Appendix
Proof
We only show the proof of Theorem 1 (a), because this proof can be applied to the proof of Theorem 1 (b) in an obvious way. When a new observation is from \(\pi _1\), the conditional distribution of U given \(\varvec{\mu }_1\), \(\varvec{\mu }_2\) and \(\varvec{\varSigma }\) is \(U | \varvec{\mu }_1, \varvec{\mu }_2, \varvec{\varSigma } , \mathbf{y} \in \pi _1 \sim N ( \varDelta ^2/2, \varDelta ^2)\) with \(\varDelta ^2 = (\varvec{\mu }_1 - \varvec{\mu }_2)' \varvec{\varSigma } ^{ - 1 } (\varvec{\mu }_1 - \varvec{\mu }_2)\). Define \( v = \varDelta ^2/c \) with \(c = N_1^{ - 1 } + N_2^{ - 1 }\). Since
$$\begin{aligned} \frac{ 1 }{ \sqrt{ c } }\varvec{\varSigma }^{-1/2}( \varvec{\mu }_1 - \varvec{\mu }_2) | \varvec{\varSigma } \sim N_p\left( \frac{ 1 }{ \sqrt{c} } \varvec{\varSigma } ^{-1/2}( \bar{ \mathbf{x} }_1 - \bar{ \mathbf{x} }_2), \varvec{I}_p \right) , \end{aligned}$$
the conditional distribution of \(v | \varvec{\varSigma } \) is the non-central \(\chi ^2\) distribution with non-centrality parameter \( \delta = (\bar{ \mathbf{x} }_1 - \bar{ \mathbf{x} }_2 )' \varvec{\varSigma } ^{ - 1 } (\bar{ \mathbf{x} }_1 - \bar{ \mathbf{x} }_2 )/c. \) This is expressed as
$$\begin{aligned} v | \varvec{\varSigma } \sim \chi ^2_p \left( \frac{Q}{nc}z \right) , \end{aligned}$$
where
$$\begin{aligned} Q = (\bar{ \mathbf{x} }_1 - \bar{ \mathbf{x} }_2 )' \varvec{S}^{ - 1 } (\bar{ \mathbf{x} }_1 - \bar{ \mathbf{x} }_2 ),~ z = \frac{(\bar{ \mathbf{x} }_1 - \bar{ \mathbf{x} }_2 )' \varvec{\varSigma } ^{ - 1 } (\bar{ \mathbf{x} }_1 - \bar{ \mathbf{x} }_2) }{(\bar{ \mathbf{x} }_1 - \bar{ \mathbf{x} }_2 )' (n\varvec{S})^{ - 1 } (\bar{ \mathbf{x} }_1 - \bar{ \mathbf{x} }_2 )}. \end{aligned}$$
Since z is distributed as \(\chi ^2_m\) from Theorem 3.2.8 of Muirhead (1982), the unconditional distribution of U is given by
$$\begin{aligned} f ( u| \mathbf{y} \in \pi _1 )&{=} \int _0^\infty \int _0^\infty \frac{ 1 }{ \sqrt{ c v } } \phi \left( \frac{ u - cv/2 }{ \sqrt{ c v } } \right) g_p( v ) \exp \left( - \frac{ Q }{ 2nc } z \right) {_0}F_1 \left( \frac{ p }{ 2 } ; \frac{ Q }{ 4nc } v z \right) \\&\quad \times g_m( z ) \mathrm{{d}}v \mathrm{{d}}z. \end{aligned}$$
Using Lemma 1.3.3 of Muirhead (1982), the desired result follows immediately. \(\square \)
Proof of Theorem 2.
Proof
We derive the first four moments of U using the expression of density given in Theorem 1. If we define
$$\begin{aligned} W( k )&= \left( \frac{ nc }{ nc + Q } \right) ^{ n / 2 } \int _0^\infty \frac{ v^{ p/2 + k - 1 } e^{ - v/2 } }{ 2^{ p/2 } \varGamma ( p/2 ) } {_1}F_1 \left( \frac{ n }{ 2 } ; \frac{ p }{ 2 } ; \frac{ Q }{ 2 ( nc + Q ) } v \right) \mathrm{{d}}v, \end{aligned}$$
(12)
for a non-negative number k, then the first four moments of U can be represented by
$$\begin{aligned} \mathrm {E}(U)&= \frac{ c }{ 2 } W( 1 ), \nonumber \\ \mathrm {E}(U^2)&= \frac{ c^2 }{ 4 } W( 2 ) + c W( 1 ), \nonumber \\ \mathrm {E}(U^3)&= \frac{ c^3 }{ 8 } W( 3 ) + \frac{ 3c^2 }{ 2 } W( 2 ), \nonumber \\ \mathrm {E}(U^4)&= \frac{ c^4 }{ 16 } W( 4 ) + \frac{ 3c^3 }{ 2 } W( 3 ) + 3c^2 W( 2 ). \end{aligned}$$
(13)
Hence, the central moments of U are expressed as
$$\begin{aligned} \mathrm {V}(U)&= \frac{ c^2 }{ 4 } ( W( 2 ) - W( 1 )^2 ) + c W( 1 ), \end{aligned}$$
(14)
$$\begin{aligned} \mathrm {E}(U - \mathrm {E}(U) )^3&= \frac{ c^3 }{ 8 } W( 3 ) + \frac{ 3c^2 }{ 2 } W( 2 ) - \frac{ 3c^3 }{ 8 } W( 2 )W( 1 ) \nonumber \\&\quad - \frac{ 3c^2 }{ 2 } W( 1 )^2 + \frac{ c^3 }{ 4 } W( 1 )^3, \end{aligned}$$
(15)
$$\begin{aligned} \mathrm {E}(U - \mathrm {E}(U) )^4&= \frac{ c^4 }{ 16 } W( 4 ) + \frac{ 3c^3 }{ 2 } W( 3 ) + 3c^2 W( 2 ) \nonumber \\&\quad - \frac{ c^4 }{ 4 } W( 3 )W( 1 ) - 3c^3 W( 2 )W( 1 ) + \frac{ 3c^4 }{ 8 } W( 2 )W( 1 )^2 \nonumber \\&\quad + \frac{ 3c^3 }{ 2 } W( 1 )^3 - \frac{ 3c^4 }{ 16 } W( 1 )^4. \end{aligned}$$
(16)
Applying Lemma 1.3.3 of Muirhead (1982) to (12),
$$\begin{aligned} W(k)&= \left( \frac{ nc }{ nc + Q } \right) ^{ n / 2 } \frac{ \varGamma ( p/2 + k ) }{ \varGamma ( p/2 ) } 2^k {_2}F_1 \left( \frac{ n }{ 2 }, \frac{ p }{ 2 } + k ; \frac{ p }{ 2 } ; \frac{ Q }{ nc + Q } \right) . \end{aligned}$$
(17)
Using equality
$$\begin{aligned} {_2}F_1 (a, b ; c ; z) = ( 1 - z )^{ -a} {_2}F_1 \left( a, c - b ; c ; \frac{ z }{ z - 1 } \right) \end{aligned}$$
in (17),
$$\begin{aligned} W(k)&= \frac{ \varGamma ( p/2 + k ) }{ \varGamma ( p/2 ) } 2^k {_2}F_1 \left( \frac{ n }{ 2 }, - k ; \frac{ p }{ 2 } ; -\frac{ Q }{ nc } \right) \nonumber \\&= \frac{ \varGamma ( p/2 + k ) }{ \varGamma ( p/2 ) } 2^k \sum _{r = 0}^k \frac{ ( n/2 )_r ( -k )_r }{ ( p/2 )_r } \frac{ 1 }{ r! } \left( - \frac{ Q }{ nc } \right) ^r. \end{aligned}$$
(18)
If we put \(k=\)1, 2, 3 and 4 into (18), then
$$\begin{aligned} W( 1 )&= p \left( 1 + \frac{ Q }{ pc } \right) , \\ W( 2 )&= p( p + 2 ) \left[ 1 + \frac{ 2Q }{ pc } + \frac{ n( n + 2 ) }{ p( p + 2 ) } \left( \frac{ Q }{ nc } \right) ^2 \right] , \\ W( 3 )&= p( p + 2 )( p + 4 ) \left[ 1 + \frac{ 3Q }{ pc } + \frac{ 3n( n + 2 ) }{ p( p + 2 ) } \left( \frac{ Q }{ nc } \right) ^2 + \frac{ n( n + 2 )( n + 4 ) }{ p( p + 2 )( p + 4 ) } \left( \frac{ Q }{ nc } \right) ^3 \right] , \\ W( 4 )&= p( p + 2 )( p + 4 )( p + 6 ) \left[ 1 + \frac{ 4Q }{ pc } + \frac{ 6n( n + 2 ) }{ p( p + 2 ) } \left( \frac{ Q }{ nc } \right) ^2 \right. \\&\quad \left. +4 \frac{ n( n + 2 )( n + 4 ) }{ p( p + 2 )( p + 4 ) } \left( \frac{ Q }{ nc } \right) ^3 + \frac{ n( n + 2 )( n + 4 )( n + 6 ) }{ p( p + 2 )( p + 4 )( p + 6 ) } \left( \frac{ Q }{ nc } \right) ^4 \right] . \end{aligned}$$
If we substitute these for (13), (14), (15) and (16), and then make arrangements of the equations, mean, variance, skewness and kurtosis of U can be obtained. \(\square \)
Proof of Theorem 3.
Proof
We only show the proof of Theorem 3\(\mathrm {(a)}\), because this proof can be applied to the proof of Theorem 3\(\mathrm {(b)}\) in an obvious way. If we integrate the posterior predictive density of U given in Theorem 1\(\mathrm {(a)}\) from \(-\infty \) to x, then
$$\begin{aligned} F( x )&= \left( \frac{ nc }{ nc + Q } \right) ^{n/2} \int \limits _0^\infty \varPhi \left( \frac{ x - cv / 2 }{ \sqrt{ cv } } \right) g_p(v){_1}F_1 \left( \frac{ n }{ 2 } ; \frac{ p }{ 2 } ; \frac{ Q }{ 2 ( nc + Q ) } v \right) \mathrm{{d}}v. \end{aligned}$$
(19)
If we apply an identity
$$\begin{aligned} {_1}F_1 ( a ; b ; x ) = e^x {_1}F_1 \left( b - a ; b ; -x \right) \end{aligned}$$
to (19), then
$$\begin{aligned} F( x )&= \left( \frac{ nc }{ nc + Q } \right) ^{n/2} \int _0^\infty \varPhi \left( \frac{ x - cv / 2 }{ \sqrt{ cv } } \right) g_p(v) \exp \left( \frac{ Q }{ 2( nc + Q ) } v \right) \\&\quad \times \sum _{k=0}^{ \infty } \frac{ \left( \frac{ p-n }{ 2 } \right) _k }{ \left( \frac{ p }{ 2 } \right) _k } \frac{ v^k }{ k! } \left( \frac{ - Q }{ 2( nc + Q ) } \right) ^k \mathrm{{d}}v \\&= \sum _{k=0}^{ \infty } \frac{ \left( \frac{ p-n }{ 2 } \right) _k }{ k! \left( \frac{ p }{ 2 } \right) _k } \left( \frac{ - Q }{ 2( nc + Q ) } \right) ^k \left( \frac{ nc }{ nc + Q } \right) ^{n/2} \\&\frac{ 1 }{ 2^{ \frac{ p }{ 2 } } \varGamma \left( \frac{ p }{ 2 } \right) } \int _0^{\infty } \varPhi \left( \frac{ x - cv / 2 }{ \sqrt{ cv } } \right) v^{ \frac{ p }{ 2 } + k - 1 } \exp \left( -\frac{ 1 }{ 2 } \left( 1 - \frac{ Q }{ nc + Q } \right) v \right) \mathrm{{d}}v. \end{aligned}$$
If we change the variable to \(nc ( nc + Q )^{ -1 } v = v^*\), then
$$\begin{aligned} F( x )&= \sum _{k=0}^{ \infty } \frac{ \left( \frac{ p-n }{ 2 } \right) _k }{ k! \left( \frac{ p }{ 2 } \right) _k } \left( \frac{ - Q }{ 2( nc + Q ) } \right) ^k \left( \frac{ nc }{ nc + Q } \right) ^{ \frac{ n - p }{ 2 } - k} \nonumber \\&\quad \frac{ 1 }{ 2^{ \frac{ p }{ 2 } } \varGamma \left( \frac{ p }{ 2 } \right) } \int _0^{\infty } \varPhi \left( \frac{ x - ( c + Q/n )v^* / 2 }{ \sqrt{ ( c + Q/n ) v^* } } \right) v^{*\frac{ p }{ 2 } + k - 1 } \exp \left( -\frac{ v^* }{ 2 } \right) \mathrm{{d}}v. \end{aligned}$$
(20)
If we apply equalities
$$\begin{aligned} (x)_n&= \frac{ \varGamma ( x + n ) }{ \varGamma ( x ) },~ (- x )_n = ( - 1)^n ( x - n + 1)_n \end{aligned}$$
to (20), then
$$\begin{aligned} F( x )&= \left( \frac{ nc }{ nc + Q } \right) ^{ - 1 } \sum _{k=0}^{ \infty } \frac{ \varGamma \left( \frac{ n - p }{ 2 } + 1 \right) }{ k! \varGamma \left( \frac{ n - p }{ 2 } - k + 1 \right) } \left( 1 - \frac{ nc }{ nc + Q } \right) ^k \left( \frac{ nc }{ nc + Q } \right) ^{\frac{ n - p }{ 2 } - k + 1} M(x), \end{aligned}$$
with
$$\begin{aligned} M(x)&= \int _0^{\infty } \varPhi \left( \frac{ x - ( c + Q/n )v / 2 }{ \sqrt{ ( c + Q/n ) v } } \right) g_{p+2k}( v ) \mathrm{{d}}v, \end{aligned}$$
where \(g_{ p + 2k } (v)\) is the \(\chi ^2\) density of degrees of freedom \(p + 2k\). To calculate F(x), we need to compute M(x). First, if we put \(x = 0\) and \(\alpha = c + Q/n\) in M(x), then
$$\begin{aligned} M(0)&= \int _0^\infty \varPhi \left( - \frac{ \sqrt{ \alpha v } }{ 2 } \right) g_{p + 2k}( v ) \mathrm{{d}}v \\&= \frac{ 1 }{ \sqrt{ 2 \pi } } \frac{ 1 }{ 2^ { \frac{ p }{ 2 } + k} \varGamma ( \frac{ p }{ 2 } + k ) } \int _0^\infty \int _{ -\infty }^{ - \sqrt{ \alpha v }/2 } v^{ \frac{ p }{ 2 } + k - 1 } \exp \left( - \frac{ z^2 }{ 2 } - \frac{ v }{ 2 } \right) \mathrm{{d}}z\mathrm{{d}}v \\&= \frac{ 1 }{ \sqrt{2 \pi } } \frac{ 1 }{ 2^ { \frac{ p }{ 2 } + k} \varGamma ( \frac{ p }{ 2 } + k ) } \int _0^\infty \int _{ \sqrt{ \alpha v }/2 }^{ \infty } v^{ \frac{ p }{ 2 } + k - 1 } \exp \left( - \frac{ z^2 }{ 2 } - \frac{ v }{ 2 } \right) \mathrm{{d}}z\mathrm{{d}}v. \end{aligned}$$
If we make the change of variable \(y = \sqrt{ v }\), then
$$\begin{aligned} M(0)&= \sqrt{ \frac{ 2 }{ \pi } } \frac{ 1 }{ 2^ { \frac{ p }{ 2 } + k} \varGamma ( \frac{ p }{ 2 } + k ) } \int _0^\infty \int _{ \sqrt{ \alpha }y/2 }^{ \infty } y^{ p + 2k - 1 } \exp \left( - \frac{ z^2 }{ 2 } - \frac{ y^2 }{ 2 } \right) \mathrm{{d}}z\mathrm{{d}}v. \end{aligned}$$
(21)
The integral (21) can be evaluated by applying the proof of Lemma D.1 given by Dalton and Dougherty (2011). Next, we consider the case \(x\ne 0\). Let \(\phi ( \cdot )\) be the standard normal density and \(G_s (\cdot )\) be the CDF of a \(\chi ^2\) variable with degrees of freedom s. If we apply the integration by parts to M(x), then
$$\begin{aligned} M(x)&= \left[ \varPhi \left( \frac{ x - \alpha v / 2 }{ \sqrt{ \alpha v } } \right) G_{ p + 2k }( v ) \right] _0^\infty \\&\quad + \frac{ 1 }{ 2 }\int _0^\infty \phi \left( \frac{ x - \alpha v / 2 }{ \sqrt{ \alpha v } } \right) \left( \frac{ x }{ \sqrt{ \alpha } } v^{ - 3/2 } + \frac{ \sqrt{ \alpha } }{ 2 } v^{- 1/2 } \right) G_{ p + 2k }( v ) \mathrm{{d}}v \\&= \frac{ 1 }{ 2 }\int _0^\infty \phi \left( \frac{ x - \alpha v / 2 }{ \sqrt{ \alpha v } } \right) \left( \frac{ x }{ \sqrt{ \alpha } } v^{ - 3/2 } + \frac{ \sqrt{ \alpha } }{ 2 } v^{- 1/2 } \right) G_{ p + 2k }( v ) \mathrm{{d}}v \\&= \frac{ e^{ x/2 } }{ 2 \sqrt{ 2\pi } } \left[ \frac{ x }{ \sqrt{ \alpha } } \int _0^\infty v^{ -3/2 } \exp \left( - \frac{ x^2 }{ 2 \alpha }\frac{ 1 }{ v } - \frac{ \alpha }{ 8 } v \right) G_{ p + 2k }( v ) \mathrm{{d}}v \right. \\&\quad + \left. \frac{ \sqrt{ \alpha } }{ 2 } \int _0^\infty v^{ -1/2 } \exp \left( - \frac{ x^2 }{ 2 \alpha }\frac{ 1 }{ v } - \frac{ \alpha }{ 8 } v \right) G_{ p + 2k }( v ) \mathrm{{d}}v \right] . \\ \end{aligned}$$
If we define
$$\begin{aligned}&N( t ) = \int _0^\infty v^{ -t/2 } \exp \left( - \frac{ x^2 }{ 2\alpha }\frac{ 1 }{ v } - \frac{ \alpha }{ 8 } v \right) G_{ p + 2k }( v ) \mathrm{{d}}v, \end{aligned}$$
(22)
then,
$$\begin{aligned} M(x)&= \frac{ e^{ x/2 } }{ 2 \sqrt{ 2\pi } } \left[ \frac{ x }{ \sqrt{ \alpha } } N( 3 ) + \frac{ \sqrt{ \alpha } }{ 2 } N( 1 ) \right] . \end{aligned}$$
(23)
Equation (22) can be evaluated using the following expressions for \(G_{ p + 2k }( v )\) given in the Appendix of Fatti (1983):
$$\begin{aligned} G_{ p + 2k }( v ) = {\left\{ \begin{array}{ll} \displaystyle 1 - \exp \left( - \frac{ v }{ 2 } \right) \sum _{ i = 0 }^{ p/2 + k - 1 } \frac{ 1 }{ i! } \left( \frac{ v }{ 2 } \right) ^i, &{} p\text { is even, } \\ \displaystyle 2 \varPhi ( \sqrt{ v } ) - 1 - \exp \left( - \frac{ v }{ 2 } \right) \sum _{ i = 0 }^{ ( p - 1 )/2 + k - 1 } \frac{ 1 }{ \varGamma ( i + 3/2 ) } \left( \frac{ v }{ 2 } \right) ^{ i + 1/2 } &{} \ p\text { is odd}. \end{array}\right. } \end{aligned}$$
When p is even, (22) becomes
$$\begin{aligned} N( t )&= 2 \left( \frac{ 2 | x | }{ \alpha } \right) ^{ 1 - t/2 } K_{ 1 - t/2 } \left( \frac{ | x | }{ 2 } \right) \nonumber \\&\quad - \sum _{ i = 0 }^{ p/2 + k - 1 } \frac{ 2 }{ 2^i i! } \left( \frac{ 2 | x | }{ \sqrt{ \alpha } \sqrt{ \alpha + 4 } } \right) ^{ i + 1 - t/2 } K_{ i + 1 - t/2 } \left( \frac{ | x | }{ 2 } \sqrt{ 1 + \frac{ 4 }{ \alpha } } \right) , \end{aligned}$$
(24)
where we applied the equality
$$\begin{aligned}&\int _0^\infty v^{ s - 1 } e^{ - t_1 v^{ - 1 } - t_2 v } \mathrm{{d}}v = 2 \left( \frac{ t_1 }{ t_2 } \right) ^{s/2} K_{s} \left( 2\sqrt{t_1 t_2} \right) , \end{aligned}$$
(25)
where \(t_1>0\) and \(t_2>0\). This equality is given by equation 9 of 3.471 in Gradshteyn and Ryzhik (2007). If we put (24) into (23), then
$$\begin{aligned} M(x)&= \frac{ e^{ x/2 } }{ 2\sqrt{ 2 \pi } } \left\{ \frac{ x }{ \sqrt{ \alpha } } \left[ 2 \left( \frac{ 2 | x | }{ \alpha } \right) ^{ -1/2 } K_{ -1/2 } \left( \frac{ | x | }{ 2 } \right) \right. \right. \nonumber \\&\quad \left. - \sum _{ i = 0 }^{ p/2 + k - 1 } \frac{ 2 }{ 2^i i! } \left( \frac{ 2 | x | }{ \sqrt{ \alpha } \sqrt{ \alpha + 4 } } \right) ^{ i - 1/2 } K_{ i - 1/2 } \left( \frac{ | x | }{ 2 } \sqrt{ 1 + \frac{ 4 }{ \alpha } } \right) \right] \nonumber \\&\quad + \frac{ \sqrt{ \alpha } }{ 2 } \left[ 2 \left( \frac{ 2 | x | }{ \alpha } \right) ^{ -1/2 } K_{ 1/2 } \left( \frac{ | x | }{ 2 } \right) \right. \nonumber \\&\quad \left. \left. - \sum _{ i = 0 }^{ p/2 + k - 1 } \frac{ 2 }{ 2^i i! } \left( \frac{ 2 | x | }{ \sqrt{ \alpha } \sqrt{ \alpha + 4 } } \right) ^{ i + 1/2 } K_{ i + 1/2 } \left( \frac{ | x | }{ 2 } \sqrt{ 1 + \frac{ 4 }{ \alpha } } \right) \right] \right\} . \end{aligned}$$
(26)
If we use the equality
$$\begin{aligned} K_{1/2} ( x ) = K_{-1/2} ( x ) = \sqrt{ \frac{ \pi }{ 2x } } \exp ( - x ) \end{aligned}$$
in (26), then the desired result for even p and \(x\ne 0\) can be obtained. When p is odd, Eq. (22) becomes
$$\begin{aligned} N( t )&= \int _0^\infty v^{ -\frac{t}{2} } \exp \left( - \frac{ x^2 }{ 2 \alpha }\frac{ 1 }{ v } - \frac{ \alpha }{ 8 } v \right) \nonumber \\&\quad \times \left[ 2 \varPhi ( \sqrt{ v } ) - 1 - \exp \left( - \frac{ v }{ 2 } \right) \sum _{ i = 0 }^{ \frac{p - 1}{2} + k - 1 } \frac{ 1 }{ \varGamma ( i + \frac{3}{2} ) } \left( \frac{ v }{ 2 } \right) ^{ i + \frac{1}{2} } \right] \mathrm{{d}}v \nonumber \\&= \int _0^\infty v^{ -\frac{t}{2} } \exp \left( - \frac{ x^2 }{ 2 \alpha }\frac{ 1 }{ v } - \frac{ \alpha }{ 8 } v \right) \nonumber \\&\quad \times \left[ 2 \int _0^{ \sqrt{v} } \phi (z) dz - \exp \left( - \frac{ v }{ 2 } \right) \sum _{ i = 0 }^{ \frac{p - 1}{2} + k - 1 } \frac{ 1 }{ \varGamma ( i + \frac{3}{2} ) } \left( \frac{ v }{ 2 } \right) ^{ i + \frac{1}{2} } \right] \mathrm{{d}}v. \end{aligned}$$
(27)
Since the error function is defined as
$$\begin{aligned} \mathrm{erf}( z ) = \frac{ 2 }{ \sqrt{ \pi } } \int _0^z e^{ -t^2 } \mathrm{{d}}t, \end{aligned}$$
we can express Eq. (27) as
$$\begin{aligned} N(t)&= \int _0^\infty v^{ -\frac{t}{2} } \exp \left( - \frac{ x^2 }{ 2\alpha }\frac{ 1 }{ v } - \frac{ \alpha }{ 8 } v \right) \nonumber \\&\quad \times \left[ \mathrm{erf}\left( \sqrt{ \frac{ v }{ 2 } } \right) - \exp \left( - \frac{ v }{ 2 } \right) \sum _{ i = 0 }^{ \frac{p-1}{2} + k - 1 } \frac{ 1 }{ \varGamma ( i + \frac{3}{2} ) } \left( \frac{ v }{ 2 } \right) ^{ i + \frac{1}{2} } \right] \mathrm{{d}}v. \end{aligned}$$
(28)
Using the equality
$$\begin{aligned} \mathrm{erf}( z ) = \frac{ 2 z e^{ - z^2 } }{ \sqrt{ \pi } } {_1} F_1 \left( 1; \frac{ 3 }{ 2 } ; z^2 \right) , \end{aligned}$$
Eq. (28) becomes
$$\begin{aligned} N(t)&= \int _0^\infty v^{ -\frac{t}{2} } \exp \left( - \frac{ x^2 }{ 2 \alpha }\frac{ 1 }{ v } - \frac{ \alpha }{ 8 } v \right) \nonumber \\&\quad \times \left[ \sqrt{ \frac{ 2 }{ \pi } } \sqrt{ v } \exp \left( - \frac{ v }{ 2 } \right) {_1}F_1 \left( 1 ; \frac{ 3 }{ 2 } ; \frac{ v }{ 2 } \right) -\quad \exp \left( - \frac{ v }{ 2 } \right) \sum _{ i = 0 }^{ \frac{p-1}{2} + k - 1 } \frac{ 1 }{ \varGamma ( i + \frac{3}{2} ) } \left( \frac{ v }{ 2 } \right) ^{ i + \frac{1}{2} } \right] \mathrm{{d}}v \nonumber \\&=\sqrt{ \frac{ 2 }{ \pi } }\int _0^\infty v^{ (1 - t)/2 } \exp \left( - \frac{ x^2 }{ 2 \alpha }\frac{ 1 }{ v } - \frac{ \alpha + 4 }{ 8 } v \right) {_1}F_1 \left( 1 ; \frac{ 3 }{ 2 } ; \frac{ v }{ 2 } \right) \mathrm{{d}}v \nonumber \\&\quad - \sum _{ i = 0 }^{ \frac{ p - 1 }{2} + k - 1 } \frac{ 1 }{ 2^{ i + 1/2 } }\frac{ 1 }{ \varGamma ( i + 3/2 ) } \int _0^\infty v^{ (1 - t)/2 + i } \exp \left( - \frac{ x^2 }{ 2 \alpha }\frac{ 1 }{ v } - \frac{ \alpha + 4 }{ 8 } v \right) \mathrm{{d}}v. \end{aligned}$$
(29)
The second term of (29) can be computed as
$$\begin{aligned} \sum _{ i = 0 }^{ \frac{ p - 1 }{2} + k - 1 } \frac{ 2 }{ 2^{ i + 1/2 } }\frac{ 1 }{ \varGamma ( i + 3/2 ) } \left( \frac{ 2 | x | }{ \sqrt{\alpha }\sqrt{ \alpha + 4 } } \right) ^{ \frac{ 1 - t }{ 2 } + 1 + i } K_{ \frac{ 1 - t }{ 2 } + 1 + i } \left( \frac{ | x | }{ 2 } \sqrt{ 1 + \frac{ 4 }{ \alpha } } \right) . \end{aligned}$$
(30)
using equality (25), while the first term of (29) as
$$\begin{aligned}&\sqrt{ \frac{ 2 }{ \pi } }\int _0^\infty v^{ (1 - t)/2 } \exp \left( - \frac{ x^2 }{ 2 \alpha }\frac{ 1 }{ v } - \frac{ \alpha + 4 }{ 8 } v \right) \int _0^\infty e^{ -w } {_0}F_1 \left( \frac{ 3 }{ 2 } ; \frac{ vw }{ 2 } \right) \mathrm{{d}}w \mathrm{{d}}v. \nonumber \\&\quad =2 \int _0^\infty v^{ - t/2 } \exp \left( - \frac{ x^2 }{ 2 \alpha }\frac{ 1 }{ v } - \frac{ \alpha }{ 8 } v \right) \nonumber \\&\qquad \times \int _0^\infty \frac{ v^{ 3/2 - 1}e^{-v/2} }{ 2^{3/2} \varGamma ( 3/2 ) }e^{ -2w/2 } {_0}F_1 \left( \frac{ 3 }{ 2 } ; \frac{ 2w }{ 4 }v \right) \mathrm{{d}}w \mathrm{{d}}v. \end{aligned}$$
(31)
Since the integrand in (31) is the noncentral chi squared density with degrees of freedom 3 and noncentrality parameter 2w and Corollary 1.3.5 of Muirhead (1982), (31) becomes
$$\begin{aligned}&2\sum _{i = 0}^\infty \int _0^\infty v^{ - t/2 } \exp \left( - \frac{ x^2 }{ 2 \alpha }\frac{ 1 }{ v } - \frac{ \alpha }{ 8 } v \right) g_{3 + 2i}(v)\mathrm{{d}}v \nonumber \\&\quad = \sum _{i = 0}^\infty \frac{ 1 }{ 2^{1/2 + i}\varGamma ( 3/2 + i )} \int _0^\infty v^{ ( 3 + 2i - t)/2 - 1 } \exp \left( - \frac{ x^2 }{ 2 \alpha }\frac{ 1 }{ v } - \frac{ \alpha + 4 }{ 8 } v \right) \mathrm{{d}}v \nonumber \\&\quad =\sum _{i = 0}^\infty \frac{ 2 }{ 2^{1/2 + i}\varGamma ( 3/2 + i )} \left( \frac{ 2 | x | }{ \sqrt{ \alpha }\sqrt{ \alpha + 4 } } \right) ^{ \frac{ 1 - t }{ 2 } + 1 + i } K_{ \frac{ 1 - t }{ 2 } + 1 + i } \left( \frac{ | x | }{ 2 } \sqrt{ 1 + \frac{ 4 }{ \alpha } } \right) . \end{aligned}$$
(32)
Putting (30) and (32) into (29),
$$\begin{aligned} N( t )&= \sum _{i = ( p - 1 )/2 + k }^\infty \frac{ 2 }{ 2^{1/2 + i}\varGamma ( 3/2 + i )} \left( \frac{ 2 | x | }{ \sqrt{ \alpha }\sqrt{ \alpha + 4 } } \right) ^{ \frac{ 1 - t }{ 2 } + 1 + i } \nonumber \\&\quad K_{ \frac{ 1 - t }{ 2 } + 1 + i } \left( \frac{ | x | }{ 2 } \sqrt{ 1 + \frac{ 4 }{ \alpha } }\right) . \end{aligned}$$
(33)
If we put (33) into (23), then we can obtain the result for odd p and \(x \ne 0\). \(\square \)
Proof
Since \((n-p)/2\) is an even number,
$$\begin{aligned}&{_1}F_1 \left( \frac{ n }{ 2 } ; \frac{ p }{ 2 } ; \frac{ Q }{ 2 ( nc + Q ) } v \right) = \exp \left( \frac{ Q }{ 2 ( nc + Q ) } v \right) {_1}F_1 \left( \frac{ p - n }{ 2 } ; \frac{ p }{ 2 } ; \frac{ Q }{ 2 ( nc + Q ) } v \right) \\&\quad = \sum _{k=0}^{ | p - n |/2 } \frac{ \left( \frac{ p-n }{ 2 } \right) _k }{ \left( \frac{ p }{ 2 } \right) _k } \frac{ v^k }{ k! } \left( \frac{ - Q }{ 2( nc + Q ) } \right) ^k. \end{aligned}$$
Thus, F(x) is represented by
$$\begin{aligned} F(x) =&\sum _{k=0}^{ |p - n|/2 } \frac{ \left( \frac{ p-n }{ 2 } \right) _k }{ k! \left( \frac{ p }{ 2 } \right) _k } \left( \frac{ - Q }{ 2( nc + Q ) } \right) ^k \left( \frac{ nc }{ nc + Q } \right) ^{n/2} \\&\frac{ 1 }{ 2^{ \frac{ p }{ 2 } } \varGamma \left( \frac{ p }{ 2 } \right) } \int _0^{\infty } \varPhi \left( \frac{ x - cv / 2 }{ \sqrt{ cv } } \right) v^{ \frac{ p }{ 2 } + k - 1 } \exp \left( -\frac{ 1 }{ 2 } \left( 1 - \frac{ Q }{ nc + Q } \right) v \right) \mathrm{{d}}v. \end{aligned}$$
The rest of the proof is the same as that of Theorem 3\(\mathrm {(a)}\). In addition, the finite representation for \(K_{i \pm 1/2}\) is found in 8.468 and 8.469 of Gradshteyn and Ryzhik (2007). \(\square \)
Proof
Since mean and variance of X/f are given by \( \mathrm { E }( X/f ) = \delta \) and \( \mathrm { V }( X/f ) = \gamma / f \) with \(\gamma = \varOmega + 2 \delta ^2\), the conditional distribution of the standardized X/f is given by
$$\begin{aligned} S_f \sim N \left( - \frac{ \sqrt{ f } \delta }{ \sqrt{ \gamma } } + \frac{ \delta }{ g \sqrt{ f \gamma } } (gW), (gW) \right) , \end{aligned}$$
where \(W \sim \chi ^2_f\) and \(g = \varOmega / (f \gamma )\). From Definition 2.1 of Barndorff-Nielsen et al. (1982), the distribution of \(S_f\) is the normal variance-mean mixture with position \(- \sqrt{ f } \delta / \sqrt{ \gamma }\), drift \(\delta / (g \sqrt{ f \gamma }),\) structure matrix 1, and mixing distribution \(gW {\mathop {=}\limits ^{\mathrm {d}}} g \chi ^2_f\). Hence, from equation (2.2) of Barndorff-Nielsen et al. (1982), the characteristic function of \(S_f\) is given by
$$\begin{aligned} {\hat{g}}( \theta ) = \exp \left( - \frac{ i \theta \sqrt{ f } \delta }{ \sqrt{ \gamma } } \right) \left( 1 - \frac{ 2 i \theta \delta }{ \sqrt{ f \gamma } } + \frac{ \theta ^2 \varOmega }{ f \gamma } \right) ^ { -f / 2 }. \end{aligned}$$
The cumulant generating function of \(S_f\) becomes
$$\begin{aligned} K( \theta )&= - \frac{ i \theta \sqrt{ f } \delta }{ \sqrt{ \gamma } } - \frac{ f }{ 2 } \log \left( 1 - \frac{ 2 i \theta \delta }{ \sqrt{ f \gamma } } + \frac{ \theta ^2 \varOmega }{ f \gamma } \right) \\&= - \frac{ i \theta \sqrt{ f } \delta }{ \sqrt{ \gamma } } - \frac{ f }{ 2 } \sum _{ r = 1 }^\infty \frac{ ( -1)^{ r + 1 } }{ r } \left( - \frac{ 2 i \theta \delta }{ \sqrt{ f \gamma } } + \frac{ \theta ^2 \varOmega }{ f \gamma } \right) ^r \\&= - \frac{ i \theta \sqrt{ f } \delta }{ \sqrt{ \gamma } } - \frac{ 1 }{ 2 } \sum _{ r = 1 }^\infty \frac{ ( -1)^{ r + 1 } }{ r } \sum _{ s = 0 }^r \left( {\begin{array}{c} r \\ s \end{array}}\right) \left( - \frac{ 2i \delta }{ \sqrt{ \gamma } } \right) ^{ r - s } \left( \frac{ \varOmega }{ \gamma } \right) ^s \left( \frac{ \theta }{ \sqrt{ f } } \right) ^{ r + s } \left( \frac{ 1 }{ \sqrt{ f } } \right) ^{ - 2 }. \end{aligned}$$
For \(l \ge 2\), the lth derivative of \(K( \theta )\) is given by
$$\begin{aligned} K^{ ( l ) } ( \theta )&= - \frac{ 1 }{ 2 } \sum _{ r = 1 }^\infty \frac{ ( - 1 )^{ r + 1 } }{ r } \sum _{ s = 0 }^r \left( {\begin{array}{c} r \\ s \end{array}}\right) \left( - \frac{ 2i \delta }{ \sqrt{ \gamma } } \right) ^{ r - s } \left( \frac{ \varOmega }{ \gamma } \right) ^s \nonumber \\&\quad \times ( r + s )( r + s - 1 ) \cdots ( r + s - l + 1 ) \left( \frac{ \theta }{ \sqrt{ f } } \right) ^{ r + s - l } \left( \frac{ 1 }{ \sqrt{ f } } \right) ^{ l - 2} . \end{aligned}$$
(34)
To compute the lth cumulant, we express (34) as
$$\begin{aligned} K^{ ( l ) } ( \theta )= & {} \sum _{ r + s = l, s\le r } \left[ - \frac{ 1 }{ 2 } \frac{ ( - 1 )^{ r + 1 } }{ r } \left( {\begin{array}{c} r \\ s \end{array}}\right) \left( - \frac{ 2i \delta }{ \sqrt{ \gamma } } \right) ^{ r - s } \left( \frac{ \varOmega }{ \gamma } \right) ^s \right. \nonumber \\&\left. \times ( r + s )( r + s - 1 ) \cdots ( r + s - l + 1 ) \left( \frac{ \theta }{ \sqrt{ f } } \right) ^{ r + s - l } \left( \frac{ 1 }{ \sqrt{ f } } \right) ^{ l - 2} \right] \nonumber \\&+ \sum _{ r + s \ne l, s\le r } \left[ - \frac{ 1 }{ 2 } \frac{ ( - 1 )^{ r + 1 } }{ r } \left( {\begin{array}{c} r \\ s \end{array}}\right) \left( - \frac{ 2i \delta }{ \sqrt{ \gamma } } \right) ^{ r - s } \left( \frac{ \varOmega }{ \gamma } \right) ^s \right. \nonumber \\&\left. \times ( r + s )( r + s - 1 ) \cdots ( r + s - l + 1 ) \left( \frac{ \theta }{ \sqrt{ f } } \right) ^{ r + s - l } \left( \frac{ 1 }{ \sqrt{ f } } \right) ^{ l - 2} \right] .\nonumber \\ \end{aligned}$$
(35)
If we put \(l=3\) in (35), then the first summation becomes
$$\begin{aligned}&- \frac{ 1 }{ 2 } \left[ \frac{ ( - 1 )^3 }{ 2 } \left( {\begin{array}{c} 2 \\ 1 \end{array}}\right) \left( - \frac{ 2i \delta }{ \sqrt{ \gamma } } \right) \left( \frac{ \varOmega }{ \gamma } \right) 3! \frac{ 1 }{ \sqrt{ f } } +\frac{ ( - 1 )^4 }{ 3 } \left( {\begin{array}{c} 3 \\ 0 \end{array}}\right) \left( - \frac{ 2i \delta }{ \sqrt{ \gamma } } \right) ^3 3! \frac{ 1 }{ \sqrt{ f } } \right] \\&\quad = -i \frac{ 2 \delta ( 3 \varOmega + 4\delta ^2 ) }{ \gamma ^{3/2} }\frac{ 1 }{ \sqrt{ f } }. \end{aligned}$$
Hence, the third cumulant of \(S_f\) is given by
$$\begin{aligned} \frac{ 2 \delta ( 3 \varOmega + 4\delta ^2 ) }{ ( 2\delta ^2 + \varOmega )^{3/2} } \frac{ 1 }{ \sqrt{ f } }. \end{aligned}$$
The fourth cumulant of \(S_f\) can be obtained by following the procedures to obtain the third cumulant of \(S_f\). The characteristic function of \(S_f\) is represented by
$$\begin{aligned} {\hat{g}}( \theta ) = \exp \left\{ -\frac{ 1 }{ 2 } \theta ^2 + \frac{ 1 }{ \sqrt{ n } } \frac{ 1 }{ 3! } {\tilde{\kappa }}_3 ( i \theta )^3 + \frac{ 1 }{ n } \frac{ 1 }{ 4! } {\tilde{\kappa }}_4 ( i \theta )^4 + \cdots \right\} , \end{aligned}$$
(36)
where
$$\begin{aligned} {\tilde{\kappa }}_3 = \frac{ 2 \delta ( 3 \varOmega + 4\delta ^2 ) }{ ( 2\delta ^2 + \varOmega )^{3/2} } \text { and } {\tilde{\kappa }}_4 = \frac{ 6 ( \varOmega ^2 + 8\varOmega \delta ^2 + 8 \delta ^4 ) }{ ( 2\delta ^2 + \varOmega )^2 }. \end{aligned}$$
Since Eq. (36) is identical to equation (2.7) of Hall (1992), the Edgeworth expansion of \(S_f\) is followed immediately. \(\square \)
