Explicit formula for Gaussian moments in two dimensions
In this Appendix, we present the derivation of the explicit formula for 2D central moments of the Gaussian probability density function \(f_{N}(\mathbf {x})\) with the covariance matrix
$$\begin{aligned} \varSigma = \left( \begin{array}{cc} \sigma _1 &{}\quad \rho \\ \rho &{}\quad \sigma _2 \end{array} \right) . \end{aligned}$$
It holds for the inverse matrix \(\varSigma ^{-1}\) and its determinant
$$\begin{aligned} \varSigma ^{-1} = \frac{1}{|\varSigma |}\left( \begin{array}{rr} \sigma _2 &{} \quad -\rho \\ - \rho &{}\quad \sigma _1 \end{array} \right) \equiv \left( \begin{array}{cc} a &{}\quad b \\ b &{}\quad c \end{array} \right) , \quad |\varSigma ^{-1}| = ac-b^2 = \frac{1}{|\varSigma |}. \end{aligned}$$
If \(m+n\) is odd, the moments vanish due to the symmetry
$$\begin{aligned} \mathrm {m}_{mn}^{(f_{N})} = 0. \end{aligned}$$
For \(m+n\) even we have
$$\begin{aligned} \mathrm {m}_{mn}^{(f_{N})}&= \frac{1}{2\pi \sqrt{|\varSigma |}} \iint \limits _{{\mathbb {R}}^2} x^m y^n \mathrm {e}^{-\frac{1}{2}(ax^2+2bxy+cy^2)} \mathrm {\,d}x \mathrm {\,d}y \\&= \frac{1}{2\pi \sqrt{|\varSigma |}} \iint \limits _{{\mathbb {R}}^2} x^m \mathrm {e}^{-\frac{1}{2}x^2\left( a-\frac{b^2}{c}\right) } y^n \mathrm {e}^{-\frac{1}{2}\left( y+\frac{b}{c}x\right) ^2c} \mathrm {\,d}x \mathrm {\,d}y =\\&= \left| \begin{array}{rl} y+\frac{b}{c}x &{} = u \\ x &{} = v \end{array} \right| = \frac{1}{2\pi \sqrt{|\varSigma |}} \iint \limits _{{\mathbb {R}}^2} v^m \mathrm {e}^{-\frac{1}{2}v^2\left( a-\frac{b^2}{c}\right) } \left( u-\frac{b}{c}v\right) ^n \mathrm {e}^{-\frac{1}{2}u^2c}\mathrm {\,d}u \mathrm {\,d}v . \end{aligned}$$
We can separate the integrals and use the formula for 1D moments of Gaussian function:
$$\begin{aligned}&= \frac{1}{2\pi \sqrt{|\varSigma |}} \sum _{k=0}^{n} \left( {\begin{array}{c}n\\ k\end{array}}\right) \left( -\frac{b}{c}\right) ^{n-k} \int \limits _{\mathbb {R}}u^k \mathrm {e}^{-\frac{1}{2}u^2c} \mathrm {\,d}u \int \limits _{\mathbb {R}}v^{m+n-k} \mathrm {e}^{-\frac{1}{2}v^2\left( a-\frac{b^2}{c}\right) } \mathrm {\,d}v \nonumber \\&= \frac{1}{\sqrt{|\varSigma |}} \sum _{\begin{array}{c} k=0,\\ k \ \mathrm {even} \end{array}}^{n} \left( {\begin{array}{c}n\\ k\end{array}}\right) \left( \frac{-b}{c}\right) ^{n-k} \left( \frac{1}{{a-\frac{b^2}{c}}}\right) ^{\frac{m+n-k+1}{2}} (m+n-k-1)!! \left( \frac{1}{{c}}\right) ^{\frac{k+1}{2}} (k-1)!! \nonumber \\&= \sum _{\begin{array}{c} k=0,\\ k \ \mathrm {even} \end{array}}^{n} \left( {\begin{array}{c}n\\ k\end{array}}\right) \left( \frac{-b}{ |\varSigma ^{-1}| }\right) ^{n-k} \left( \frac{c}{ |\varSigma ^{-1}| }\right) ^{\frac{m-n}{2}} |\varSigma |^{k/2} (m+n-k-1)!! (k-1)!! \nonumber \\&= \sum _{i=0}^{\lfloor \frac{n}{2}\rfloor } \left( {\begin{array}{c}n\\ 2i\end{array}}\right) \rho ^{n-2i} \sigma _1^{\frac{m-n}{2}} \left( \sigma _1\sigma _2-\rho ^2\right) ^{i} (m+n-2i-1)!! (2i-1)!! \nonumber \\&= \sum _{i=0}^{\lfloor \frac{n}{2}\rfloor } \sum _{j=0}^{i} (-1)^{i-j} \left( {\begin{array}{c}n\\ 2i\end{array}}\right) \left( {\begin{array}{c}i\\ k\end{array}}\right) (m+n-2i-1)!! (2i-1)!! \rho ^{n-2j} \sigma _1^{\frac{m-n}{2}+j} \sigma _2^{j}. \end{aligned}$$
(24)
We may reduce the quadratic form in the exponent to a sum of squares in the following way
$$\begin{aligned} \frac{1}{2\pi \sqrt{|\varSigma |}}\iint \limits _{{\mathbb {R}}^2} y^n \mathrm {e}^{-\frac{1}{2}y^2\left( c-\frac{b^2}{a}\right) } x^m \mathrm {e}^{-\frac{1}{2}\left( x+\frac{b}{a}y\right) ^2a}\mathrm {\,d}x \mathrm {\,d}y. \end{aligned}$$
Then using the substitution
$$\begin{aligned} \begin{array}{rl} x+\frac{b}{a}y &{}= u \\ y &{} = v \end{array} \end{aligned}$$
another formula for moments of bivariate Gaussian distribution is obtained
$$\begin{aligned} \mathrm {m}_{mn}^{(f_{N})} = \sum _{i=0}^{\lfloor \frac{m}{2}\rfloor }\sum _{j=0}^{i}(-1)^{i-j}\left( {\begin{array}{c}m\\ 2i\end{array}}\right) \left( {\begin{array}{c}i\\ j\end{array}}\right) (m+n-2i-1)!! (2i-1)!!\rho ^{m-2j}\sigma _1^{j}\sigma _2^{\frac{n-m}{2}+j}. \end{aligned}$$
(25)
When we compare these two results, it is obvious that the coefficients of negative powers must be zero. Hence, moments are composed of positive powers of the elements of covariance matrix only
$$\begin{aligned} \mathrm {m}_{mn}^{(f_{N})} = \mathop {\sum _{i=0}^{\lfloor \frac{m}{2}\rfloor }\sum _{j=0}^{i}}_{\begin{array}{c} j\ge \frac{m-n}{2} \end{array}}(-1)^{i-j}\left( {\begin{array}{c}m\\ 2i\end{array}}\right) \left( {\begin{array}{c}i\\ j\end{array}}\right) (m+n-2i-1)!!\rho ^{m-2j}\sigma _1^{j} \sigma _2^{\frac{n-m}{2}+j}. \end{aligned}$$
(26)
Proof of the equivalence
Let us show that Formulas (22) and (23) for convolution invariants are equivalent. The proof is done by induction.
Proof
\(A_{00} = 1\) in Formula (22) as well as in Formula (23).
Let us assume \((m,n),\, m+n>0\). From the induction assumption, the explicit formula is valid for all indices (p, q), where \(p\le m, \ q \le n\) and \((p,q)\ne (m,n)\).
$$\begin{aligned} A_{mn}&= m_{mn} - \mathop {\sum _{l=0}^{m} \sum _{k=0}^{n}}_{\begin{array}{c} l+k\ne 0,\\ l+k \ \mathrm {even} \end{array}} \left( {\begin{array}{c}m\\ l\end{array}}\right) \left( {\begin{array}{c}n\\ k\end{array}}\right) \mathop {\sum _{i=0}^{\lfloor \frac{k}{2} \rfloor } \sum _{j=0}^{i}}_{\begin{array}{c} j\ge \frac{k-l}{2} \end{array}} (-1)^{i-j} \left( {\begin{array}{c}k\\ 2i\end{array}}\right) \left( {\begin{array}{c}i\\ j\end{array}}\right) (l+k-2i-1)!! (2i-1)!! \\&\quad \cdot m_{11}^{k-2j} m_{20}^{\frac{l-k}{2}+j} m_{02}^{j}A_{m-l,n-k} = \\&= m_{mn} - \mathop {\sum _{l=0}^{m} \sum _{k=0}^{n}}_{\begin{array}{c} l+k\ne 0,\\ l+k \ \mathrm {even} \end{array}} \left( {\begin{array}{c}m\\ l\end{array}}\right) \left( {\begin{array}{c}n\\ k\end{array}}\right) \mathop {\sum _{i=0}^{\lfloor \frac{k}{2} \rfloor } \sum _{j=0}^{i}}_{\begin{array}{c} j \ge \frac{k-l}{2} \end{array}} (-1)^{i-j} \left( {\begin{array}{c}k\\ 2i\end{array}}\right) \left( {\begin{array}{c}i\\ j\end{array}}\right) \\&\quad \cdot (l+k-2i-1)!! (2i-1)!!\, m_{11}^{k-2j} m_{20}^{\frac{l-k}{2}+j} m_{02}^{j} \\&\quad \cdot \mathop {\sum _{s=0}^{n-k} \sum _{t=0}^{m-l}}_{ s+t \ \mathrm {even}} (-1)^{\frac{s+t}{2}} \left( {\begin{array}{c}m-l\\ t\end{array}}\right) \left( {\begin{array}{c}n-k\\ s\end{array}}\right) \mathop {\sum _{\alpha =0}^{\lfloor \frac{s}{2} \rfloor } \sum _{\beta =0}^{\alpha }}_{\begin{array}{c} \beta \ge \frac{s-t}{2} \end{array}} (-1)^{\alpha -\beta } \left( {\begin{array}{c}s\\ 2\alpha \end{array}}\right) \left( {\begin{array}{c}\alpha \\ \beta \end{array}}\right) \\&\quad (2\alpha -1)!! (s+t-2\alpha -1)!! \\&\quad \cdot m_{11}^{s-2\beta } m_{20}^{\frac{t-s}{2}+\beta } m_{02}^{\beta }m_{m-l-t,n-k-s} = \\&= m_{mn} - \mathop {\sum _{l = 0}^{m} \sum _{k=0}^{n}}_{\begin{array}{c} l+k \ne 0,\\ l+k \ \mathrm {even} \end{array}} \frac{m!}{l!(m-l)!} \frac{n!}{k!(n-k)!} \mathop {\sum _{i = 0}^{\lfloor \frac{k}{2} \rfloor } \sum _{j=0}^{i}}_{\begin{array}{c} j \ge \frac{k-l}{2} \end{array}} (-1)^{i-j} \left( {\begin{array}{c}k\\ 2i\end{array}}\right) \left( {\begin{array}{c}i\\ j\end{array}}\right) (2i-1)!!\\&\quad \cdot (l+k-2i-1)!! \\&\quad \cdot m_{11}^{k-2j} m_{20}^{\frac{l-k}{2}+j} m_{02}^{j} \mathop {\sum _{s = 0}^{n-k} \sum _{t = 0}^{m-l}}_{ s+t \ \mathrm {even}} (-1)^{\frac{s+t}{2}} \frac{(m-l)!}{t!(m-l-t)!} \frac{(n-k)!}{s!(n-k-s)!} \\&\quad \cdot \mathop {\sum _{\alpha =0}^{\lfloor \frac{s}{2}\rfloor }\sum _{\beta =0}^{\alpha }}_{\begin{array}{c} \beta \ge \frac{s-t}{2} \end{array}}(-1)^{\alpha -\beta }\left( {\begin{array}{c}s\\ 2\alpha \end{array}}\right) \left( {\begin{array}{c}\alpha \\ \beta \end{array}}\right) (2\alpha -1)!!(s+t-2\alpha -1)!! m_{11}^{s-2\beta } m_{20}^{\frac{t-s}{2}+\beta }\\&\quad \cdot m_{02}^{\beta }m_{m-l-t,n-k-s} \\ \end{aligned}$$
$$\begin{aligned}&= \left| \begin{array}{c} p = k + s \\ q = t + l \end{array}\right| = m_{mn} - \mathop {\sum _{l = 0}^{m} \sum _{k=0}^{n}}_{\begin{array}{c} l+k \ne 0,\\ l + k \ \mathrm {even} \end{array}} \frac{m!}{l!} \frac{n!}{k!} \mathop {\sum _{i=0}^{\lfloor \frac{k}{2} \rfloor } \sum _{j=0}^{i}}_{\begin{array}{c} j\ge \frac{k-l}{2} \end{array}} (-1)^{i-j} \left( {\begin{array}{c}k\\ 2i\end{array}}\right) \left( {\begin{array}{c}i\\ j\end{array}}\right) \\&\quad \cdot (l+k-2i-1)!! (2i-1)!! \\&\quad \cdot m_{11}^{k-2j} m_{20}^{\frac{l-k}{2}+j} m_{02}^{j} \mathop {\sum _{p=k}^{n}\sum _{q=l}^{m}}_{ p+q \ \mathrm {even}} \frac{(-1)^{\frac{p+q-k-l}{2}}}{(q-l)!(m-q)!} \frac{1}{(p-k)!(n-p)!} \mathop {\sum _{\alpha = 0}^{\lfloor \frac{p-k}{2} \rfloor } \sum _{\beta =0}^{\alpha }}_{\begin{array}{c} \beta \ge \frac{p-q+l-k}{2} \end{array}} (-1)^{\alpha -\beta } \\&\quad \cdot \left( {\begin{array}{c}p-k\\ 2\alpha \end{array}}\right) \left( {\begin{array}{c}\alpha \\ \beta \end{array}}\right) (2\alpha -1)!!\\&\quad \cdot (p+q-k-l-2\alpha -1)!! \, m_{11}^{p-k-2\beta } m_{20}^{\frac{q-p+k-l}{2}+\beta } m_{02}^{\beta }m_{m-q,n-p} = \\&= m_{mn} - \mathop {\sum _{l = 0}^{m} \sum _{k=0}^{n}}_{\begin{array}{c} l+k \ne 0,\\ l+k \ \mathrm {even} \end{array}} \mathop {\sum _{p = k}^{n} \sum _{q = l}^{m}}_{ p+q \ \mathrm {even}} \left( {\begin{array}{c}m\\ q\end{array}}\right) \left( {\begin{array}{c}q\\ l\end{array}}\right) \left( {\begin{array}{c}n\\ p\end{array}}\right) \left( {\begin{array}{c}p\\ k\end{array}}\right) (-1)^{\frac{p+q-k-l}{2}} m_{m-q,n-p}\\&\quad \cdot \mathop {\sum _{i = 0}^{\lfloor \frac{k}{2} \rfloor } \sum _{j = 0}^{i}}_{\begin{array}{c} j \ge \frac{k-l}{2} \end{array}} (-1)^{i-j} \left( {\begin{array}{c}k\\ 2i\end{array}}\right) \left( {\begin{array}{c}i\\ j\end{array}}\right) \\&\quad \cdot (l+k-2i-1)!! (2i-1)!! \mathop {\sum _{\alpha = 0}^{\lfloor \frac{p-k}{2} \rfloor } \sum _{\beta = 0}^{\alpha }}_{\begin{array}{c} \beta \ge \frac{p-q+l-k}{2} \end{array}} (-1)^{\alpha -\beta } \left( {\begin{array}{c}p-k\\ 2\alpha \end{array}}\right) \left( {\begin{array}{c}\alpha \\ \beta \end{array}}\right) \\&\quad \cdot (2\alpha -1)!! (p+q-k-l-2\alpha -1)!! \\&\quad \cdot m_{11}^{p-2j-2\beta } m_{20}^{\frac{q-p}{2}+j+\beta } m_{02}^{j+\beta } = \\&= m_{mn} - \mathop {\sum _{p = 0}^{n} \sum _{k = 0}^{p} \sum _{q = 0}^{m} \sum _{l = 0}^{q}}_{\begin{array}{c} l + k \ne 0, p+q \ne 0\\ l+k \ \mathrm {even}, p+q \ \mathrm {even} \end{array}} \left( {\begin{array}{c}m\\ q\end{array}}\right) \left( {\begin{array}{c}q\\ l\end{array}}\right) \left( {\begin{array}{c}n\\ p\end{array}}\right) \left( {\begin{array}{c}p\\ k\end{array}}\right) (-1)^{\frac{p+q-k-l}{2}} m_{m-q,n-p}\\&\quad \cdot \mathop {\sum _{i=0}^{\lfloor \frac{k}{2}\rfloor }\sum _{j=0}^{i}}_{\begin{array}{c} j\ge \frac{k-l}{2} \end{array}}(-1)^{i-j} \left( {\begin{array}{c}k\\ 2i\end{array}}\right) \left( {\begin{array}{c}i\\ j\end{array}}\right) \\&\quad \cdot (l+k-2i-1)!! (2i-1)!! \mathop {\sum _{\alpha = 0}^{\lfloor \frac{p-k}{2} \rfloor } \sum _{\beta = 0}^{\alpha }}_{\begin{array}{c} \beta \ge \frac{p-q+l-k}{2} \end{array}} (-1)^{\alpha -\beta } \left( {\begin{array}{c}p-k\\ 2\alpha \end{array}}\right) \left( {\begin{array}{c}\alpha \\ \beta \end{array}}\right) (2\alpha -1)!!\\&\quad \cdot (p+q-k-l-2\alpha -1)!! \\&\quad \cdot m_{11}^{p-2j-2\beta } m_{20}^{\frac{q-p}{2}+j+\beta } m_{02}^{j+\beta } = \\ \end{aligned}$$
$$\begin{aligned}&= m_{mn} - \mathop {\sum _{p = 0}^{n} \sum _{q = 0}^{m}}_{\begin{array}{c} p+q \ne 0,\\ p+q \ \mathrm {even} \end{array}} (-1)^{\frac{p+q}{2}} \left( {\begin{array}{c}m\\ q\end{array}}\right) \left( {\begin{array}{c}n\\ p\end{array}}\right) m_{m-q,n-p} \mathop {\sum _{k = 0}^{p} \sum _{l = 0}^{q}}_{\begin{array}{c} k+l \ \mathrm {even}\\ k+l \ne 0 \end{array}} (-1)^{\frac{k+l}{2}} \left( {\begin{array}{c}q\\ l\end{array}}\right) \left( {\begin{array}{c}p\\ k\end{array}}\right) \\&\quad \cdot \mathop {\sum _{i = 0}^{\lfloor \frac{k}{2} \rfloor } \sum _{j=0}^{i}}_{\begin{array}{c} j \ge \frac{k-l}{2} \end{array}} (-1)^{i-j} \left( {\begin{array}{c}k\\ 2i\end{array}}\right) \left( {\begin{array}{c}i\\ j\end{array}}\right) \\&\quad \cdot (2i-1)!! (l+k-2i-1)!! m_{11}^{k-2j} m_{20}^{\frac{l-k}{2}+j} m_{02}^{j} \mathop {\sum _{\alpha = 0}^{\lfloor \frac{p-k}{2} \rfloor } \sum _{\beta = 0}^{\alpha }}_{\begin{array}{c} \beta \ge \frac{p-q-k+l}{2} \end{array}} (-1)^{\alpha -\beta } \left( {\begin{array}{c}p-k\\ 2\alpha \end{array}}\right) \left( {\begin{array}{c}\alpha \\ \beta \end{array}}\right) \\&\quad \cdot (2\alpha -1)!! (p+q-k-l-2\alpha -1)!! \, m_{11}^{p-k-2\beta } m_{20}^{\frac{q-p+k-l}{2}+\beta } m_{02}^{\beta } = \\&= m_{mn} - \mathop {\sum _{p = 0}^{n} \sum _{q = 0}^{m}}_{\begin{array}{c} p+q \ne 0,\\ p+q \ \mathrm {even} \end{array}} (-1)^{\frac{p+q}{2}} \left( {\begin{array}{c}m\\ q\end{array}}\right) \left( {\begin{array}{c}n\\ p\end{array}}\right) m_{m-q,n-p} \\&\quad \cdot \Bigg [ \mathop {\sum _{ k =0}^{p} \sum _{l = 0}^{q}}_{\begin{array}{c} k+l \ \mathrm {even}\\ \end{array}} (-1)^{\frac{k+l}{2}} \left( {\begin{array}{c}p\\ k\end{array}}\right) \left( {\begin{array}{c}q\\ l\end{array}}\right) m_{l,k}^{(G)} m_{q-l,p-k}^{(G)} \\&\quad - \mathop {\sum _{\alpha = 0}^{\lfloor \frac{p}{2} \rfloor } \sum _{\beta = 0}^{\alpha }}_{\begin{array}{c} \beta \ge \frac{p-q}{2} \end{array}} (-1)^{\alpha -\beta } \left( {\begin{array}{c}p\\ 2\alpha \end{array}}\right) \left( {\begin{array}{c}\alpha \\ \beta \end{array}}\right) (2\alpha - 1)!! (p + q - 2\alpha - 1)!! \, m_{11}^{p-2\beta } m_{20}^{\frac{q-p}{2}+\beta } m_{02}^{\beta } \Bigg ] = \\&= \mathop {\sum _{l = 0}^{m} \sum _{k = 0}^{n}}_{ l+k \ \mathrm {even}} (-1)^{\frac{k+l}{2}} \left( {\begin{array}{c}m\\ l\end{array}}\right) \left( {\begin{array}{c}n\\ k\end{array}}\right) \mathop {\sum _{i = 0}^{\lfloor \frac{k}{2} \rfloor } \sum _{j = 0}^{i}}_{\begin{array}{c} j \ge \frac{k-l}{2} \end{array}} (-1)^{i-j} \left( {\begin{array}{c}k\\ 2i\end{array}}\right) \left( {\begin{array}{c}i\\ j\end{array}}\right) (l+k-2i-1)!! (2i-1)!! \\&\quad \cdot m_{11}^{k-2j} m_{20}^{\frac{l-k}{2}+j} m_{02}^{j}m_{m-l,n-k} . \end{aligned}$$
It remains to prove that for \(p+q>0,\, p+q \) even, it holds
$$\begin{aligned} \mathop {\sum _{k = 0}^{p} \sum _{l = 0}^{q}}_{\begin{array}{c} k+l \ \mathrm {even} \end{array}} (-1)^{\frac{k+l}{2}} \left( {\begin{array}{c}p\\ k\end{array}}\right) \left( {\begin{array}{c}q\\ l\end{array}}\right) m_{l,k}^{(f_{N})} m_{q-l,p-k}^{(f_{N})} = 0. \end{aligned}$$
(27)
For \(\frac{p+q}{2}\) being odd, the proof is trivial because every combination is present twice with the opposite signs. Thus, all terms vanish.
$$\begin{aligned} k = a, l= b\Rightarrow & {} (-1)^{\frac{a+b}{2}}\left( {\begin{array}{c}p\\ a\end{array}}\right) \left( {\begin{array}{c}q\\ b\end{array}}\right) m_{b,a}m_{q-b,p-a} \nonumber \\ k = p-a, l= q-b\Rightarrow & {} (-1)^{\frac{p+q-(a+b)}{2}}\left( {\begin{array}{c}p\\ p-a\end{array}}\right) \left( {\begin{array}{c}q\\ q-b\end{array}}\right) m_{q-b,p-a}m_{b,a} = \\&\quad -\left[ (-1)^{\frac{a+b}{2}}\left( {\begin{array}{c}p\\ a\end{array}}\right) \left( {\begin{array}{c}q\\ b\end{array}}\right) m_{b,a}m_{q-b,p-a}\right] \end{aligned}$$
For \(\frac{p+q}{2}\) even we have
$$\begin{aligned}&\mathop {\sum _{k = 0}^{p} \sum _{l = 0}^{q}}_{\begin{array}{c} k+l \ \mathrm {even} \end{array}} (-1)^{\frac{k+l}{2}} \left( {\begin{array}{c}p\\ k\end{array}}\right) \left( {\begin{array}{c}q\\ l\end{array}}\right) m_{l,k}^{(f_{N})} m_{q-l,p-k}^{(f_{N})} = \nonumber \\&\quad = \mathop {\sum _{k = 0}^{p} \sum _{l = 0}^{q}}_{\begin{array}{c} k+l \ \mathrm {even} \end{array}} (-1)^{\frac{k+l}{2}} \left( {\begin{array}{c}p\\ k\end{array}}\right) \left( {\begin{array}{c}q\\ l\end{array}}\right) \mathop {\sum _{i = 0}^{\lfloor \frac{k}{2} \rfloor } \sum _{t = 0}^{i}}_{\begin{array}{c} t \ge \frac{k-l}{2} \end{array}} (-1)^{i-t} \left( {\begin{array}{c}k\\ 2i\end{array}}\right) \left( {\begin{array}{c}i\\ t\end{array}}\right) (l+k-2i-1)!!\nonumber \\&\qquad \cdot (2i-1)!! \, m_{11}^{k-2t} m_{20}^{\frac{l-k}{2}+t} m_{02}^{t}\nonumber \\&\qquad \cdot \mathop {\sum _{s = 0}^{\lfloor \frac{p-k}{2} \rfloor } \sum _{r = 0}^{s}}_{\begin{array}{c} r \ge \frac{p-q-k+l}{2} \end{array}} (-1)^{s-r} \left( {\begin{array}{c}p-k\\ 2s\end{array}}\right) \left( {\begin{array}{c}s\\ r\end{array}}\right) (p+q-l-k-2s-1)!!\nonumber \\&\qquad \cdot (2s-1)!! \, m_{11}^{p-k-2r} m_{20}^{\frac{q-l-p+k}{2}+r} m_{02}^{r} = \nonumber \\&\qquad = \mathop {\sum _{k = 0}^{p} \sum _{l = 0}^{q} \sum _{i = 0}^{\lfloor \frac{k}{2} \rfloor } \sum _{t = 0}^{i} \sum _{s = 0}^{\lfloor \frac{p-k}{2} \rfloor } \sum _{r = 0}^{s}}_{\begin{array}{c} k+l \ \mathrm {even}\ \wedge \ t \ge \frac{k-l}{2}\ \wedge \ r \ge \frac{p-q-k+l}{2} \end{array}} \left( {\begin{array}{c}p\\ k\end{array}}\right) \left( {\begin{array}{c}q\\ l\end{array}}\right) \left( {\begin{array}{c}k\\ 2i\end{array}}\right) \left( {\begin{array}{c}i\\ t\end{array}}\right) \left( {\begin{array}{c}p-k\\ 2s\end{array}}\right) \left( {\begin{array}{c}s\\ r\end{array}}\right) (2i-1)!! (2s-1)!! \nonumber \\&\qquad \cdot (l+k-2i-1)!! (p+q-l-k-2s-1)!! (-1)^{\frac{k+l}{2} +i-t+s-r}\nonumber \\&\quad \quad \cdot m_{11}^{p-2t-2r} m_{20}^{\frac{q-p}{2} +r+t} m_{02}^{r+t} \nonumber \\&\quad = \mathop {\sum _{k = 0}^{p} \sum _{l = 0}^{q} \sum _{t = 0}^{\lfloor \frac{k}{2} \rfloor } \sum _{i = t}^{\lfloor \frac{k}{2} \rfloor } \sum _{s = 0}^{\lfloor \frac{p-k}{2} \rfloor } \sum _{r = 0}^{s}}_{\begin{array}{c} k+l \ \mathrm {even}\ \wedge \ t \ge \frac{k-l}{2} \ \wedge \ r \ge \frac{p-q-k+l}{2} \end{array}} \left( {\begin{array}{c}p\\ k\end{array}}\right) \left( {\begin{array}{c}q\\ l\end{array}}\right) \left( {\begin{array}{c}k\\ 2i\end{array}}\right) \left( {\begin{array}{c}i\\ t\end{array}}\right) \left( {\begin{array}{c}p-k\\ 2s\end{array}}\right) \left( {\begin{array}{c}s\\ r\end{array}}\right) (2i-1)!! (2s-1)!!\nonumber \\&\quad \quad \cdot (l+k-2i-1)!! (p+q-l-k-2s-1)!! (-1)^{\frac{k+l}{2} +i-t+s-r} m_{11}^{p-2t-2r} m_{20}^{\frac{q-p}{2} +r+t} m_{02}^{r+t}\nonumber \\ \end{aligned}$$
$$\begin{aligned}&\quad =\left| \begin{matrix} k = 0 : p,\ t = 0: \lfloor \frac{k}{2} \rfloor \Rightarrow \\ t = 0 : \lfloor \frac{p}{2} \rfloor , \ k = 2t : p \end{matrix} \right| \left. \begin{matrix} s = 0 : \lfloor \frac{p-k}{2} \rfloor ,\ r = 0 : s \Rightarrow \\ r = 0 : \lfloor \frac{p-k}{2} \rfloor ,\ s = r : \lfloor \frac{p-k}{2} \rfloor \end{matrix} \right| =\nonumber \\&\quad = \sum _{t = 0}^{\lfloor \frac{p}{2} \rfloor } \mathop {\sum _{k = 2t}^{p} \sum _{l = 0}^{q} \sum _{i = t}^{\lfloor \frac{k}{2} \rfloor } \sum _{r = 0}^{\lfloor \frac{p-k}{2} \rfloor } \sum _{s = r}^{\lfloor \frac{p-k}{2} \rfloor }}_{\begin{array}{c} k+l \ \mathrm {even}\ \wedge \ t \ge \frac{k-l}{2} \ \wedge \ r \ge \frac{p-q-k+l}{2} \end{array}} \left( {\begin{array}{c}p\\ k\end{array}}\right) \left( {\begin{array}{c}q\\ l\end{array}}\right) \left( {\begin{array}{c}k\\ 2i\end{array}}\right) \left( {\begin{array}{c}i\\ t\end{array}}\right) \left( {\begin{array}{c}p-k\\ 2s\end{array}}\right) \left( {\begin{array}{c}s\\ r\end{array}}\right) (2i-1)!! (2s-1)!! \nonumber \\&\qquad \cdot (l+k-2i-1)!! (p+q-l-k-2s-1)!! (-1)^{\frac{k+l}{2} +i-t+s-r} m_{11}^{p-2t-2r}\nonumber \\&\qquad \cdot m_{20}^{\frac{q-p}{2} +r+t} m_{02}^{r+t} \nonumber \\&\quad = \left| \begin{matrix} k = 2t : p,\ r = 0 : \lfloor \frac{p-k}{2} \rfloor \Rightarrow \\ r = 0 : \lfloor \frac{p-2t}{2} \rfloor ,\ k = 2t : p-2r \end{matrix} \right| =\nonumber \\&\quad = \mathop {\sum _{t = 0}^{\lfloor \frac{p}{2} \rfloor } \sum _{r = 0}^{\lfloor \frac{p-2t}{2} \rfloor } \sum _{k = 2t}^{p-2r} \sum _{l = 0}^{q} \sum _{i = t}^{\lfloor \frac{k}{2} \rfloor } \sum _{s = r}^{\lfloor \frac{p-k}{2} \rfloor }}_{\begin{array}{c} k+l \ \mathrm {even} \ \wedge \ t \ge \frac{k-l}{2} \ \wedge \ r \ge \frac{p-q-k+l}{2} \end{array}} \left( {\begin{array}{c}p\\ k\end{array}}\right) \left( {\begin{array}{c}q\\ l\end{array}}\right) \left( {\begin{array}{c}k\\ 2i\end{array}}\right) \left( {\begin{array}{c}i\\ t\end{array}}\right) \left( {\begin{array}{c}p-k\\ 2s\end{array}}\right) \left( {\begin{array}{c}s\\ r\end{array}}\right) (2i-1)!! (2s-1)!! \nonumber \\&\quad \cdot (l+k-2i-1)!! (p+q-l-k-2s-1)!! (-1)^{\frac{k+l}{2} +i-t+s-r} m_{11}^{p-2t-2r}\nonumber \\&\quad \quad \cdot m_{20}^{\frac{q-p}{2}+r+t} m_{02}^{r+t}\nonumber \\&= \sum _{t = 0}^{\lfloor \frac{p}{2} \rfloor } \sum _{r = 0}^{\lfloor \frac{p-2t}{2} \rfloor } m_{11}^{p-2t-2r} m_{20}^{\frac{q-p}{2}+r+t} m_{02}^{r+t} \mathop {\sum _{k = 2t}^{p-2r} \sum _{l = 0}^{q} \sum _{i = t}^{\lfloor \frac{k}{2} \rfloor } \sum _{s = r}^{\lfloor \frac{p-k}{2} \rfloor }}_{\begin{array}{c} k+l \ \mathrm {even} \ \wedge \ t \ge \frac{k-l}{2} \ \wedge \ r \ge \frac{p-q-k+l}{2} \end{array}} (-1)^{\frac{k+l}{2} +i-t+s-r} \left( {\begin{array}{c}p\\ k\end{array}}\right) \left( {\begin{array}{c}q\\ l\end{array}}\right) \nonumber \\&\quad \quad \cdot \left( {\begin{array}{c}k\\ 2i\end{array}}\right) \left( {\begin{array}{c}i\\ t\end{array}}\right) \left( {\begin{array}{c}p-k\\ 2s\end{array}}\right) \left( {\begin{array}{c}s\\ r\end{array}}\right) (2i-1)!! (2s-1)!! (l{+}k{-}2i{-}1)!! (p+q-l-k-2s-1)!!\nonumber \\&\quad = \left| \begin{matrix} t + r = N \Rightarrow r = N - t \\ t = 0 : \lfloor \frac{p}{2} \rfloor \Rightarrow \\ N = t : \lfloor \frac{p}{2} \rfloor \end{matrix} \right| = \nonumber \\ \end{aligned}$$
$$\begin{aligned}&\quad = \sum _{t = 0}^{\lfloor \frac{p}{2} \rfloor } \sum _{N = t}^{\lfloor \frac{p}{2} \rfloor } m_{11}^{p-2N} m_{20}^{\frac{q{-}p}{2} {+}N} m_{02}^{N} \mathop {\sum _{k = 2t}^{p-2N+2t} \sum _{l = 0}^{q} \sum _{i = t}^{\lfloor \frac{k}{2} \rfloor } \sum _{s = N-t}^{\lfloor \frac{p-k}{2} \rfloor }}_{\begin{array}{c} k+l \ \mathrm {even} \wedge N-\frac{p-k-q+l}{2} {\ge } t \ge \frac{k-l}{2} \end{array}} ({-}1)^{\frac{k+l}{2} +i+s-N} \left( {\begin{array}{c}p\\ k\end{array}}\right) \left( {\begin{array}{c}q\\ l\end{array}}\right) \left( {\begin{array}{c}k\\ 2i\end{array}}\right) \left( {\begin{array}{c}i\\ t\end{array}}\right) \nonumber \\&\quad \quad \cdot \left( {\begin{array}{c}p-k\\ 2s\end{array}}\right) \left( {\begin{array}{c}s\\ N-t\end{array}}\right) (2i-1)!! (2s-1)!! (l+k-2i-1)!! (p+q-l-k-2s-1)!! \nonumber \\&\quad = \left| \begin{matrix} t = 0 : \lfloor \frac{p}{2} \rfloor , N = t : \lfloor \frac{p}{2} \rfloor \Rightarrow \\ N = 0 : \lfloor \frac{p}{2} \rfloor , t = 0 : N \end{matrix} \right| = \nonumber \\&\quad = \sum _{N = 0}^{\lfloor \frac{p}{2} \rfloor } \sum _{t = 0}^{N} m_{11}^{p-2N} m_{20}^{\frac{q-p}{2}+N} m_{02}^{N} \mathop {\sum _{k = 2t}^{p-2N+2t} \sum _{l = 0}^{q} \sum _{i = t}^{\lfloor \frac{k}{2} \rfloor } \sum _{s = N-t}^{\lfloor \frac{p-k}{2} \rfloor }}_{\begin{array}{c} k+l \ \mathrm {even} \ \wedge \ N-\frac{p-k-q+l}{2} \ge t \ge \frac{k-l}{2} \end{array}} (-1)^{\frac{k+l}{2} +i+s-N} \left( {\begin{array}{c}p\\ k\end{array}}\right) \left( {\begin{array}{c}q\\ l\end{array}}\right) \left( {\begin{array}{c}i\\ t\end{array}}\right) \nonumber \\&\qquad \cdot \left( {\begin{array}{c}k\\ 2i\end{array}}\right) \left( {\begin{array}{c}p-k\\ 2s\end{array}}\right) \left( {\begin{array}{c}s\\ N-t\end{array}}\right) (2i-1)!! (2s-1)!! (l{+}k-2i{-}1)!! (p+q{-}l{-}k{-}2s-1)!!= \nonumber \\&= \left| \begin{matrix} k - 2t = m \Rightarrow k = m + 2t \\ k = 2t : p - 2N + 2t \Rightarrow \\ m = 0 : p - 2N \end{matrix} \right| =\nonumber \\&= \sum _{N = 0}^{\lfloor \frac{p}{2} \rfloor } m_{11}^{p-2N} m_{20}^{\frac{q-p}{2} +N} m_{02}^{N} \mathop {\sum _{t = 0}^{N} \sum _{m = 0}^{p-2N} \sum _{l = 0}^{q} \sum _{i = t}^{\lfloor \frac{m+2t}{2} \rfloor } \sum _{s = N-t}^{\lfloor \frac{p-m-2t}{2} \rfloor }}_{\begin{array}{c} m+l \ \mathrm {even} \ \wedge \ N-\frac{p-q}{2} \ge \frac{l-m}{2} \ge 0 \end{array}} (-1)^{\frac{m+l}{2} +t+i+s-N}\nonumber \\&\quad \cdot \left( {\begin{array}{c}p\\ m+2t\end{array}}\right) \left( {\begin{array}{c}q\\ l\end{array}}\right) \left( {\begin{array}{c}m+2t\\ 2i\end{array}}\right) \left( {\begin{array}{c}i\\ t\end{array}}\right) \nonumber \\&\quad \cdot \left( {\begin{array}{c}s\\ N-t\end{array}}\right) \left( {\begin{array}{c}p-m-2t\\ 2s\end{array}}\right) (2i-1)!! (2s-1)!!\nonumber \\&\quad \cdot (l+m + 2t-2i-1)!! (p+q-l-m - 2t-2s-1)!! \nonumber \\&= \left| \begin{matrix} i - t {=} j \Rightarrow i = j + t \\ i {=} t : \lfloor \frac{m+2t}{2} \rfloor \Rightarrow \\ j {=} 0 : \lfloor \frac{m}{2} \rfloor \end{matrix} \right| = \nonumber \\ \end{aligned}$$
$$\begin{aligned}&= \sum _{N = 0}^{\lfloor \frac{p}{2} \rfloor } m_{11}^{p-2N} m_{20}^{\frac{q-p}{2} +N} m_{02}^{N} \mathop {\sum _{t = 0}^{N} \sum _{m = 0}^{p-2N} \sum _{l = 0}^{q} \sum _{j = 0}^{\lfloor \frac{m}{2} \rfloor } \sum _{s = N-t}^{\lfloor \frac{p-m-2t}{2} \rfloor }}_{\begin{array}{c} m+l \ \mathrm {even} \ \wedge \ N-\frac{p-q}{2} \ge \frac{l-m}{2} \ge 0 \end{array}} (-1)^{\frac{m+l}{2} +j+2t+s-N}\nonumber \\&\quad \cdot \left( {\begin{array}{c}p\\ m+2t\end{array}}\right) \left( {\begin{array}{c}q\\ l\end{array}}\right) \left( {\begin{array}{c}m+2t\\ 2j+2t\end{array}}\right) \left( {\begin{array}{c}j+t\\ t\end{array}}\right) \nonumber \\&\quad \cdot \left( {\begin{array}{c}s\\ N-t\end{array}}\right) \left( {\begin{array}{c}p-m-2t\\ 2s\end{array}}\right) (2j+2t-1)!! (2s-1)!! (l+m -2j-1)!!\nonumber \\&\quad \cdot (p+q-l-m - 2t-2s-1)!! \nonumber \\&= \left| \begin{matrix} s - N + t = k \Rightarrow s = N - t + k \\ s = N - t : \lfloor \frac{p-2m-2t}{2} \rfloor \Rightarrow \\ k = 0 : \lfloor \frac{p-2m-2N}{2} \rfloor \end{matrix} \right| = \nonumber \\&= \sum _{N = 0}^{\lfloor \frac{p}{2} \rfloor } m_{11}^{p-2N} m_{20}^{\frac{q-p}{2} +N} m_{02}^{N} \mathop {\sum _{t = 0}^{N} \sum _{m = 0}^{p-2N} \sum _{l = 0}^{q} \sum _{j = 0}^{\lfloor \frac{m}{2} \rfloor } \sum _{k = 0}^{\lfloor \frac{p-m-2N}{2} \rfloor }}_{\begin{array}{c} m+l \ \mathrm {even} \ \wedge \ N-\frac{p-q}{2} \ge \frac{l-m}{2} \ge 0 \end{array}} (-1)^{\frac{m+l}{2} +j+k+t} \left( {\begin{array}{c}p\\ m+2t\end{array}}\right) \left( {\begin{array}{c}q\\ l\end{array}}\right) \nonumber \\&\quad \cdot \left( {\begin{array}{c}m+2t\\ 2j+2t\end{array}}\right) \left( {\begin{array}{c}j+t\\ t\end{array}}\right) \left( {\begin{array}{c}N-t+k\\ N-t\end{array}}\right) \left( {\begin{array}{c}p-m-2t\\ 2N-2t+2k\end{array}}\right) (2j+2t-1)!! (2N-2t+2k-1)!! \nonumber \\&\quad \cdot (l+m -2j-1)!! (p+q-l-m - 2N-2k-1)!! \nonumber \\&= \sum _{N = 0}^{\lfloor \frac{p}{2} \rfloor } m_{11}^{p-2N} m_{20}^{\frac{q-p}{2} +N} m_{02}^{N} \mathop {\sum _{t = 0}^{N} \sum _{m = 0}^{p-2N} \sum _{l = 0}^{q} \sum _{j = 0}^{\lfloor \frac{m}{2} \rfloor } \sum _{k = 0}^{\lfloor \frac{p-m-2N}{2} \rfloor }}_{\begin{array}{c} m+l \ \mathrm {even} \ \wedge \ N-\frac{p-q}{2} \ge \frac{l-m}{2} \ge 0 \end{array}} (-1)^{\frac{m+l}{2} +j+k+t} \left( {\begin{array}{c}q\\ l\end{array}}\right) p! \nonumber \\&\quad \cdot \frac{(2j+2t-1)!!}{(2j+2t)!(m-2j)!} \frac{(j+t)!}{t!j!} \frac{(N-t+k)!}{(N-t)!k!} \frac{(2N-2t+2k-1)!!}{(2N-2t+2k)!(p-m-2N-2k)!} \nonumber \\&\quad \cdot (l+m -2j-1)!! (p+q-l-m - 2N-2k-1)!! \nonumber \\&= \sum _{N = 0}^{\lfloor \frac{p}{2} \rfloor } m_{11}^{p-2N} m_{20}^{\frac{q-p}{2} +N} m_{02}^{N} \mathop {\sum _{t = 0}^{N} \sum _{m = 0}^{p-2N} \sum _{l = 0}^{q} \sum _{j = 0}^{\lfloor \frac{m}{2} \rfloor } \sum _{k = 0}^{\lfloor \frac{p-m-2N}{2} \rfloor }}_{\begin{array}{c} m+l \ \mathrm {even} \ \wedge \ N-\frac{p-q}{2} \ge \frac{l-m}{2} \ge 0 \end{array}} (-1)^{\frac{m+l}{2} +j+k+t} \left( {\begin{array}{c}q\\ l\end{array}}\right) p! \nonumber \\&\quad \cdot \frac{1}{(2j+2t)!!(m-2j)!} \frac{(j+t)!}{t!j!} \frac{(N-t+k)!}{(N-t)!k!} \frac{1}{(2N-2t+2k)!!(p-m-2N-2k)!} \end{aligned}$$
$$\begin{aligned}&\quad \cdot (l+m -2j-1)!! (p+q-l-m - 2N-2k-1)!! \nonumber \\&= \sum _{N = 0}^{\lfloor \frac{p}{2} \rfloor } m_{11}^{p-2N} m_{20}^{\frac{q-p}{2} +N} m_{02}^{N} \mathop {\sum _{t = 0}^{N} \sum _{m = 0}^{p-2N} \sum _{l = 0}^{q} \sum _{j = 0}^{\lfloor \frac{m}{2} \rfloor } \sum _{k = 0}^{\lfloor \frac{p-m-2N}{2} \rfloor }}_{\begin{array}{c} m+l \ \mathrm {even} \ \wedge \ N-\frac{p-q}{2} \ge \frac{l-m}{2} \ge 0 \end{array}} (-1)^{\frac{m+l}{2} +j+k+t} \left( {\begin{array}{c}q\\ l\end{array}}\right) p! \nonumber \\&\quad \cdot \frac{1}{2^{j+t}(j+t)!(m-2j)!} \frac{(j+t)!}{t!j!} \frac{(N-t+k)!}{(N-t)!k!} \frac{1}{2^{N-t+k}(N-t+k)!(p-m-2N-2k)!} \nonumber \\&\quad \cdot (l+m -2j-1)!! (p+q-l-m - 2N-2k-1)!! \nonumber \\&= \sum _{N = 0}^{\lfloor \frac{p}{2} \rfloor } m_{11}^{p-2N} m_{20}^{\frac{q-p}{2} +N} m_{02}^{N} \mathop {\sum _{t = 0}^{N} \sum _{m = 0}^{p-2N} \sum _{l = 0}^{q} \sum _{j = 0}^{\lfloor \frac{m}{2} \rfloor } \sum _{k = 0}^{\lfloor \frac{p-m-2N}{2} \rfloor }}_{\begin{array}{c} m+l \ \mathrm {even} \ \wedge \ N-\frac{p-q}{2} \ge \frac{l-m}{2} \ge 0 \end{array}} (-1)^{\frac{m+l}{2} +j+k+t} \left( {\begin{array}{c}q\\ l\end{array}}\right) p! \nonumber \\&\quad \cdot \frac{1}{2^{j}j!(m-2j)!} \frac{1}{2^{N+k}k!(p-m-2N-2k)!} \frac{1}{t!(N-t)!} \nonumber \\&\quad \cdot (l+m -2j-1)!! (p+q-l-m - 2N-2k-1)!! \nonumber \\&= \sum _{N = 0}^{\lfloor \frac{p}{2} \rfloor } m_{11}^{p-2N} m_{20}^{\frac{q-p}{2} +N} m_{02}^{N} \mathop {\sum _{m = 0}^{p-2N} \sum _{l = 0}^{q} \sum _{j = 0}^{\lfloor \frac{m}{2} \rfloor } \sum _{k = 0}^{\lfloor \frac{p-m-2N}{2} \rfloor }}_{\begin{array}{c} m+l \ \mathrm {even} \ \wedge \ N-\frac{p-q}{2} \ge \frac{l-m}{2} \ge 0 \end{array}} (-1)^{\frac{m+l}{2} +j+k} \left( {\begin{array}{c}q\\ l\end{array}}\right) p! \nonumber \\&\quad \cdot \frac{1}{(2j)!!(m-2j)!} \frac{1}{(2k)!!(p-m-2N-2k)!2^{N}N!} \nonumber \\&\quad \cdot (l+m -2j-1)!! (p+q-l-m - 2N-2k-1)!! \sum _{t = 0}^{N} (-1)^t \left( {\begin{array}{c}N\\ t\end{array}}\right) = \end{aligned}$$
(28)
$$\begin{aligned}&= \sum _{N = 0}^{\lfloor \frac{p}{2} \rfloor } m_{11}^{p-2N} m_{20}^{\frac{q-p}{2} +N} m_{02}^{N} \mathop {\sum _{m = 0}^{p-2N} \sum _{l = 0}^{q} \sum _{j = 0}^{\lfloor \frac{m}{2} \rfloor } \sum _{k = 0}^{\lfloor \frac{p-m-2N}{2} \rfloor }}_{\begin{array}{c} m+l \ \mathrm {even} \ \wedge \ N-\frac{p-q}{2} \ge \frac{l-m}{2} \ge 0 \end{array}} (-1)^{\frac{m+l}{2} +j+k} \left( {\begin{array}{c}q\\ l\end{array}}\right) p! \nonumber \\&\quad \cdot \frac{(2j-1)!!}{(2j)!(m-2j)!} \frac{(2k-1)!!}{(2k)!(p-m-2N-2k)!(2N)!!} \nonumber \\&\quad \cdot (l+m -2j-1)!! (p+q-l-m - 2N-2k-1)!! (1-1)^N . \end{aligned}$$
(29)
All the terms of (29) are zero if \(N>0\). If \(N=0\), there remains the last term only
$$\begin{aligned}&\mathop {\sum _{m = 0}^{p} \sum _{l = 0}^{q} \sum _{j = 0}^{\lfloor \frac{m}{2} \rfloor } \sum _{k = 0}^{\lfloor \frac{p-m}{2} \rfloor }}_{\begin{array}{c} m+l \ \mathrm {even} \ \wedge \ q-p \ge l-m \ge 0 \end{array}} (-1)^{\frac{m+l}{2} +j+k} \left( {\begin{array}{c}p\\ m\end{array}}\right) \left( {\begin{array}{c}q\\ l\end{array}}\right) \left( {\begin{array}{c}m\\ 2j\end{array}}\right) \left( {\begin{array}{c}p-m\\ 2k\end{array}}\right) \nonumber \\&\quad \cdot (2j-1)!! (2k-1)!! (l+m -2j-1)!! (p+q-l-m-2k-1)!!. \end{aligned}$$
(30)
Now we prove that this term is zero as well. This term is equivalent to
$$\begin{aligned}&\mathop {\sum _{m = 0}^{p} \sum _{l = 0}^{q}}_{\begin{array}{c} m+l \ \mathrm {even} \\ q-p \ge l-m \ge 0 \end{array}} (-1)^{\frac{m+l}{2}} \left( {\begin{array}{c}p\\ m\end{array}}\right) \left( {\begin{array}{c}q\\ l\end{array}}\right) \sum _{j = 0}^{\lfloor \frac{m}{2} \rfloor } (-1)^j \left( {\begin{array}{c}m\\ 2j\end{array}}\right) (l+m-2j-1)!! (2j-1)!! \nonumber \\&\quad \cdot \sum _{k = 0}^{\lfloor \frac{p-m}{2} \rfloor } (-1)^k \left( {\begin{array}{c}p-m\\ 2k\end{array}}\right) (p+q-l-m-2k-1)!! (2k-1)!! = \varXi . \end{aligned}$$
(31)
It can be shown for \( l \ge m\) using the method of generating functions described in Gould and Quaintance (2012) that
$$\begin{aligned} \sum _{j = 0}^{\lfloor \frac{m}{2} \rfloor } (-1)^j \left( {\begin{array}{c}m\\ 2j\end{array}}\right) (l+m-2j-1)!! (2j-1)!! = \frac{l!}{(l-m)!!}. \end{aligned}$$
(32)
We adopt the notation from Gould and Quaintance (2012) for double factorial binomial coefficients and we recall \((p+q)/2\) is even. The previous expression can be rewritten
$$\begin{aligned} \varXi&= \mathop {\sum _{m = 0}^{p} \sum _{l = 0}^{q}}_{\begin{array}{c} m+l \ \mathrm {even} \\ q-p \ge l-m \ge 0 \end{array}} (-1)^{\frac{m+l}{2}} \left( {\begin{array}{c}p\\ m\end{array}}\right) \left( {\begin{array}{c}q\\ l\end{array}}\right) \frac{l!}{(l-m)!!} \frac{(q-l)!}{(q-l-p+m)!!} = \nonumber \\&= \frac{q!}{(q-p)!!} \mathop {\sum _{m = 0}^{p} \sum _{l = m}^{q-p+m}}_{\begin{array}{c} m+l \ \mathrm {even} \\ q-p \ge l-m \ge 0 \end{array}} (-1)^{\frac{m+l}{2}} \left( {\begin{array}{c}p\\ m\end{array}}\right) \left( \left( {\begin{array}{c}q-p\\ l-m\end{array}}\right) \right) = \left| \begin{matrix} l - m = 2j \\ j = 0 : \frac{q-p}{2} \end{matrix} \right| = \nonumber \\&= \frac{q!}{(q-p)!!} \mathop {\sum _{m = 0}^{p} \sum _{j = 0}^{\frac{q-p}{2}}} (-1)^{j+m} \left( {\begin{array}{c}p\\ m\end{array}}\right) \left( \left( {\begin{array}{c}q-p\\ 2j\end{array}}\right) \right) = \nonumber \\&= \frac{q!}{(q-p)!!} \mathop {\sum _{m = 0}^{p}} (-1)^{m} \left( {\begin{array}{c}p\\ m\end{array}}\right) \sum _{j = 0}^{\frac{q-p}{2}} (-1)^j \left( \left( {\begin{array}{c}q-p\\ 2j\end{array}}\right) \right) \end{aligned}$$
(33)
The inner sum is zero if \(q > p\)
$$\begin{aligned}&\sum _{j = 0}^{\frac{q-p}{2}} (-1)^j \left( \left( {\begin{array}{c}q-p\\ 2j\end{array}}\right) \right) = \sum _{j = 0}^{\frac{q-p}{2}} (-1)^j \frac{(q-p)!!}{(2j)!!(q-p-2j)!!}\nonumber \\&\quad = \sum _{j = 0}^{\frac{q-p}{2}} (-1)^j \left( {\begin{array}{c}\frac{q-p}{2}\\ j\end{array}}\right) = (1-1)^{\frac{q-p}{2}} = 0. \end{aligned}$$
(34)
For the case \(q = p\) (\(q - p\) must be non-negative) the inner sum equals 1 and the expression (33) is
$$\begin{aligned} p! \mathop {\sum _{m = 0}^{p}} (-1)^{m} \left( {\begin{array}{c}p\\ m\end{array}}\right) = p! (1-1)^p \end{aligned}$$
(35)
which completes the proof because it is zero whenever \(p > 0\).
The formula
$$\begin{aligned} \mathop {\sum _{k = 0}^{p} \sum _{l = 0}^{q}}_{\begin{array}{c} k+l \ \mathrm {even} \end{array}} (-1)^{\frac{k+l}{2}} \left( {\begin{array}{c}p\\ k\end{array}}\right) \left( {\begin{array}{c}q\\ l\end{array}}\right) m_{l,k}^{(f_{N})} m_{q-l,p-k}^{(f_{N})} = 0 \end{aligned}$$
(36)
holds not only for \(p + q\) even but for all p and q. If \(p + q\) is odd, then \(m_{q-l,p-k}^{(f_{N})}\) is Gaussian moment of the odd order and all the terms in summation are zero. \(\square \)
