第7講確率過程・時系列解析

ランダムウォーク

次の確率変数 $X_i~(i=1,\ldots,t),~Z_t~(t > 0)$ を考える．

X_{i} = {\begin{cases} 1 \\ - 1 \end{cases}

$\begin{equation} X_i = \begin{cases} 1 \\ -1 \end{cases} \end{equation}$

P (X = 1) = \frac{1}{2}

$\begin{equation} P(X = 1) = \frac{1}{2} \end{equation}$

P (X = - 1) = \frac{1}{2}

$\begin{equation} P(X = -1) = \frac{1}{2} \end{equation}$

Z_{t} = X_{1} + \dots + X_{t}

$\begin{equation} Z_t = X_1 + \cdots + X_t \end{equation}$

このとき，

E [X] = 1 \times \frac{1}{2} + (- 1) \times \frac{1}{2} = 0

$\begin{equation} E[X] = 1 \times \frac{1}{2} + (-1) \times \frac{1}{2} = 0 \end{equation}$

E [X^{2}] = 1^{2} \times \frac{1}{2} + (- 1)^{2} \times \frac{1}{2} = 1

$\begin{equation} E[X^2] = 1^2 \times \frac{1}{2} + (-1)^2 \times \frac{1}{2} = 1 \end{equation}$

V [X] = E [X^{2}] - (E [X])^{2} = 1

$\begin{equation} V[X] = E[X^2] - (E[X])^2 = 1 \end{equation}$

なので，

E [Z_{t}] = E [X_{1}] + \dots + E [X_{t}] = 0

$\begin{equation} E[Z_t] = E[X_1] + \cdots + E[X_t] = 0 \end{equation}$

V [Z_{t}] = V [X_{1}] + \dots + V [X_{t}] = t

$\begin{equation} V[Z_t] = V[X_1] + \cdots + V[X_t] = t \end{equation}$

また，

0 < s < t

$0 < s < t$ とすると，

\begin{aligned} E [Z_{s} Z_{t}] & = E [Z_{s} (Z_{s} + Z_{t} - Z_{s})] \\ = E [{Z_{s}}^{2}] + E [Z_{s} (Z_{t} - Z_{s})] \\ = V [Z_{s}] + (E [Z_{s}])^{2} + E [Z_{s}] E [Z_{t} - Z_{s}] \\ = V [Z_{s}] + (E [Z_{s}])^{2} + E [Z_{s}] E [Z_{t - s}] \\ = s + 0^{2} + 0 \times 0 \\ = s \end{aligned}

$\begin{align} E[Z_sZ_t] &= E[Z_s(Z_s + Z_t - Z_s)] \\ &= E[{Z_s}^2] + E[Z_s(Z_t - Z_s)] \\ &= V[Z_s] + (E[Z_s])^2 + E[Z_s]\,E[Z_t - Z_s] \\ &= V[Z_s] + (E[Z_s])^2 + E[Z_s]\,E[Z_{t-s}] \\ &= s + 0^2 + 0 \times 0 \\ &= s \end{align}$

\begin{aligned} C o v [Z_{s}, Z_{t}] & = E [Z_{s} Z_{t}] - E [Z_{s}] E [Z_{t}] \\ = s - 0 \times 0 \\ = s \end{aligned}

$\begin{align} Cov[Z_s, Z_t] &= E[Z_sZ_t] - E[Z_s]E[Z_t] \\ &= s - 0 \times 0 \\ &= s \end{align}$

\begin{aligned} M_{Z_{t}} (α) & = E [e^{α Z_{t}}] \\ = E [e^{α (X_{1} + \dots + X_{t})}] \\ = E [e^{α X_{1}}] \dots E [e^{α X_{t}}] \\ = {(E [e^{α X_{1}}])}^{t} \\ = {(e^{α \times 1} \times \frac{1}{2} + e^{α \times (- 1)} \times \frac{1}{2})}^{t} \\ = {(\frac{e^{α} + e^{- α}}{2})}^{t} \end{aligned}

$\begin{align} M_{Z_t}(\alpha) &= E[e^{\alpha Z_t}] \\ &= E[e^{\alpha (X_1 + \cdots + X_t)}] \\ &= E[e^{\alpha X_1}] \cdots E[e^{\alpha X_t}] \\ &= \left(E[e^{\alpha X_1}]\right)^t \\ &= \left(e^{\alpha \times 1} \times \frac{1}{2} + e^{\alpha \times (-1)} \times \frac{1}{2}\right)^t \\ &= \left(\frac{e^{\alpha} + e^{-\alpha}}{2}\right)^t \end{align}$

t

$t$ 回の試行のうち，

\frac{t}{2} + \frac{k}{2}

$\ds\frac{t}{2} + \frac{k}{2}$ 回

X_{i} = 1

$X_i = 1$ となれば，

Z_{t} = k

$Z_t = k$ となるので，

P (X_{t} = k) = {\begin{cases} _{t} C_{\frac{t + k}{2}} {(\frac{1}{2})}^{t} & (- k \leq t \leq k, t + k = e v e n) \\ 0 & (o t h e r) \end{cases}

$\begin{equation} P(X_t = k) = \begin{cases} \ds{}_tC_{\frac{t+k}{2}}\left(\frac{1}{2}\right)^t & (-k \le t \le k,~ t + k = \mathrm{even}) \\ 0 & (\mathrm{other}) \end{cases} \end{equation}$

第7講 確率過程・時系列解析

ランダムウォーク

第7講確率過程・時系列解析