**Lemma 2:**

Then, the probability given in equation (3) will increase when a better population is selected. In the next theorem, we will prove that when the number of decision-making stages goes to

*B*(*αi*,*<sup>k</sup>* , *βi*,*<sup>k</sup>* ) = *Bi* =1.

for 1,2,3,...

å (5)

*j*=1 *l*

*Pj* =1 , and *Pj* >0 Then, if

as *k* goes to infinity we

(6)

*k*

*P k*

*j* .

<sup>ï</sup> <sup>=</sup> <sup>î</sup>

are different positive constants, ∑

for 0

*<sup>t</sup>* >0. Taking the limit on *Rk* , *<sup>j</sup>*

1,


æ ö ç ÷

ç ÷ è ø å å

 () *j k j jj kj j l l k k*

*l l s s t t <sup>s</sup> s ii t t ii l l i i*

*c l c l <sup>l</sup> c cl l c cl c l c l* = =

= Þ= = Þ= å å

ç ÷ = = <sup>=</sup> ç ÷

1, 1 1


1 1

å å (7)

*cR cl*

*i k i ii i i*

*cR cl*

infinity this probability converges to one for the best population.

128 Dynamic Programming and Bayesian Inference, Concepts and Applications

In order to prove the theorem first we prove the following two lemmas.

, 1, 1

=

*R c R*

ï

), there exist at most one non-zero *l*

,

*Lim R l Lim*

®¥ ®¥

*<sup>t</sup>* >0, then by equation (6) we have

1 1

= =

In other words, we conclude *cs* =*ct*, which is a contradiction.

*i i i i i i*

and

ì

*c R*

; *j* =1, 2, ..., *l*} as

1,


<sup>ï</sup> <sup>=</sup> <sup>ï</sup> <sup>=</sup> <sup>í</sup>

*jk j l k j ik i i*


*j*

*<sup>s</sup>* >0 and *l*

*th* population is the best, then *Limk*→*<sup>∞</sup>*

Define a recursive sequence {*Rk* , *<sup>j</sup>*

where *c*1, *c*2, ..., and *cl*

Suppose there are two nonzero *l*

*<sup>s</sup>* >0 and *l*

(*Rk* , *<sup>j</sup>*

*l <sup>j</sup>* <sup>=</sup> *Limk*→*<sup>∞</sup>*

**Proof:**

have

Now since *l*

**Theorem 1**

**Lemma 1:**

If the *i*

Sequence *Rk* , *<sup>j</sup>* converges to one for *j* = *g* and converges to zero for *j* ≠ *g*, where *g* is an index for the maximum value of *cj* .
