The complex multinormal distribution, quadratic forms in complex random vectors and an omnibus goodness-of-fit test for the complex normal distribution

Abstract

This paper first reviews some basic properties of the (noncircular) complex multinormal distribution and presents a few characterizations of it. The distribution of linear combinations of complex normally distributed random vectors is then obtained, as well as the behavior of quadratic forms in complex multinormal random vectors. We look into the problem of estimating the complex parameters of the complex normal distribution and give their asymptotic distribution. We then propose a virtually omnibus goodness-of-fit test for the complex normal distribution with unknown parameters, based on the empirical characteristic function. Monte Carlo simulation results show that our test behaves well against various alternative distributions. The test is then applied to an fMRI data set and we show how it can be used to “validate” the usual hypothesis of normality of the outside-brain signal. An R package that contains the functions to perform the test is available from the authors.

Appendix

Proof of Theorem 1. From Mardia et al. (1979, p. 33), $Z_{1},\ldots ,Z_{d}$ are independent if and only if

$$\begin{aligned} \varphi _{(Z_{1},\ldots ,Z_{d})}^{}(v_{1},\ldots ,v_{d}) = \prod _{k=1}^{d} \varphi _{Z_{k}}^{}(v_{k}). \end{aligned}$$

(26)

We proceed by induction starting with $d=2$. Let $Z = (Z_{1},Z_{2})^{{\mathsf {T}}}\sim CN_{2}(\mu ,\varGamma ,P)$. If $\mu = (\mu _{1},\mu _{2})^{{\mathsf {T}}},\, v=(v_1,v_2)^{{\mathsf {T}}}$ and $\varGamma = \mathrm{diag }(\gamma _{11},\gamma _{22}),\, P = \mathrm{diag }(p_{11},p_{22})$, then, from (7),

$$\begin{aligned} \varphi _{(Z_{1},Z_{2})}(v)&= \exp \left\{ \jmath \mathrm{Re }\left( v_{1}^{*}\mu _{1}\right) +\jmath \mathrm{Re }\left( v_{2}^{*}\mu _{2} \right) \right. \\&\ \ \ - \left. \frac{1}{4}\left[ |v_{1}|^{2}\gamma _{11} + |v_{2}|^{2}\gamma _{22} + \mathrm{Re }\left( (v_{1}^{*})^{2}p_{11}+(v_{2}^{*})^{2}p_{22} \right) \right] \right\} \\&= \varphi _{Z_{1}}(v_{1})\varphi _{Z_{2}}(v_{2}). \end{aligned}$$

Inversely, assume $Z_{1},\, Z_{2}$ are independent so that (26) is verified. Letting $\varGamma =(\gamma _{ij})$ and $P=(p_{ij})$, we have for all $v=(v_1,v_2)^{{\mathsf {T}}}\in \mathbb {C}^2$

$$\begin{aligned} -4\ln (\varphi _{(Z_{1},Z_{2})}(v)) = -4\ln (\varphi _{Z_{1}}(v_{1})\varphi _{Z_{2}}(v_{2})), \end{aligned}$$

so that

$$\begin{aligned}&-4\jmath \mathrm{Re }(v^{\mathcal {H}}\mu ) + v^{\mathcal {H}}\varGamma v + \mathrm{Re }(v^{\mathcal {H}} P v^{*}) = -4\jmath (\mathrm{Re }(v_{1}^{*}\mu _{1})+\mathrm{Re }(v_{2}^{*}\mu _{2}))\\&\qquad + |v_{1}|^{2}\gamma _{11} + |v_{2}|^{2}\gamma _{22} + \mathrm{Re }\left( (v_{1}^{*})^2p_{11} + (v_{2}^{*})^2p_{22}\right) , \end{aligned}$$

which shows that

$$\begin{aligned} \mathrm{Re }(v_{1}^{*}v_{2}\gamma _{12}) +\mathrm{Re }( (v_{1}v_{2})^{*}p_{12}) = 0. \end{aligned}$$

With $v_{1}^{*} = a_{1}-\jmath b_{1}$ and $v_{2} = a_{2}+\jmath b_{2}$, simple algebra leads to the equivalent equation:

$$\begin{aligned}&(a_{1}a_{2}+b_{1}b_{2})\mathrm{Re }(\gamma _{12})+(a_{1}a_{2}-b_{1}b_{2})\mathrm{Re }(p_{12}) \nonumber \\ -&(a_{1}b_{2}-b_{1}a_{2})\mathrm{Im }(\gamma _{12})+ (b_{1}a_{2}+a_{1}b_{2})\mathrm{Im }(p_{12})= 0, \end{aligned}$$

(27)

which must hold for all $a_{1},a_{2},b_{1},b_{2}$. In particular, they must hold for the following cases:

$$\begin{aligned} a_{1}&= a_{2} = b_{1} = b_{2} = \lambda \\ a_{1}&= a_{2} = b_{1} = -b_{2} = \lambda \\ b_{1}&= b_{2} = 0, \ a_{1} = a_{2} = \lambda \\ a_{2}&= b_{1} = 0, \ a_{1} = b_{2} = \lambda \end{aligned}$$

for some $\lambda \in \mathbb {R}\backslash \{0\}$. This leads to the linear system $\lambda ^2AX = 0$ with

$X = (\mathrm{Re }(\gamma _{12}),\mathrm{Re }(p_{12}),\mathrm{Im }(\gamma _{12})),\mathrm{Im }(p_{12})))^{{\mathsf {T}}}$ and

$$\begin{aligned} A = \begin{pmatrix} 2 &{} 0 &{} 0 &{} 2 \\ 0 &{} 2 &{} 2 &{} 0 \\ 1 &{} 1 &{} 0 &{} 0 \\ 0 &{} 0 &{} -1 &{} 1 \end{pmatrix}. \end{aligned}$$

Since $\det (A)\ne 0$, the only solution is $\mathrm{Re }(\gamma _{21})=\mathrm{Re }(p_{12})=\mathrm{Im }(\gamma _{21})=\mathrm{Im }(p_{12}) = 0$, which proves that $\varGamma$ and $P$ must be diagonal.

Now, for a given $d > 2$, let

$$\begin{aligned} Z_{(d+1)} = (Z_{1},\ldots ,Z_{d},Z_{d+1})^{{\mathsf {T}}}\sim CN_{d+1}(\mu _{d+1},\varGamma _{d+1},P_{d+1}), \end{aligned}$$

where $Z_{1},\ldots ,Z_{d}$ are independent, which, by the induction hypothesis, implies $\varGamma _{d} = \mathrm{diag }(\gamma _{11},\ldots ,\gamma _{dd})$ and $P_{d} = \mathrm{diag }(p_{11},\ldots ,p_{dd})$. Let

with

$$\begin{aligned} \varGamma _{(\cdot ,d+1)} = \begin{pmatrix} \gamma _{1,d+1} \\ \vdots \\ \gamma _{d,d+1} \end{pmatrix} = \varGamma _{(d+1,\cdot )}^{\mathcal {H}}, \end{aligned}$$

and similarly

with

$$\begin{aligned} P_{(\cdot ,d+1)} = \begin{pmatrix} p_{1,d+1} \\ \vdots \\ p_{d,d+1} \end{pmatrix} = P_{(d+1,\cdot )}^{{\mathsf {T}}}. \end{aligned}$$

The proof is that

$$\begin{aligned} \varphi _{(Z_{1},\cdots ,Z_{d+1})^{{\mathsf {T}}}}( (\nu _{1}, \cdots , \nu _{d+1}) ) = \prod _{k=1}^{d+1} \varphi _{Z_{k}}(\nu _{k}) \end{aligned}$$

when $\varGamma _{(.,d+1)}=0$ and $P_{(.,d+1)}=0$ is straightforward. We focus on the converse. Consider the equality

$$\begin{aligned} -4\ln \left( \varphi _{(Z_{1},\ldots ,Z_{d+1})}(v_{1},\ldots ,v_{d+1})\right) = -4\ln \left( \,\prod _{k=1}^{d+1}\varphi _{Z_{k}}(v_{k})\right) , \end{aligned}$$

which should hold for any values of $v_1,\ldots ,v_{d+1}$. It is easy to see that the right hand side of this equation is

$$\begin{aligned} -4\jmath \mathrm{Re }(v_{(d+1)}^{\mathcal {H}}\mu _{d+1})+ \sum _{k=1}^{d+1} |v_{k}|^{2}\gamma _{kk} + \mathrm{Re }\left( \,\sum _{k=1}^{d+1} (v_{k}^{*})^{2}p_{kk}\right) , \end{aligned}$$

(28)

with $v_{(d+1)} = (v_{1},\ldots ,v_{d+1})^{{\mathsf {T}}}$. As for the left hand side, we have from (7)

$$\begin{aligned} -4\ln (\varphi _{Z_{(d+1)}}(v_{(d+1)}))&= -4 \jmath \mathrm{Re }(v_{(d+1)}^{\mathcal {H}}\mu _{d+1}) + v_{(d+1)}^{\mathcal {H}}\varGamma _{d+1}v_{(d+1)} \\&+ \mathrm{Re }(v_{(d+1)}^{\mathcal {H}}P_{d+1}v_{(d+1)}^{*}) \\&= -4\jmath \mathrm{Re }( v_{(d+1)}^{\mathcal {H}}\mu _{d+1}) + \sum _{k=1}^{d+1}\left( |v_{k}|^{2}\gamma _{kk} \right) \\&+ \mathrm{Re }\left( \,\sum _{k=1}^{d+1} (v_{k}^{*})^{2}p_{kk} \right) \\&+ 2\mathrm{Re }\left( \,\sum _{k=1}^{d} v_{k}^{*}v_{d+1}\gamma _{k,d+1} + (v_{d+1}v_{k})^{*}p_{k,d+1}\right) . \end{aligned}$$

Comparing this with (28) shows that we must have:

$$\begin{aligned} \mathrm{Re }\left( \,\sum _{k=1}^{d+1} v_{k}^{*}v_{d+1}\gamma _{k,d+1} + (v_{d+1}v_{k})^{*}p_{k,d+1}\right) = 0. \end{aligned}$$

Now, if there is only one nonzero $v_{j}$ among $\{v_{1},\ldots ,v_{d}\}$ while $v_{d+1} \ne 0$, then

$$\begin{aligned} \mathrm{Re }( v_{j}^{*}v_{d+1}\gamma _{j,d+1}) +\mathrm{Re }( (v_{j}v_{d+1})^{*}p_{j,d+1})=0. \end{aligned}$$

The argument used in the case $d=2$ shows that $\gamma _{j,d+1} = p_{j,d+1} = 0$. Because $j$ is arbitrary, $\varGamma _{(\cdot ,d+1)} = P_{(\cdot ,d+1)} = 0$ and, thus, $\varGamma _{(d+1)}$ and $P_{(d+1)}$ are diagonal. This concludes the proof. $\square$

Proof of Corollary 1. Without loss of generality, assume $\mu _{1} = \mu _{2} = 0$. The proof that $Z_{1}$ and $Z_{2}$ are independent when $\varGamma _{12} = P_{12} = 0$ follows from noticing that this entails $\varphi _{Z}(\nu _{1},\nu _{2}) = \varphi _{Z_{1}}(\nu _{1})\varphi _{Z_{2}}(\nu _{2})$.

Inversely, let $Z_{1,k} \sim CN_{1}(0,\gamma _{k,k},p_{k,k})$ be the $k$-th component of $Z_{1}$ and $Z_{2,\ell } \sim CN_{1}(0,\gamma _{\ell ,\ell },p_{\ell ,\ell })$ the $\ell$-th component of $Z_{2}$.

Assuming that $Z_{1},\, Z_{2}$ are independent, $Z_{1, k}$ and $Z_{2,\ell }$ are independent and Theorem 1 ensures that the $(k,\ell )$ coefficients in $\varGamma _{12}$ and $P_{12}$ are equal to zero.

Because $k, \ell$ are abitrary, the proof follows. $\square$

Proof of Theorem 2. From (12), where $\mathcal {X} \sim N_{2d}(\mu _{\mathcal {X}},\varSigma _{\mathcal {X}})$. Let $S^{1/2}$ be a real $2d\times 2d$ matrix such that $(S^{1/2})^{{\mathsf {T}}}S^{1/2} = S$. Using an eigenvalue decomposition, we have $S^{1/2}\varSigma _{\mathcal {X}} (S^{1/2})^{{\mathsf {T}}}=O\Lambda O^{{\mathsf {T}}}$ where $\Lambda$ is the diagonal matrix of the real eigenvalues $\lambda _{1}\ge \lambda _{2}\ge \cdots \ge \lambda _{2d}\ge 0$ and $O$ is orthogonal. Hence, $O^{{\mathsf {T}}}S^{1/2}\mathcal {X}\sim N_{2d}(O^{{\mathsf {T}}}S^{1/2}\mu _{\mathcal {X}},\Lambda )$. This entails that

where $q$ is the number of nonzero eigenvalues $\lambda _k$, the $\mathcal {N}_{k}$ are independent $N(0,1)$ random variables and $\tau _{k}$ is the $k$-th component of $O^{{\mathsf {T}}}S^{1/2}\mu _{\mathcal {X}}$.

Taking $S^{1/2} = MR^{1/2}M^{-1}$, we get , so that where $e_{k}$ is the $k$-th vector in the canonical basis for $\mathbb {R}^{2d}$. We then have

where $\delta _{k}^{2} = \frac{\tau _{k}^{2}}{\lambda _{k}}$ ($1\le k\le q$), which vanishes if $\mu = 0$.

Now, if $x_k$ is the $k$th eigenvector of $\varGamma _PR$ corresponding to the eigenvalue $\alpha _k$, we have $\varGamma _PRx_k=\alpha _kx_k$. This gives

$$\begin{aligned} \varGamma _PM^{-1}SMx_k&= \alpha _kx_k,\\ S^{1/2}M\varGamma _PM^{-1}(S^{1/2})^{{\mathsf {T}}}S^{1/2}Mx_k&= \alpha _kS^{1/2}Mx_k. \end{aligned}$$

which, from (3) and after taking the conjugate, gives

$$\begin{aligned} S^{1/2}\varSigma _{\mathcal {X}}(S^{1/2})^{{\mathsf {T}}}(S^{1/2}M^*x_k^*)=\frac{\alpha _k}{2}(S^{1/2}M^*x_k^*). \end{aligned}$$

We thus have $S^{1/2}M^*x_k^*=Oe_k$ and $\alpha _k/2=\lambda _k$. We finally obtain

$\square$

Proof of Theorem 3. First, we recall the following property of the Moore–Penrose pseudo-inverse (Penrose 1955). Let $A$ be $m\times n$ complex and $B$ be $n\times p$ complex, if $AA^{\mathcal {H}}=I_{m}$ or $B^{\mathcal {H}}B=I_{p}$, then the Moore–Penrose pseudo-inverse $(AB)^{+}$ of $AB$ satisfies $(AB)^{+}=B^{+}A^{+}$.

From (1) and equation above (1), $Q = 2\mathcal {X}^{{\mathsf {T}}}M\varGamma _{P}^{+}M^{-1}\mathcal {X}$, where $\mathcal {X} \sim N_{2d}(0,\varSigma _{\mathcal {X}})$. From (3) and the above property,

$$\begin{aligned} \varGamma _{P}^{+}&= \left( M^{-1}\varSigma _{X}(M^{\mathcal {H}})^{-1}\right) ^{+} = \frac{1}{2}M^{-1}\varSigma _{\mathcal {X}}^{+}M. \end{aligned}$$

Therefore, $Q=\mathcal {X}^{{\mathsf {T}}}\varSigma _{\mathcal {X}}^{+}\mathcal {X}$. Now, from (Searle 1971, Corollary 2s.2, p.69), $Q\sim \chi _{\mathrm{tr }(\varSigma _{\mathcal {X}}^{+}\varSigma _{\mathcal {X}})}^{2}$ and (Rao 1973, Proposition (ii)b p. 25) ensures that $\mathrm{tr }(\varSigma _{\mathcal {X}}^{+}\varSigma _{\mathcal {X}})=\mathrm{rank }(\varSigma _{\mathcal {X}}) = \mathrm{rank }(\varGamma _{P})$. When $\varGamma _{P}$ is of full rank, $\varGamma _{P}^{+} = \varGamma _{P}^{-1}$ and $\mathrm{rank }(\varGamma _{P}) = 2d$.

$\square$

Proof of Theorem 4. Because we were not able to find a proof of this theorem in the literature, we provide one here.

In view of (1), we have . Hence,

using a result in Feldman (1965) and the standard C.L.T. for real r.v. But

from (8). $\square$

Proof of Proposition 1. From (6), for all $\nu ,\,\varphi _{AX_n}(\nu )=\varphi _{X_n}(A^{\mathcal {H}}\nu ) \underset{n\rightarrow \infty }{\longrightarrow }\varphi _{X}(A^{\mathcal {H}}\nu ) =\varphi _{AX}(\nu )$. $\square$

Proof of Corollary 2. The result follows from Theorem 4, Proposition 1 and (10). $\square$

Proof of Proposition 2. The result is obvious regarding the method of moments estimators. As for the m.l.e., the likelihood function can we written from (4) as

where is a $2d \times n$ matrix of a.c.r.v. and $e_{} = \begin{pmatrix} 1,\ldots , 1 \end{pmatrix}^{{\mathsf {T}}}$ is $n$-dimensional. With , we notice that . Moreover,

where . Thus,

Obviously, $L(\mu ,\varGamma _{P})$ is maximized in $\mu$ when . Moreover,

$$\begin{aligned} L(\hat{\mu },\varGamma _{P})&= \frac{1}{\pi ^{nd}\left| \varGamma _{P}\right| ^{n/2}}\exp \left\{ -\frac{n}{2}\mathrm{tr }\left( \varGamma _{P}^{-1}\hat{\varGamma }_{P}\right) \right\} \\&\quad \le \frac{1}{\pi ^{nd}\left| \hat{\varGamma }_{P}\right| ^{n/2}}\exp \left\{ -dn\right\} \\&= L(\hat{\mu },\hat{\varGamma }_{P}), \end{aligned}$$

where the inequality holds in view of Srivastava and Khatri (1979, Theorem 1.10.4). This gives the m.l.e. for $\varGamma _{P}$. We recover the corresponding estimators for $\varGamma$ and $P$ by extracting the corresponding terms in $\hat{\varGamma }_{P}$. $\square$