Pricing a Piecewise Inverse-Square Option with a Change of Measure

# Option with payoff $K^2/S^2$


# Option with payoff $K^2/S^2$












Given the dynamics of the risky asset ( with dividend $q$ ),

$$ \frac{dS_t}{S_t}=(\mu-q)dt + \sigma dW_t^P $$

Consider a european option with payoff,

$$ P_0(S) = \begin{cases} 1, & \text{if $S\le K$} \\ \frac{K^2}{S^2}, & \text{if $S\gt K$} \end{cases} $$

I'm supposed to show that the value of the option is given by,

$$ V(S, t) = \left( \frac{K^2}{S^2}\right)e^{(3\sigma^2+2q-3r)(T-t)}\mathcal{N}(\hat{d_1})+e^{-r(T-t)}\mathcal{N}(-\hat{d_2}) $$ where $$ \begin{align} \hat{d_1}&=\frac{\text{log}(S/K)+(r-q-\frac{5}{2}\sigma^2)(T-t)}{\sigma \sqrt{T-t}}\\ \hat{d_2}&=\frac{\text{log}(S/K)+(r-q-\frac{1}{2}\sigma^2)(T-t)}{\sigma \sqrt{T-t}}\\ \end{align} $$

I got pretty close by doing the following,

$$ \begin{align} V(S, t) &= e^{-r(T-t) }\mathbb{E}_t^\mathbb{Q}\left[\mathbb{1}_{S_T \le K} + \frac{K^2}{S^2}\mathbb{1}_{S_T\ge K}\right]\\ &= e^{-r(T-t) }\mathbb{Q}(S_T \le K) + \mathbb{E}_t^\mathbb{Q}\left[\frac{K^2}{S_T^2}\mathbb{1}_{S_T\ge K}\right]\\ &= e^{-r(T-t)}\mathcal{N}(-\hat{d_2})+ \frac{K^2}{S_t^2}\mathbb{E}_t^\mathbb{Q}\left[\frac{S_t^2}{S_T^2}\mathbb{1}_{S_T\ge K}\right]\\ &= e^{-r(T-t)}\mathcal{N}(-\hat{d_2})+ \frac{K^2}{S^2}\mathbb{E}_t^\mathbb{Q}\left[\text{exp}(-2(r-q-\frac{1}{2}\sigma^2)(T-t)-2\sigma W^Q_{T-t}) \cdot\mathbb{1}_{S_T\ge K} \right]\\ &= e^{-r(T-t)}\mathcal{N}(-\hat{d_2})+ \frac{K^2}{S^2}e^{(\sigma^2+2q-3r)}\mathbb{E}_t^\mathbb{Q}\left[\text{exp}(-2\sigma W^Q_{T-t}) \cdot\mathbb{1}_{S_T\ge K}\right]\\ \end{align} $$

and then i'm not quite sure how to proceed from here.

Any help is greatly appreciated!

## Answer by Danny (score 2, accepted)

https://quant.stackexchange.com/a/19528

I managed to figure out another way of doing it via change of measure as follows...

We know that the dynamics of $S_t^2$ is given by,

$$ \begin{align} S_t^2&=S_0^2 \text{exp}\left( \left( 2r-2q-\sigma^2\right)t+2\sigma^2 W_t^Q \right)\\ \Rightarrow \frac{1}{S_t^2}&=\frac{1}{S_0^2} \text{exp}\left( -\left( 2r-2q-\sigma^2\right)t-2\sigma^2 W_t^Q \right)\\ \Rightarrow \text{exp}\left( \left( 2r-2q-3\sigma^2\right)t \right)\frac{S_0^2}{S_t^2} &= \text{exp}\left( -2\sigma^2 t -2\sigma W_t^Q \right)\\ &=D_t \end{align} $$

where $D_t$ is a change of measure.

By Girsanov, we have,

$$ D_t=\frac{d\mathbb{Q}^{S^2} }{d\mathbb{Q}}=\text{exp}\left( -2\sigma^2 t -2\sigma W_t^Q \right)\\ $$ and $$ W^{S^2}_t=W_t^Q+2\sigma t $$

is a $Q^{S^2}$-Brownian Motion.

Using the above we have,

$$ \begin{align} e^{-r(T-t)}E_t^Q\left[\frac{K^2}{S_T^2}\mathbb{1}_{\{S_T\gt K\}}\right]&=K^2 e^{-r(T-t)} \left[ \frac{1}{E_t^{Q^{S^2}}\left[\frac{1}{D_T}\right]} E_t^{Q^{S^2}} \left[\frac{1}{D_T S_T^2}\mathbb{1}_{\{S_T\gt K\}}\right]\right]\\ &=K^2 e^{-r(T-t)}D_t E_t^{Q^{S^2}} \left[e^{- \left( 2r-2q-3\sigma^2\right)t}\frac{S_T^2}{S^2_0}\frac{1}{S_T^2}\mathbb{1}_{\{S_T\gt K\}}\right]\\ &=K^2 e^{-r(T-t)}e^{ \left( 2r-2q-3\sigma^2\right)} \frac{S_0^2}{S_t^2} E_t^{Q^{S^2}} \left[e^{- \left( 2r-2q-3\sigma^2\right)T}\frac{1}{S_0^2}\mathbb{1}_{\{S_T\gt K\}}\right]\\ &=\frac{K^2}{S^2} e^{ \left( 3\sigma^2 + 2q -3r \right)(T-t)} E_t^{Q^{S^2}} \left[\mathbb{1}_{\{S_T\gt K\}}\right]\\ &=\frac{K^2}{S^2} e^{ \left( 3\sigma^2 + 2q -3r \right)(T-t)} Q^{S^2} \left( S_T\gt K\right)\\ &=\frac{K^2}{S^2} e^{ \left( 3\sigma^2 + 2q -3r \right)(T-t)} Q^{S^2} \left( S_t^2 e^{\left( r-q-\frac{1}{2}\sigma^2\right)(T-t)+\sigma W_{T-t}^Q } \gt K \right)\\ &=\frac{K^2}{S^2} e^{ \left( 3\sigma^2 + 2q -3r \right)(T-t)} Q^{S^2} \left( \left( r-q-\frac{1}{2}\sigma^2\right)(T-t)+\sigma\left( W_{T-t}^{Q^{S^2}}-2\sigma (T-t) \right) \gt \text{log}\frac{K}{S} \right)\\ &=\frac{K^2}{S^2} e^{ \left( 3\sigma^2 + 2q -3r \right)(T-t)} Q^{S^2} \left( \sigma\left( W_{T-t}^{Q^{S^2}}\right) \gt \text{log}\frac{K}{S} - \left( r-q-\frac{5}{2}\sigma^2\right)(T-t)\right)\\ &=\frac{K^2}{S^2} e^{ \left( 3\sigma^2 + 2q -3r \right)(T-t)} Q^{S^2} \left( -W_{T-t}^{Q^{S^2}} \lt \frac{\text{log}\frac{S}{K} + \left( r-q-\frac{5}{2}\sigma^2\right)(T-t)}{\sigma }\right)\\ &=\frac{K^2}{S^2} e^{ \left( 3\sigma^2 + 2q -3r \right)(T-t)} Q^{S^2} \left( \frac{1}{\sqrt{T-t}}W_{T-t}^{Q^{S^2}} \lt \frac{\text{log}\frac{S}{K} + \left( r-q-\frac{5}{2}\sigma^2\right)(T-t)}{\sigma \sqrt{T-t}}\right)\\ &=\frac{K^2}{S^2} e^{ \left( 3\sigma^2 + 2q -3r \right)(T-t)} N(\hat{d_1})\\ \end{align} $$

Combining it with the earlier part gives the required result.

## Answer by Mark Joshi (score 6)

https://quant.stackexchange.com/a/19519

Let $$I= \mathbb{E}_t^\mathbb{Q}\left[\text{exp}(-2\sigma W_{T-t}) \cdot\mathbb{1}_{S_T\ge K}\right] = \frac{1}{\sqrt{2\pi}} \int_{\hat{d}_2}^{\infty} e^{-2\sigma x} e^{-x^2/2} dx.$$ So $$ I = \frac{1}{\sqrt{2\pi}} \int_{\hat{d}_2}^{\infty} e^{-(x-2\sigma)^2/2} dx \, e^{2\sigma^2}. $$ Change variables $y = x-2\sigma$ and you are done.