$X \sim N[m,v^2]   => E[X] = m ,  Var[X] = v^2$

$Y = e^X  \sim LogN[m, v^2] => E[Y] = e^{m+0.5v^2}   , Var[Y] = E[Y]^2(e^{v^2}-1)$

$\alpha $ = Continuously compounded annual return on the stock

$\overline r $ = Mean of monthly interest rates

$\overline{\sigma}$ = Sample Standard deviation of monthly rates

$\alpha = \frac{\overline r}{h} + \delta + 0.5 [\overline{\sigma}\sqrt{h}]^2$

$n=(\alpha – \delta + 0.5\sigma^2)(T-t)$  $m=(\alpha – \delta – 0.5\sigma^2)(T-t)$  + * mminus |$v^2 = \sigma^2(T-t)$

$\hat{d_1} = \frac{ln[\frac{S_t}{K}]+n}{v}$  + *Note $d_1 = \frac{ln[\frac{F^P(S)}{F^P(K)}]+0.5v^2}{v}$ and $n=(\alpha – \delta + 0.5\sigma^2)(T-t)$| $\hat{d_2} = \frac{ln[\frac{S_t}{K}]+m}{v}$ + *Note $d_2 = \frac{ln[\frac{F^P(S)}{F^P(K)}] – 0.5v^2}{v}$ and $m=(\alpha – \delta – 0.5\sigma^2)(T-t)$ | $\hat{d_2} = \hat{d_1} – v$

$ln[\frac{S_T}{S_t}] \sim N[m, v^2]$

$lnS_T \sim N[m+lnS_t, v^2]$

$S_T = S_t e^{m+vZ}$

  • $ E[S_T] = S_te^{(\alpha-\delta)(T-t)} $  |  $Var[S_T] = E[S_T]^2(e^{v^2}-1)$
  • $Median = S_te^m$
  • $P[S_T < K] = N(-\hat{d_2}) $  |  $P[S_T > K] = N(+\hat{d_2})$
  • Conditional and Partial Expectation
    • $E[S_T | S_T < K] = \frac{E[S_T]N(-\hat{d_1}) }{N(-\hat{d_2})}$ *Note that < and – go together
    • $E[S_T | S_T > K] = \frac{E[S_T]N(+\hat{d_1}) }{N(+\hat{d_2})}$ *Note that > and + go together
  • x %lognormal confidence interval of $S_T$ => Area under the curve = 1-p = x %
    • Calculate p, Calculate $z^L = N^{-1}(\frac{p}{2})$ and $z^U = -z^L$
    • Calculate $S_T^L = S_te^{m+vz^L}$ and $S_T^U = S_te^{m+vz^U}$
  • $Cov(S_T,S_t) = E[\frac{S_T}{S_t}]Var[S_t | S_0] = e^{(\alpha-\delta)(T-t)} E[S_T |S_0]^2(e^{v^2}-1)$

Note: If the price of a stock can be observed over a time interval, no matter how short the interval is, then $\sigma$ is revealed immediately by determining the quadratic variation of the logarithm of the stock price.

$\underset{n \to \infty}{Lim}\sum _{j=1}^n\left \{ ln[S(jT/n)/S(j-1)T/n)] \right \}^2 = \sigma ^2T$

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