**Lemma 2.1** [6]

*Let* \{{X}_{n},n\ge 1\} *be a sequence of arbitrary random variables*. *Let* {\mathcal{F}}_{n}=\sigma ({X}_{1},{X}_{2},\dots ,{X}_{n}) *and* {\mathcal{F}}_{-n}=\{\varphi ,\mathrm{\Omega}\}, n\ge 1. *Let* \{{\phi}_{n},n\ge 1\} *be a sequence of non*-*negative even functions of* *x*,

\frac{{\phi}_{n}(x)}{|x|}\uparrow ,\phantom{\rule{2em}{0ex}}\frac{{\phi}_{n}(x)}{{x}^{2}}\downarrow ,

(1)

*and let* \{{a}_{n},n\ge 0\} *be an increasing sequence of positive numbers*. *If*

\sum _{n=1}^{\mathrm{\infty}}\frac{E[{\phi}_{n}({X}_{n})]}{{\phi}_{n}({a}_{n})}<+\mathrm{\infty},

(2)

*and* {a}_{n}\uparrow \mathrm{\infty}, *then* \mathrm{\forall}m\ge 1,

\underset{n\to \mathrm{\infty}}{lim}\frac{1}{{a}_{n}}\sum _{k=1}^{n}\{{X}_{k}-E[{X}_{k}|{\mathcal{F}}_{k-m}]\}=0\phantom{\rule{1em}{0ex}}\mathit{\text{a.e.}}

(3)

**Lemma 2.2** *Let* \{{X}_{n},n\ge 1\} *be the* *m*-*dimension stock relative price sequence as given above and* \{{b}_{n},n\ge 1\} *be the* *m*-*dimension investment strategy sequence as given above*. *Suppose that* {b}_{n} *is measurable with respect to* {\mathcal{F}}_{n-1}=\sigma ({X}_{1},{X}_{2},\dots ,{X}_{n-1}) *and* \{{\phi}_{n},n\ge 1\} *is defined as in Lemma * 2.1. *Let* {Y}_{n}=log({b}_{n}^{T}{X}_{n}), *if*

\sum _{n=1}^{\mathrm{\infty}}\frac{E[{\phi}_{n}({Y}_{n})]}{{\phi}_{n}(n)}<+\mathrm{\infty},

(4)

*then* \mathrm{\forall}m\ge 1,

\underset{n\to \mathrm{\infty}}{lim}\frac{1}{n}\sum _{k=1}^{n}\{log\left({b}_{k}^{T}{X}_{k}\right)-E[log\left({b}_{k}^{t}{X}_{k}\right)|{\mathcal{F}}_{k-m}]\}=0\phantom{\rule{1em}{0ex}}\mathit{\text{a.e.}},

(5)

*where* {\mathcal{F}}_{-n}=\{\varphi ,\mathrm{\Omega}\}, n\ge 1.

*Proof* Because \{{Y}_{n},{\mathcal{F}}_{n},n\ge 1\} is a stochastic adapted sequence, from Lemma 2.1, this lemma holds. □

**Definition 2.3** Suppose that \{{X}_{n},n\ge 1\} is *m*-dimension vector sequence and {\mathcal{F}}_{n}^{l}=\sigma ({X}_{n},\dots ,{X}_{l}). If there exists a sequence of numbers \{\varphi (n),n\ge N\} that satisfies {lim}_{n\to \mathrm{\infty}}\varphi (n)=0, such that when n\ge N, \mathrm{\forall}l\ge 1, \mathrm{\forall}A\in {\mathcal{F}}_{1}^{l}, B\in {\mathcal{F}}_{n+l}^{+\mathrm{\infty}}=\sigma ({X}_{n+l},{X}_{n+l+1},\dots ), we have

|P(AB)-P(A)P(B)|\le \varphi (n)P(A)P(B),

(6)

then \{{X}_{n},n\ge 1\} is called a ∗-*mixing sequence*.

We easily see that (3) is equivalent to the following equation: \mathrm{\forall}B\in {\mathcal{F}}_{n+l}^{+\mathrm{\infty}},

|P(B|{\mathcal{F}}_{1}^{l})-P(B)|\le \varphi (n)P(B)\phantom{\rule{1em}{0ex}}\text{a.e.}

(7)

**Lemma 2.4** *Let* \{{X}_{n},n\ge 1\} *be a* ∗-*mixing sequence defined by Definition * 2.3. *Suppose* 1-*dimension random variable* Y\in {\mathcal{F}}_{n+l}^{+\mathrm{\infty}} *and* E|Y|<+\mathrm{\infty}, *then* \mathrm{\forall}l\ge 1,

|E[Y|{\mathcal{F}}_{1}^{l}]-E[Y]|\le \varphi (n)E|Y|\phantom{\rule{1em}{0ex}}\mathit{\text{a.e.}}

*Proof* The proof procedure is almost similar to [7], p.139. □

**Theorem 2.5** *Let* \{{X}_{n},n\ge 1\} *be an* *m* *stock relative price sequence and a* ∗-*mixing sequence*. \{{b}_{n},n\ge 1\} *is the optimized investment strategy sequence*. *Suppose that* N\ge 1, {b}_{n}\in \sigma ({X}_{n-N},{X}_{n-N+1},\dots ,{X}_{n-1}) *and* \mathrm{\forall}n\ge 1, E|log({b}_{n}^{T}{X}_{n})|\le k<+\mathrm{\infty}. *If* (4) *holds*, *then*

\underset{n\to \mathrm{\infty}}{lim}\frac{1}{n}(log{S}_{n}-E[log{S}_{n}])=0\phantom{\rule{1em}{0ex}}\mathit{\text{a.e.}}

(8)

*Proof* From Lemma 2.2, we know that \mathrm{\forall}m\ge N+1,

\underset{n\to \mathrm{\infty}}{lim}\frac{1}{n}\sum _{k=1}^{n}\{log\left({b}_{k}^{T}{X}_{k}\right)-E[log\left({b}_{k}^{T}{X}_{k}\right)|{\mathcal{F}}_{k-m}]\}=0\phantom{\rule{1em}{0ex}}\text{a.e.},

(9)

then from Lemma 2.4, we have

\begin{array}{c}\left|\frac{1}{n}\sum _{k=1}^{n}\{E[log\left({b}_{k}^{T}{X}_{k}\right)|{\mathcal{F}}_{k-m}]-E[log\left({b}_{k}^{T}{X}_{k}\right)]\}\right|\hfill \\ \phantom{\rule{1em}{0ex}}\le \frac{1}{n}\sum _{k=1}^{n}|E[log\left({b}_{k}^{T}{X}_{k}\right)|{\mathcal{F}}_{k-m}]-E[log\left({b}_{k}^{T}{X}_{k}\right)]|\hfill \\ \phantom{\rule{1em}{0ex}}\le \frac{1}{n}\sum _{k=1}^{n}E|log\left({b}_{k}^{T}{X}_{k}\right)|\varphi (m-N)\hfill \\ \phantom{\rule{1em}{0ex}}\le k\varphi (m-N).\hfill \end{array}

(10)

Since when \varphi (m-N)\to 0 (m\to \mathrm{\infty}), we can choose *m* such that k\varphi (m-N) is sufficiently small, from (9) and (10), we know that (8) holds. □

**Remark 2.6** In fact, a ∗-mixing sequence means that the sequence is gradually independent. In this paper, we suppose the relative price sequence satisfying the ∗-mixing condition, which means that the relative stock prices in two periods gradually become independent when the interval between two periods is longer and longer. In the last section, we will give two numerical examples, which show that a market satisfying the ∗-mixing condition exists. Therefore, Theorem 2.5 is meaningful.

**Remark 2.7** In Theorem 2.5, we suppose that {b}_{n} is measurable with respect to \sigma ({X}_{n-N},{X}_{n-N+1},\dots ,{X}_{n-1}). It means that the investment strategy in the *n* th period is totally decided by the information of the stock prices of the *N* periods before it. Because we consider the long term behavior of a sequence investment, this assumption is meaningful for the financial market.

**Remark 2.8** The equality (8) shows that the average return of the long term behavior of a sequence investment converges to the average of the expectation return in every period in probability 1 under the conditions in Theorem 2.5.

**Corollary 2.9** *Let* \{{X}_{n},n\ge 1\} *be the* *m* *stock relative price sequence and be the* ∗-*mixing sequence*. *Let* \{{b}_{n},n\ge 1\} *be the optimized investment strategy sequence*. *Suppose that* N\ge 1, {b}_{n}\in \sigma ({X}_{n-N},{X}_{n-N+1},\dots ,{X}_{n-1}) *and* \mathrm{\forall}n\ge 1, E|log({b}_{n}^{T}{X}_{n})|\le k, *where* k>0 *is a constant*. *If*

\sum _{n=1}^{\mathrm{\infty}}\frac{E{|{Y}_{n}|}^{p}}{{n}^{p}}<+\mathrm{\infty},\phantom{\rule{1em}{0ex}}1\le p\le 2,

(11)

*then*

\underset{n\to \mathrm{\infty}}{lim}\frac{1}{n}(log{S}_{n}-E[log{S}_{n}])=0\phantom{\rule{1em}{0ex}}\mathit{\text{a.e.}}

(12)

*Proof* Letting {\phi}_{n}(x)={|x|}^{p} in Theorem 2.5, this corollary follows. □

**Theorem 2.10** *Let* \{{X}_{n},n\ge 1\} *and* \{{b}_{n},n\ge 1\} *be defined as in Theorem * 2.5 *and let condition* (1) *in Lemma * 2.1 *be replaced by the following condition*: *as* |x| *increases*,

{\phi}_{n}(x)\uparrow ,\phantom{\rule{2em}{0ex}}\frac{{\phi}_{n}(x)}{|x|}\downarrow .

(13)

*If* (4) *holds*, *then*

\underset{n\to \mathrm{\infty}}{lim}\frac{1}{n}log{S}_{n}=0\phantom{\rule{1em}{0ex}}\mathit{\text{a.e.}}

(14)

*Proof* Let {Y}_{n}=log({b}_{n}^{T}{X}_{n}) and {Z}_{n}={Y}_{n}{I}_{\{|{Y}_{n}|\le n\}}. By (13), we have

\begin{array}{rl}E\left(\sum _{n=1}^{\mathrm{\infty}}\frac{|{Z}_{n}|}{n}\right)& =E\left(\sum _{n=1}^{\mathrm{\infty}}\frac{|{Y}_{n}|{I}_{\{|{Y}_{n}|\le n\}}}{n}\right)\\ \le E\left(\sum _{n=1}^{\mathrm{\infty}}\frac{{\phi}_{n}({Y}_{n}){I}_{\{|{Y}_{n}|\le n\}}}{{\phi}_{n}(n)}\right)\\ \le E\left(\sum _{n=1}^{\mathrm{\infty}}\frac{{\phi}_{n}({Y}_{n})}{{\phi}_{n}(n)}\right)\\ <\mathrm{\infty}.\end{array}

So

\sum _{n=1}^{\mathrm{\infty}}\frac{{Z}_{n}}{n}\phantom{\rule{1em}{0ex}}\text{converges a.e.}

(15)

In addition, from (13),

\begin{array}{rl}\sum _{n=1}^{\mathrm{\infty}}P({Z}_{n}\ne {Y}_{n})& =\sum _{n=1}^{\mathrm{\infty}}P(|{Y}_{n}|>n)\\ \le E\left(\sum _{n=1}^{\mathrm{\infty}}\frac{{\phi}_{n}({Y}_{n}){I}_{\{|{Y}_{n}|\le n\}}}{{\phi}_{n}(n)}\right)\\ \le E\left(\sum _{n=1}^{\mathrm{\infty}}\frac{{\phi}_{n}({Y}_{n})}{{\phi}_{n}(n)}\right)\\ <\mathrm{\infty},\end{array}

thus, according to the Borel-Cantelli lemma, we have

\sum _{n=1}^{\mathrm{\infty}}\frac{{Z}_{n}-{Y}_{n}}{n}\phantom{\rule{1em}{0ex}}\text{converges a.e.}

(16)

By (15) and (16), we find that the conclusion holds. □