(t)∼i.i.dN(0,1n−3).

13. For i = 1, …, k:

14. Compute vi(t)=|rni(t)+zi(t)|.

15. End for.

16. Compute the set of indexes which are associated with the u highest values of the vector (v1(t),…,vk(t)), denote it by S₂ and compute ct=|S1∩S2|u.

17. End for.

18. Compute d¯=1B∑t=1Bdt and c¯=1B∑t=1Bct

19. Return d¯, c¯, 1B∑t=1B(ct−c¯)2 and 1B∑t=1B(dt−d¯)2