is a holomorphic function of z on$_{C}$. Consequently, d ( x ; µ ) is a continuous function of x on$_{R}$. Proof: It can be shown that ∫ ∞ −∞ φ ( y − z ) log φ ( y − z )d y is a constant, thus a holomorphic function of z on$_{C}$. Therefore, it remains to prove that is a holomorphic function of z on$_{C}$. First, by Jensen’s inequality, we have where a = − E$_{µ}$ { X } and b = 1 2 $^{(}$E$_{µ}$ $^{{}$X 2 $^{}}$+ log(2 π ) ) are real numbers due to the fact that µ ∈ Λ( q ) . Thus, p$_{Y}$ ( y ; µ ) ∈ [ e − 1 $_{2}$y $^{2}$− ay − $^{b}$, 1] , i.e., As a result, we have which is integrable. (Here Re ( z ) and Im ( z ) represent the real and imaginary parts of z , respectively.) It follows that ξ ( z ) given by (24) exists for any µ ∈ Λ( q ) and z ∈$_{C}$ . Suppose U is an open and bounded subset of $_{C}$. There exists an r > 0 such that | Re ( z ) | ≤ r and | Im ( z ) | ≤ r for all z ∈ U . It is easy to check that e − ( y − Re ( z )) 2 2 ≤ e − y 2 2 + | yr | (31) ≤ e − y 2 2 + yr + e − y 2 2 − yr (32) = e r 2 2 $^{[}$e − 1 $_{2}$( y − r ) 2 + e − 1 $_{2}$( y + r ) 2 $^{]}$. (33) 11 | φ ( y − z ) log p$_{Y}$ ( y ; µ ) | ≤ 1 √ 2 π ∣ $_{∣}$e − ( y − z ) 2 2 ∣ ∣ $^{(}$1 2 y 2 + ay + b$^{)}$ (29) = 1 √ 2 π e − ( y − Re ( z )) 2 − Im 2 ( z ) 2 $^{(}$1 2 y 2 + ay + b$^{)}$ , (30) p$_{Y}$ ( y ; µ ) = E$_{µ}$$^{{}$ 1 √ 2 π e − ( y − X ) 2 2 } (25) ≥ 1 √ 2 π e − 1 $_{2}$E$_{µ}$ { ( y − X ) $^{2}$} (26) = e − 1 $_{2}$y $^{2}$− ay − b (27) | log P$_{Y}$ ( y ; µ ) | ≤ 1 2 y 2 + ay + b. (28) ξ ( z ) = ∫ ∞ −∞ φ ( y − z ) log p$_{Y}$ ( y ; µ )d y (24)