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!


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.

  • $\begingroup$ this is almost there except that it is $\hat{d}_2$ instead of $\hat{d}_1$ required by the answer. that's the strange thing i don't understand. I don't see how i can get the $\frac{5}{2}\sigma^2$ thing $\endgroup$ – Danny Aug 28 '15 at 8:39
  • $\begingroup$ oh hang on, maybe i didn't read properly $\endgroup$ – Danny Aug 28 '15 at 8:41

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.

  • $\begingroup$ Very good. However, the original question asked how to finish the original proof not to find a new one! $\endgroup$ – Mark Joshi Aug 30 '15 at 4:32
  • $\begingroup$ Haha yea that's true. I was just tossing ideas around to see what works. Do you know if any existing material details the change of measure approach? Maybe there's more I can learn .... $\endgroup$ – Danny Sep 1 '15 at 15:25
  • $\begingroup$ well change of numeraire is a standard way to do this sort of thing but wouldn't easily apply here. There is some discussion of it, and also of doing drift changes to price things in my book particularly for barrier options. (Concepts and practice of mathematical finance.) $\endgroup$ – Mark Joshi Sep 1 '15 at 21:26
  • $\begingroup$ Cool thanks! I will check it out the book:) $\endgroup$ – Danny Sep 2 '15 at 21:57

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