Ratios option pricing

To find the initial price $V_0$ of the ratios option, we need to compute the expected discounted payoff under the risk-neutral measure $Q$. The payoff at time $T_1$ is given by:

$$H_{T_1} = \frac{S_{T_1}}{S_{T_0}}$$

The initial price is the expected present value of this payoff:

$$V_0 = E^Q\left[ e^{-r T_1} H_{T_1} \right] = e^{-r T_1} E^Q\left[ \frac{S_{T_1}}{S_{T_0}} \right]$$

Under the risk-neutral measure, the stock price $S_t$ follows the geometric Brownian motion:

$$dS_t = r S_t dt + \sigma S_t dW_t^Q$$

This implies that:

$$S_{T_1} = S_{T_0} e^{(r - \frac{1}{2} \sigma^2)(T_1 - T_0) + \sigma (W_{T_1} - W_{T_0})}$$

Therefore, the ratio $\frac{S_{T_1}}{S_{T_0}}$ simplifies to:

$$\frac{S_{T_1}}{S_{T_0}} = e^{(r - \frac{1}{2} \sigma^2)(T_1 - T_0) + \sigma (W_{T_1} - W_{T_0})}$$

Since $W_{T_1} - W_{T_0}$ is normally distributed with mean 0 and variance $T_1 - T_0$, we can compute the expected value:

$$E^Q\left[ e^{\sigma (W_{T_1} - W_{T_0})} \right] = e^{\frac{1}{2} \sigma^2 (T_1 - T_0)}$$

Thus, the expected ratio becomes:

$$E^Q\left[ \frac{S_{T_1}}{S_{T_0}} \right] = e^{(r - \frac{1}{2} \sigma^2)(T_1 - T_0)} \cdot e^{\frac{1}{2} \sigma^2 (T_1 - T_0)} = e^{r (T_1 - T_0)}$$

Plugging back into $V_0$:

$$V_0 = e^{-r T_1} \cdot e^{r (T_1 - T_0)} = e^{-r T_0}$$

Answer: An explicit expression: V₀ = e^( – r × T₀ )