Assume the Black-Scholes framework. You are given:
•A stock S pays no dividend
•The continuously compounded risk-free interest rate is 8.5%
•A contingent claim Y pays $\frac{S(4)}{S(2)}$ at time 4.
Calculate the time-$0$ price of Y
This is an IFM question from an Actex study manual. So far I have that
$$\frac{S(4)}{S(2)} = \frac{S_0e^{R(0,4)}}{S_0e^{R(0,2)}} = \frac{e^{R(0,4)}}{e^{R(0,2)}}$$
Since returns are additive, $R(0,4) = R(0,2) + R(2,4)$. Thus, $$\frac{S(4)}{S(2)} = e^{R(2,4)}$$
Since I don't know what S(4) or S(2) will be in the future, I must take their expected value and discount it back to time 0 with continuous rate r=8.5%.
$$E[\frac{S(4)}{S(2)}] = E[e^{R(2,4)}]e^{(-.085)(4)}$$
I know that $E[e^{R(2,4)}] = e^{m + v^2/2}$ where $m$ and $v^2$ are the mean and variance for $R(2,4)$. I know that $R(2,4)$ ~ $N((\alpha - \frac{\sigma^2}{2})t, \sigma^2 t)$ but I don't have the values for $\alpha$ or $\sigma$ (the volatility)
Can anyone help explain this?
In this framework,
$$Y_0=E\left[e^{-rT_4}\frac{S_{T_4}}{S_{T_2}}\right]=E\left[\frac{e^{-rT_2}}{e^{-rT_2}}e^{-rT_4}\frac{S_{T_4}}{S_{T_2}}\right]=e^{-rT_2}E\left[\frac{S_{T_4}e^{-rT_4}}{S_{T_2}e^{-rT_2}}\right]=e^{-rT_2}E\left[\frac{M_{T_4}}{M_{T_2}}\right]$$
where $M_t=S_{t}e^{-rt}$.
$M_t$ is a martingale in this framework, therefore :
$$E\left[\frac{M_{T_4}}{M_{T_2}}\right]=E\left[\frac{E\left[M_{T_4}|\mathcal{F}_{T_2}\right]}{M_{T_2}}\right]=E\left[\frac{M_{T_2}}{M_{T_2}}\right]=1$$
Finally, $$Y_0=e^{-rT_2}$$