On quantitative trait locus mapping with an interference phenomenon

Abstract

We consider the likelihood ratio test (LRT) process related to the test of the absence of QTL (a QTL denotes a gene with quantitative effect on a trait) on the interval [0, T] representing a chromosome. The observation is the trait and the composition of the genome at some locations called “markers”. We focus on the interference phenomenon, i.e. a recombination event inhibits the formation of another recombination event nearby. We give the asymptotic distribution of the LRT process under the null hypothesis that there is no QTL on [0, T] and under local alternatives with a QTL at $t^{\star}$ on [0, T]. We show that the LRT process is asymptotically the square of a “linear interpolated and normalized process”. We prove that under the null hypothesis, the distribution of the maximum of the LRT process is the same for a model with or without interference. However, the powers of detection are totally different between the two models.

Appendix: Proofs of theoretical results

1.1 Proof of Theorem 1

As mentioned before, we consider values of t and $t^{\star},$ distinct of marker locations and the result can be prolonged by continuity on markers.

1.1.1 Study of the score process under the null hypothesis

The study is based on the key lemma:

Lemma 2

$$ u(t)=\alpha(t) X(t_1) + \beta(t) X(t_2) $$

with $\alpha(t)= \frac{t_{2}-t}{t_{2}-t_{1}}$ and $\beta(t)= \frac{t-t_{1}}{t_{2}-t_{1}}.$

To prove this lemma, use formula (5) and check that both coincide whatever the value of X(t ₁),  X(t ₂) is. Now using formula (7), we have

$$ \frac{\partial l^n_t}{ \partial q}\mid_{\theta_{0}} = \sum_{j=1}^n \frac{Y_j -\mu }{ \sigma^2 } u_j(t) = 1/\sigma \sum_{j=1}^n \varepsilon_j u_j(t) = \frac{\alpha(t)}{\sigma} \sum_{j=1}^n \varepsilon_j X_j(t_1) +\frac{ \beta(t) }{\sigma} \sum_{j=1}^n \varepsilon_j X_j(t_2) $$

(10)

this proves the interpolation. On the other hand

$$ S_n(t_k) = \sum_{j=1} ^n \frac{ \varepsilon_j X_j(t_k) }{\sqrt{n}} \quad k=1,2 $$

and a direct application of central limit theorem implies that these two variables have a limit distribution which is Gaussian centered distribution with variance

$$ \left( \begin{array}{ll} 1 & \exp\,( -2|t_2-t_1|) \\ \exp\,( -2|t_2-t_1|) & 1\\ \end{array} \right). $$

This proves the expression of the covariance. The weak convergence of the score process, S _n (.), is then a direct consequence of (10), the convergence of (S _n (t ₁),S _n (t ₂)) and the continuous mapping theorem.

1.1.2 Study of the score process under the local alternative

Under the alternative

$$ S_n(t) = \frac{a}{n \sigma} \sum_{j=1} ^n \frac{ U_j(t^*) u_j(t)}{\sqrt{\hbox{Var}\left\{u(t)\right\}}} + \frac{1}{\sqrt{n}} \sum_{j=1} ^n \varepsilon_j \frac{ u_j(t)}{\sqrt{\hbox{Var}\left\{u(t)\right\}}}. $$

The second term has the same distribution as under the null hypothesis and the first one gives the expectation. We have

$$ {\mathbb{E}} \left\{S_n(t)\right\} = \frac{a\,{\mathbb{E}}\left\{ U(t^*) u(t) \right\}}{\sigma\,\sqrt{\hbox{Var}\left\{u(t)\right\}}}. $$

According to Lemma 2, we have:

$$ {\mathbb{E}} \left\{U(t^*)u(t)\right\}= \alpha(t)\,{\mathbb{E}}\left\{X(t_{1})U(t^*)\right\}\,+\,\beta(t)\,{\mathbb{E}}\left\{U(t^*)X(t_{2})\right\}. $$

So, we need now to calculate ${\mathbb{E}\left\{X(t_{1})U(t^*)\right\}}$ and ${\mathbb{E}\left\{U(t^*)X(t_{2})\right\}. }$ We have

$$ \begin{aligned} {\mathbb{P}}\left\{X(t_{1})U(t^{\star})=-1\right\}&= {\mathbb{P}}\left\{U(t^{\star})=1\mid X(t_{1})=-1,X(t_{2})=1\right\}{\mathbb{P}}\left\{X(t_{1})=-1,X(t_{2})=1\right\}\\ &+ {\mathbb{P}}\left\{U(t^{\star})=1\mid X(t_{1})=-1,X(t_{2})=-1\right\}{\mathbb{P}}\left\{X(t_{1})=-1,X(t_{2})=-1\right\}\\ &+ {\mathbb{P}}\left\{U(t^{\star})=-1\mid X(t_{1})=1,X(t_{2})=1\right\}{\mathbb{P}}\left\{X(t_{1})=1,X(t_{2})=1\right\}\\ &+ {\mathbb{P}}\left\{U(t^{\star})=-1\mid X(t_{1})=1,X(t_{2})=-1\right\}{\mathbb{P}}\left\{X(t_{1})=1,X(t_{2})=-1\right\}\\ &= \frac{\beta(t^{\star})r(t_1,t_2)}{2} + 0 + 0 + \frac{\beta(t^{\star})r(t_1,t_2)}{2} = \beta(t^{\star})r(t_1,t_2). \end{aligned} $$

As a consequence,

$$ {\mathbb{P}}\left\{X(t_{1})U(t^{\star})=1\right\}= 1-\beta(t^{\star})r(t_1,t_2). $$

As a result,

$$ {\mathbb{E}}\left\{X(t_{1})U(t^{\star})\right\}= 1-2\beta(t^{\star})r(t_1,t_2)= \alpha(t^{\star})+\beta(t^{\star}) \rho(t_1,t_2)\quad \hbox {with}\;\rho(t_1,t_2)= e^{-2\mid t_1-t_2 \mid}. $$

In the same way, we obtain

$$ {\mathbb{E}}\left\{U(t^{\star})X(t_{2})\right\}= \alpha(t^{\star})\rho(t_1,t_2)+\beta(t^{\star}) . $$

This gives the result.

1.1.3 About the LRT process

Since the model with t fixed is regular, it is easy to prove that for fixed t

$$ \Uplambda_n (t) = S_n^2(t) + o_{P} (1) $$

(11)

under the null hypothesis.

Let us consider a local alternative defined by t* and $q = a/\sqrt{n}.$ The model with t* fixed is differentiable in quadratic mean, this implies that the alternative defines a contiguous sequence of alternatives. By Le Cam’s first Lemma, relation (11) remains true under the alternative. This gives the result for the convergence of finite-dimensional distribution. Concerning the study of the supremum of the LRT process, the proof is exactly the same as in Azaïs et al. (2012) which is based on results of Azaïs et al. (2006), (2009) and Gassiat (2002). □

1.2 Proof of Theorem 2

We recall that we consider values t or $t^{\star}$ of the parameters that are distinct of the markers positions, and the result will be prolonged by continuity at the markers positions.

The proof of the theorem is the same as the proof of Theorem 1 as soon as we can limit our attention to the interval (t ^ℓ, t ^r) when considering a unique instant t. So, under H ₀, the result is straightforward. However, under the local alternative, the proof is more complicated than the proof of Theorem 1. Indeed, the location $t^{\star}$ of the QTL and the location t, can belong to a different marker interval.

According to the proof of Theorem 1, under the alternative

$$ S_n(t) = \frac{a}{n \sigma} \sum_{j=1} ^n \frac{ U_j(t^*) u_j(t)}{\sqrt{\hbox{Var}\left\{u(t)\right\}}} + \frac{1}{\sqrt{n}} \sum_{j=1} ^n \varepsilon_j \frac{ u_j(t)}{\sqrt{\hbox{Var}\left\{u(t)\right\}}}. $$

As previously, the second term has the same distribution as under the null hypothesis and the first one gives the expectation. We have

$$ {\mathbb{E}} \left\{S_n(t)\right\} = \frac{a\,{\mathbb{E}}\left\{ U(t^*) u(t) \right\}}{\sigma\,\sqrt{\hbox{Var}\left\{u(t)\right\}}}. $$

We notice that we have ${u(t^{\star})=\mathbb{E}\left\{U(t^{\star})\mid X(t^{\star \ell})X(t^{\star r})\right\}. }$ Besides, u(t) is a function of X(t ^ℓ) and X(t ^r). As a consequence, by the properties of conditional expectancy, we have

$$ {\mathbb{E}}\left\{ U(t^*) u(t) \right\}= {\mathbb{E}}\left\{ u(t^*) u(t) \right\}. $$

According to Lemma 2,

$$ \begin{aligned} {\mathbb{E}}\left\{ u(t^*) u(t) \right\}&= \alpha(t^{\star})\alpha(t) {\mathbb{E}}\left\{X(t^{\star\ell})X(t^{\ell})\right\} +\beta(t^{\star})\alpha(t) {\mathbb{E}}\left\{X(t^{\star r})X(t^{\ell})\right\}\\ &+ \alpha(t^{\star})\beta(t) {\mathbb{E}}\left\{X(t^{\star\ell})X(t^{r})\right\} +\beta(t^{\star})\beta(t){\mathbb{E}}\left\{X(t^{\star r})X(t^{r})\right\}\\ &= \alpha(t^{\star})\alpha(t)\rho(t^{\ell},t^{\star \ell}) +\beta(t^{\star})\alpha(t) \rho(t^{\ell},t^{\star r})\\ &+ \alpha(t^{\star})\beta(t)\rho(t^{\star \ell},t^{r}) +\beta(t^{\star})\beta(t)\rho(t^{r},t^{\star r}). \end{aligned} $$

In order to obtain ${\mathbb{E}\left\{ u(t^*) u(t^{\ell}) \right\}, }$ we just have to use the dominated convergence theorem. As a result

$$ {\mathbb{E}}\left\{ u(t^*) u(t^{\ell})\right\}= \alpha(t^{\star})\rho(t^{\ell},t^{\star \ell})+\beta(t^{\star})\rho(t^{\ell},t^{\star r}) . $$

To conclude the proof, we just have to notice that

$$ \begin{aligned} {\mathbb{E}}\left\{ u(t^*) u(t^{\ell})\right\}&= \rho(t^{\ell},t^{\star \ell}) \left\{\alpha(t^{\star}) + \beta(t^{\star}) \rho(t^{\star \ell},t^{\star r})\right\}\quad\hbox {if} \; t^\star>t^{\ell}\\ &= \rho(t^{\ell},t^{\star r}) \left\{\alpha(t^{\star})\rho(t^{\star r},t^{\star \ell}) + \beta(t^{\star}) \right\}\quad\hbox {if}\; t^\star<t^{\ell}. \end{aligned} $$

In order to obtain ${\mathbb{E}\left\{ u(t^*) u(t^{r})\right\}, }$ we just have to replace t ^ℓ by t ^r. This gives the result. □