(a) 对 log S t \log S_t log S t 用伊藤公式。对 f ( s ) = log s f(s)=\log s f ( s ) = log s 有 f ′ ( s ) = 1 / s , f ′ ′ ( s ) = − 1 / s 2 f'(s)=1/s,\ f''(s)=-1/s^2 f ′ ( s ) = 1/ s , f ′′ ( s ) = − 1/ s 2 ,因此
d ( log S t ) = 1 S t d S t − 1 2 1 S t 2 ( d S t ) 2 = ( α − 1 2 σ 2 ) d t + σ d B t . d(\log S_t)=\frac{1}{S_t}dS_t-\frac{1}{2}\frac{1}{S_t^2}(dS_t)^2
=(\alpha-\tfrac{1}{2}\sigma^2)dt+\sigma\, dB_t. d ( log S t ) = S t 1 d S t − 2 1 S t 2 1 ( d S t ) 2 = ( α − 2 1 σ 2 ) d t + σ d B t .
积分得到
log S t = log S 0 + ( α − 1 2 σ 2 ) t + σ B t ⇒ S t = S 0 exp ( ( α − 1 2 σ 2 ) t + σ B t ) . \log S_t=\log S_0+(\alpha-\tfrac{1}{2}\sigma^2)t+\sigma B_t
\Rightarrow S_t=S_0\exp\Big((\alpha-\tfrac{1}{2}\sigma^2)t+\sigma B_t\Big). log S t = log S 0 + ( α − 2 1 σ 2 ) t + σ B t ⇒ S t = S 0 exp ( ( α − 2 1 σ 2 ) t + σ B t ) .
(b) 由几何布朗运动性质
E [ S t ] = S 0 e α t . \mathbb{E}[S_t]=S_0 e^{\alpha t}. E [ S t ] = S 0 e α t .
对任意 t > 0 t>0 t > 0 都满足 E [ S t ] > S 0 \mathbb{E}[S_t]>S_0 E [ S t ] > S 0 当且仅当 α > 0 \alpha>0 α > 0 。
(c) 由 (a)
log S t S 0 = ( α − 1 2 σ 2 ) t + σ B t . \log\frac{S_t}{S_0}=(\alpha-\tfrac{1}{2}\sigma^2)t+\sigma B_t. log S 0 S t = ( α − 2 1 σ 2 ) t + σ B t .
其中 B t ∼ N ( 0 , t ) B_t\sim N(0,t) B t ∼ N ( 0 , t ) 。因此 log ( S t / S 0 ) \log(S_t/S_0) log ( S t / S 0 ) 为正态,均值 ( α − 1 2 σ 2 ) t (\alpha-\tfrac{1}{2}\sigma^2)t ( α − 2 1 σ 2 ) t ,标准差 σ t \sigma\sqrt{t} σ t 。
当 σ > 0 \sigma>0 σ > 0 时,正态变量大于 0 的概率超过 1 / 2 1/2 1/2 当且仅当其均值为正,即
α − 1 2 σ 2 > 0. \alpha-\tfrac{1}{2}\sigma^2>0. α − 2 1 σ 2 > 0.
所以条件为 α > 1 2 σ 2 \boxed{\alpha>\tfrac{1}{2}\sigma^2} α > 2 1 σ 2 (若 σ = 0 \sigma=0 σ = 0 则退化为确定性过程,条件为 α > 0 \alpha>0 α > 0 )。
英文解析
(a) * * Use the Ito formula for log S t \log S_t log S t . f ′ ( s ) = 1 / s , \f ′ ′ ( s ) = − 1 / s 2 f'(s)=1/s,\f''(s)=-1/s^2 f ′ ( s ) = 1/ s , \f ′′ ( s ) = − 1/ s 2 for f ( s ) = log s f(s)=\log s f ( s ) = log s , so
d ( log S t ) = 1 S t d S t − 1 2 1 S t 2 ( d S t ) 2 = ( α − 1 2 σ 2 ) d t + σ d B t . d(\log S_t)=\frac{1}{S_t}dS_t-\frac{1}{2}\frac{1}{S_t^2}(dS_t)^2
=(\alpha-\tfrac{1}{2}\sigma^2)dt+\sigma\, dB_t. d ( log S t ) = S t 1 d S t − 2 1 S t 2 1 ( d S t ) 2 = ( α − 2 1 σ 2 ) d t + σ d B t .
Points earned
log S t = log S 0 + ( α − 1 2 σ 2 ) t + σ B t ⇒ S t = S 0 exp ( ( α − 1 2 σ 2 ) t + σ B t ) . \log S_t=\log S_0+(\alpha-\tfrac{1}{2}\sigma^2)t+\sigma B_t
\Rightarrow S_t=S_0\exp\Big((\alpha-\tfrac{1}{2}\sigma^2)t+\sigma B_t\Big). log S t = log S 0 + ( α − 2 1 σ 2 ) t + σ B t ⇒ S t = S 0 exp ( ( α − 2 1 σ 2 ) t + σ B t ) .
(b) * * Nature by Geometric Brownian Motion
E [ S t ] = S 0 e α t . \mathbb{E}[S_t]=S_0 e^{\alpha t}. E [ S t ] = S 0 e α t .
Any t > 0 t>0 t > 0 would qualify for E [ S t ] > S 0 \mathbb{E}[S_t]>S_0 E [ S t ] > S 0 if and only if α > 0 \alpha>0 α > 0 .
log S t S 0 = ( α − 1 2 σ 2 ) t + σ B t . \log\frac{S_t}{S_0}=(\alpha-\tfrac{1}{2}\sigma^2)t+\sigma B_t. log S 0 S t = ( α − 2 1 σ 2 ) t + σ B t .
B t ∼ N ( 0 , t ) B_t\sim N(0,t) B t ∼ N ( 0 , t ) , solog ( S t / S 0 ) \log(S_t/S_0) log ( S t / S 0 ) is normal, with a mean of( α − 1 2 σ 2 ) t (\alpha-\tfrac{1}{2}\sigma^2)t ( α − 2 1 σ 2 ) t and a standard deviation ofσ t \sigma\sqrt{t} σ t .
When σ > 0 \sigma>0 σ > 0 , the probability that the normal variable is greater than 0 exceeds 1 / 2 1/2 1/2 if and only if its mean is positive, i.e.
α − 1 2 σ 2 > 0. \alpha-\tfrac{1}{2}\sigma^2>0. α − 2 1 σ 2 > 0.
So the condition is α > 1 2 σ 2 \boxed{\alpha>\tfrac{1}{2}\sigma^2} α > 2 1 σ 2 (if σ = 0 \sigma=0 σ = 0 it degenerates into a deterministic process, the condition is α > 0 \alpha>0 α > 0 ).