In the literature
see Proposition 4.3 in page 1405 in: Letac, Gérard; Massam, Hélène, The noncentral Wishart as an exponential family, and its moments, J. Multivariate Anal. 99, No. 7, 1393-1417 (2008). ZBL1140.62043.
there is a result on the expected value of an inverse of a non-central Wishart matrix. Unfortunately the result is very complicated, i.e. it is an infinite series over partitions of integers with coefficients that are expressed as certain integrals over the space of positive definite square matrices with the integrand involving derivatives of zonal polynomials. As such it is unclear, at least for me, how to compute that.
Now, by using integration by parts I managed to simplify the expressions and reduce them to the following integral below.
Let $y $ and ${\mathfrak a}$ be positive definite square matrices of dimension $N \ge 1$ and let$T \ge (N+1)/2$. Then let $\kappa:= (k_1,k_2,\cdots,k_N)$ be a partition of an integer $k$. This means that we have $k= k_1+k_2+\cdots+k_N$ and $k_1 \ge k_2 \ge \cdots \ge k_N \ge 0 $ . Finally let $C_\kappa(y)$ denote the zonal polynomial of the matrix $y$. Then the question is to compute the following integral:
\begin{equation} {\mathcal I}_N({\mathfrak a}) := \int\limits_{y>0} C_\kappa(y) \cdot \det(y)^{T-\frac{N+1}{2}} \cdot (y^{-1}) \cdot e^{-Tr[y\cdot {\mathfrak a}^{-1}]} dy \tag{1} \end{equation}
Now, by using the canonical parametrization of positive definite matrices $y = L^T \cdot L$ with $L$ being positive and upper triangular we have computed (with a little help of Mathematica) the integral above for $N=2$. Let $(a_1,a_2)$ denote the eigenvalues of the matrix ${\mathfrak a}^{-1}$. Then let ${\mathfrak P}_{n_1,n_2,n_3}$ denote the coefficient in the expansion of the zonal polynomial $C_\kappa(L^T L )$ at the powers $l_1^{n_1} l_2^{n_2} l_3^{n_3}$ with $L = [l_1, l_2]; [0, l_3]$.
Then the result reads:
\begin{eqnarray} {\mathcal I}_2({\mathfrak a}) &=& \sum\limits_{\begin{array}{l} n_1+n_2+n_3 = 2 k \\ n_1,n_2,n_3 \ge 0 \end{array}} {\mathfrak P}_{n_1,n_2,n_3} \frac{1}{2} \frac{ a_1^{-\frac{n_1}{2}} a_2^{\frac{1}{2} \left(-n_2-n_3\right)} }{ \left(a_1 a_2\right){}^{T} } \Gamma \left(T+\frac{n_1}{2}-1\right) \Gamma \left(T+\frac{n_3}{2}-\frac{3}{2}\right) \cdot \Gamma(\frac{n_2+1}{2}) \cdot \left( \begin{array}{cc} a_1 \left(\frac{1}{2} \left(n_2+n_3\right)+T-1\right) & -\frac{\sqrt{a_1 a_2} \Gamma \left(\frac{n_2}{2}+1\right) \left(T+\frac{n_1}{2}-1\right)_{\frac{1}{2}}}{\Gamma \left(\frac{1}{2} \left(n_2+1\right)\right)} \\ -\frac{\sqrt{a_1 a_2} \Gamma \left(\frac{n_2}{2}+1\right) \left(T+\frac{n_1}{2}-1\right)_{\frac{1}{2}}}{\Gamma \left(\frac{1}{2} \left(n_2+1\right)\right)} & a_2 \left(\frac{n_1}{2}+T-1\right) \\ \end{array} \right) \tag{2} \end{eqnarray}
As always, the code snippet below validates the result above. Here we go:
In[1227]:=
Clear[l];
NN = 2; T = 10;
NN2 = (NN + 1) NN/2;
k = 5;
myParts = IntegerPartitions[k, {NN}];
\[Kappa] = myParts[[RandomInteger[{1, Length[myParts]}]]];
A = DiagonalMatrix[RandomReal[{1, 2}, {NN}]];
(The parametrization of positively definite matrices)
myRoutine[l_List, NN_] := Module[{L, S, count, i, j},
L = ConstantArray[0, {NN, NN}];
count = 1;
Do[L[[i, j]] = l[[count]]; count++;, {i, 1, NN}, {j, i, NN}];
S = Transpose[L] . L;
{S, L}
];
(The actual integral in the canonical parametrization.)
myFunction[n1_, n2_, n3_, a1_, a2_] := (a1 a2)^-T a1^(-(n1/2)) a2^(
1/2 (-n2 - n3)) 1/
2 Gamma[-(3/2) + n3/2 + T] Gamma[-1 + n1/2 + T] Gamma[(1 + n2)/
2] {{a1 (-1 + (n2 + n3)/2 + T) , - Sqrt[a1 a2] Gamma[
1 + n2/2]/
Gamma[(1 + n2)/2] Pochhammer[-1 + n1/2 + T,
1/2] }, {-Sqrt[a1 a2] Gamma[1 + n2/2]/
Gamma[(1 + n2)/2] Pochhammer[-1 + n1/2 + T, 1/2] ,
a2 (-1 + n1/2 + T)}};
Clear[l, ll];
(Zonal polynomial in canonical parametrization.)
ll = Table[l[i], {i, 1, NN2}];
tmp = myRoutine[ll, NN];
S = tmp[[1]]; L = tmp[[2]];
myList = {# /. l[_] :> 1, Exponent[#, ll]} & /@
List @@ Expand[Simplify[ZonalPolynomial[[Kappa], Eigenvalues[S]]]];
(Compute integral numericaly.)
(Include the Jacobian as in page 250 in book by Muirhead)
NIntegrate[
With[{ll = Table[l[i], {i, 1, NN2}]},
With[{tmp = myRoutine[ll, NN]},
With[{S = tmp[[1]], L = tmp[[2]]},
ZonalPolynomial[[Kappa], Eigenvalues[S]] Det[S]^(T - (NN + 1)/2)
Inverse[
S] Exp[-Tr[S . A]] (2^NN Product[
L[[i, i]]^(NN + 1 - i), {i, 1, NN}])
]]],
Evaluate[Sequence @@ Table[{l[i], 0, Infinity}, {i, 1, NN2}]]
]
(Compute integral analytically.)
Sum[myList[[i, 1]] myFunction @@
Join[myList[[i, 2]], Diagonal[A]], {i, 1, Length[myList]}]
In the same way we managed to obtain the result for $N=3$. It reads:
\begin{eqnarray} &&{\mathcal I}_3({\mathfrak a}) = \sum\limits_{\begin{array}{l} n_1+n_2+n_3+n_4+n_5+n_6 = 2 k \\ n_1,n_2,n_3,n_4,n_5,n_6 \ge 0 \end{array}} {\mathfrak P}_{n_1,n_2,n_3,n_4,n_5,n_6} \cdot \frac{1}{4} \frac{a_1^{-\frac{n_1}{2}} a_2^{-\frac{n_2+n_4}{2}} a_3^{-\frac{n_3+n_5+n_6}{2}} }{(a_1 a_2 a_3)^T} \cdot \Gamma(T + \frac{n_1}{2}-1) \Gamma(T + \frac{n_4}{2}-\frac{3}{2}) \Gamma(T + \frac{n_6}{2}-2) \cdot \Gamma(\frac{1+n_2}{2}) \Gamma(\frac{1+n_3}{2}) \Gamma(\frac{1+n_5}{2}) \cdot \left( \begin{array}{ccc} \frac{1}{8} a_1 \left(\left(n_4+2 T-3\right) \left(n_3+n_6+2 T-3\right)+\left(n_2+1\right) \left(n_5+n_6+2 T-3\right)-\frac{8 \Gamma \left(\frac{n_2}{2}+1\right) \Gamma \left(\frac{n_3}{2}+1\right) \Gamma \left(\frac{n_5}{2}+1\right) \Gamma \left(T+\frac{n_4}{2}-1\right)}{\Gamma \left(\frac{1}{2} \left(n_2+1\right)\right) \Gamma \left(\frac{1}{2} \left(n_3+1\right)\right) \Gamma \left(\frac{1}{2} \left(n_5+1\right)\right) \Gamma \left(T+\frac{1}{2} \left(n_4-3\right)\right)}\right) & \frac{\sqrt{a_1} \sqrt{a_2} \Gamma \left(T+\frac{1}{2} \left(n_1-1\right)\right) \left(\frac{2 \Gamma \left(\frac{n_3}{2}+1\right) \Gamma \left(\frac{n_5}{2}+1\right) \Gamma \left(T+\frac{n_4}{2}-1\right)}{\Gamma \left(\frac{1}{2} \left(n_3+1\right)\right) \Gamma \left(\frac{1}{2} \left(n_5+1\right)\right) \Gamma \left(T+\frac{1}{2} \left(n_4-3\right)\right)}-\frac{\left(n_5+n_6+2 T-3\right) \Gamma \left(\frac{n_2}{2}+1\right)}{\Gamma \left(\frac{1}{2} \left(n_2+1\right)\right)}\right)}{4 \Gamma \left(T+\frac{n_1}{2}-1\right)} & \frac{\sqrt{a_1} \sqrt{a_3} \Gamma \left(T+\frac{1}{2} \left(n_1-1\right)\right) \left(\frac{2 \Gamma \left(\frac{n_2}{2}+1\right) \Gamma \left(\frac{n_5}{2}+1\right) \Gamma \left(T+\frac{n_4}{2}-1\right)}{\Gamma \left(\frac{1}{2} \left(n_2+1\right)\right) \Gamma \left(\frac{1}{2} \left(n_5+1\right)\right) \Gamma \left(T+\frac{1}{2} \left(n_4-3\right)\right)}-\frac{\left(n_4+2 T-3\right) \Gamma \left(\frac{n_3}{2}+1\right)}{\Gamma \left(\frac{1}{2} \left(n_3+1\right)\right)}\right)}{4 \Gamma \left(T+\frac{n_1}{2}-1\right)} \\ \frac{\sqrt{a_1} \sqrt{a_2} \Gamma \left(T+\frac{1}{2} \left(n_1-1\right)\right) \left(\frac{2 \Gamma \left(\frac{n_3}{2}+1\right) \Gamma \left(\frac{n_5}{2}+1\right) \Gamma \left(T+\frac{n_4}{2}-1\right)}{\Gamma \left(\frac{1}{2} \left(n_3+1\right)\right) \Gamma \left(\frac{1}{2} \left(n_5+1\right)\right) \Gamma \left(T+\frac{1}{2} \left(n_4-3\right)\right)}-\frac{\left(n_5+n_6+2 T-3\right) \Gamma \left(\frac{n_2}{2}+1\right)}{\Gamma \left(\frac{1}{2} \left(n_2+1\right)\right)}\right)}{4 \Gamma \left(T+\frac{n_1}{2}-1\right)} & \frac{1}{8} a_2 \left(n_1+2 T-2\right) \left(n_5+n_6+2 T-3\right) & -\frac{\sqrt{a_2} \sqrt{a_3} \left(n_1+2 T-2\right) \Gamma \left(\frac{n_5}{2}+1\right) \Gamma \left(T+\frac{n_4}{2}-1\right)}{4 \Gamma \left(\frac{1}{2} \left(n_5+1\right)\right) \Gamma \left(T+\frac{1}{2} \left(n_4-3\right)\right)} \\ \frac{\sqrt{a_1} \sqrt{a_3} \Gamma \left(T+\frac{1}{2} \left(n_1-1\right)\right) \left(\frac{2 \Gamma \left(\frac{n_2}{2}+1\right) \Gamma \left(\frac{n_5}{2}+1\right) \Gamma \left(T+\frac{n_4}{2}-1\right)}{\Gamma \left(\frac{1}{2} \left(n_2+1\right)\right) \Gamma \left(\frac{1}{2} \left(n_5+1\right)\right) \Gamma \left(T+\frac{1}{2} \left(n_4-3\right)\right)}-\frac{\left(n_4+2 T-3\right) \Gamma \left(\frac{n_3}{2}+1\right)}{\Gamma \left(\frac{1}{2} \left(n_3+1\right)\right)}\right)}{4 \Gamma \left(T+\frac{n_1}{2}-1\right)} & -\frac{\sqrt{a_2} \sqrt{a_3} \left(n_1+2 T-2\right) \Gamma \left(\frac{n_5}{2}+1\right) \Gamma \left(T+\frac{n_4}{2}-1\right)}{4 \Gamma \left(\frac{1}{2} \left(n_5+1\right)\right) \Gamma \left(T+\frac{1}{2} \left(n_4-3\right)\right)} & \frac{1}{8} a_3 \left(n_1+2 T-2\right) \left(n_4+2 T-3\right) \\ \end{array} \right) \end{eqnarray}
Likewise ${\mathfrak P}_{n_1,n_2,n_3,n_4,n_5,n_6}$ denotes the coefficient in the expansion of the zonal polynomial $C_\kappa(L^T L )$ at the powers $l_1^{n_1} l_2^{n_2} l_3^{n_3} l_4^{n_4} l_5^{n_5} l_6^{n_6}$ with $L = [l_1, l_2, l_3]; [0, l_4, l_5], [0, 0, l_6]$.
Again the relevant Mathematica code is below:
NN = 3; T =.;
n1 =.; n2 =.; n3 =.; n4 =.; n5 =.; n6 =.; a1 =.; a2 =.; a3 =.; l1 =.; \
l2 =.; l3 =.; l4 =.; l5 =.; l6 =.;
A = DiagonalMatrix[{a1, a2, a3}];
myRoutine[l_List, NN_] := Module[{L, S, count, i, j},
L = ConstantArray[0, {NN, NN}];
count = 1;
Do[L[[i, j]] = l[[count]]; count++;, {i, 1, NN}, {j, i, NN}];
S = Transpose[L] . L;
{S, L}
];
tmp = myRoutine[{l1, l2, l3, l4, l5, l6}, 3];
S = tmp[[1]]; L = tmp[[2]];
(*Parametrize the integrand. Don't forget about the Jacobian.*)
mat = Simplify[
Det[S]^(T - (NN + 1)/2)
Inverse[
S] Exp[-Tr[S . A]] (2^NN Product[
L[[i, i]]^(NN + 1 - i), {i, 1, NN}]), Assumptions -> T > 3] //
PowerExpand // Simplify;
(*Include the zonal polynomial do do the integration in \
thel-variables.*)
(*Here n1+n2+n3+n4+n5+n6==2 k*)
t0 = TimeUsed[];
myoutput =
Integrate[
l1^n1 l2^n2 l3^n3 l4^n4 l5^n5 l6^n6 mat, {l1, 0,
Infinity}, {l2, 0, Infinity}, {l3, 0, Infinity}, {l4, 0,
Infinity}, {l5, 0, Infinity}, {l6, 0, Infinity},
Assumptions ->
n1 >= 0 && n2 >= 0 && n3 >= 0 && n4 >= 0 && n5 >= 0 && n6 >= 0 &&
T > 4 && a1 > 0 && a2 > 0 && a3 > 0] // Simplify;
Print["Symbolic integration done in time=", TimeUsed[] - t0];
myoutput = ((myoutput/(1/4 (a1 a2 a3)^(-T) a1^(-(n1/2)) a2^(
1/2 (-n2 - n4)) a3^(1/2 (-n3 - n5 - n6))
Gamma[-1 + n1/2 + T] Gamma[-3/2 + 1/2 (n4) +
T] Gamma[-2 + n6/2 + T] Gamma[(1 + n2)/2] Gamma[(
1 + n3)/2] Gamma[(1 + n5)/2])) // Simplify //
PowerExpand // FullSimplify);
myoutput // MatrixForm
Having said all this my question is obvious.
How do we compute the integral in question for generic $N \in {\mathbb N} $.
