content_image/295834.jpg content_image/295833.jpg content_image/295832.jpg content_image/295831.jpg content_image/295830.jpg content_image/295828.jpg content_image/295829.jpg
Lemma 4: Let θ, a, b ∈ (0 , 1] . Define 

then 

Proof: It is easy to check that f (1) = b log b $_{a}$, f (0 $^{+}$) = b − a and 

Define g ( θ ) = θ ( b − a ) − a log $^{(}$1 − θ + b $_{a}$θ ) for θ ∈ (0 , 1] , then we have 

<img src='content_image/295834.jpg'>

<img src='content_image/295833.jpg'>

<img src='content_image/295832.jpg'>

<img src='content_image/295831.jpg'>

Since g (0 $^{+}$) = 0 , g ( θ ) ≥ 0 for all θ ∈ (0 , 1] . According to (42), we have f $^{′}$( θ ) = g ( θ ) θ 2 ≥ 0 . It follows that for all θ ∈ (0 , 1] , 

and hence 

Lemma 4 is thus established. 

Applying Lemma 4 with a = p$_{Y}$ ( y ; µ$_{0}$ ) and b = p$_{Y}$ ( y ; µ ) , we have 

<img src='content_image/295830.jpg'>

<img src='content_image/295828.jpg'>

<img src='content_image/295829.jpg'>

where the right hand side is an integrable function of y by the result that − ∫ ∞ −∞ p$_{Y}$ ( y ; µ$_{2}$ ) log p$_{Y}$ ( y ; µ$_{1}$ )d y < ∞ for any µ$_{1}$, µ$_{2}$ ∈ Λ( q ) . In fact, as in the proof of Lemma 2 (see (28)), there exist a, b ∈$_{R}$ such that 

