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Daneel Olivaw
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Pricing a call option with payoff function max{$S_T - S_{T/2}, 0$}$C=\max\{S_T - S_{T/2}, 0\}$, where $S_T$ is geometric brownian motion. AppreciateI appreciate any help! Please close this question if this is a duplicated question. Thanks all!

My approach is to take out $S_{T/2}$,:

C = $S_{T/2}$ max{$\frac{S_T}{S_{T/2}} - 1, 0$}$$C = S_{T/2} \max\left\{\frac{S_T}{S_{T/2}} - 1, 0\right\}$$

thenThen we can define a risk neuralneutral measure:

E[C] = E[$S_{T/2}$ max{$\frac{S_T}{S_{T/2}} - 1, 0$}] = $\tilde{E}$[max{$\frac{S_T}{S_{T/2}} - 1, 0$}]$$\begin{align} E[C] & = E\left[S_{T/2} \max\left\{\frac{S_T}{S_{T/2}} - 1, 0\right\}\right] \\[3pt] & = \tilde{E}\left[\max\left\{\frac{S_T}{S_{T/2}} - 1, 0\right\}\right] \end{align}$$

where we trickuse $S_{T/2}$ as the numeraire. Then plug into the call option BSBlack-Scholes formula with $S=1$, $K=1$ and $r=0$. 

Is this approach correct? My main concern is that, can Iwe use $S_{T/2}$ as the numeraire?

Pricing a call option with payoff function max{$S_T - S_{T/2}, 0$} where $S_T$ is geometric brownian motion. Appreciate any help! Please close this question if this is a duplicated question. Thanks all!

My approach is to take out $S_{T/2}$,

C = $S_{T/2}$ max{$\frac{S_T}{S_{T/2}} - 1, 0$}

then we can define a risk neural measure

E[C] = E[$S_{T/2}$ max{$\frac{S_T}{S_{T/2}} - 1, 0$}] = $\tilde{E}$[max{$\frac{S_T}{S_{T/2}} - 1, 0$}]

where we trick $S_{T/2}$ as the numeraire. Then plug into call option BS formula with $S=1$, $K=1$ and $r=0$. Is this approach correct? My concern is that, can I use $S_{T/2}$ as the numeraire?

Pricing a call option with payoff function $C=\max\{S_T - S_{T/2}, 0\}$, where $S_T$ is geometric brownian motion. I appreciate any help! Please close this question if this is a duplicated question. Thanks all!

My approach is to take out $S_{T/2}$:

$$C = S_{T/2} \max\left\{\frac{S_T}{S_{T/2}} - 1, 0\right\}$$

Then we can define a risk neutral measure:

$$\begin{align} E[C] & = E\left[S_{T/2} \max\left\{\frac{S_T}{S_{T/2}} - 1, 0\right\}\right] \\[3pt] & = \tilde{E}\left[\max\left\{\frac{S_T}{S_{T/2}} - 1, 0\right\}\right] \end{align}$$

where we use $S_{T/2}$ as the numeraire. Then plug into the call option Black-Scholes formula with $S=1$, $K=1$ and $r=0$. 

Is this approach correct? My main concern is, can we use $S_{T/2}$ as the numeraire?

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Pandaaaaaaa
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Pricing a call option with payoff function max{$S_T - S_{T/2}, 0$} where $S_T$ is geometric brownian motion. Appreciate any help! Please close this question if this is a duplicated question. Thanks all!

My approach is to take out $S_{T/2}$,

C = $S_{T/2}$ max{$\frac{S_T}{S_{T/2}} - 1, 0$}

then we can define a risk neural measure

E[C] = E[$S_{T/2}$ max{$\frac{S_T}{S_{T/2}} - 1, 0$}] = $\tilde{E}$[max{$\frac{S_T}{S_{T/2}} - 1, 0$}]

where we trick $S_{T/2}$ as the numeraire. Then plug into call option BS formula with $S=1$, $K=1$ and $r=0$. Is this approach correct? My concern is that, can I use $S_{T/2}$ as the numeraire?

Pricing a call option with payoff function max{$S_T - S_{T/2}, 0$} where $S_T$ is geometric brownian motion. Appreciate any help! Please close this question if this is a duplicated question. Thanks all!

Pricing a call option with payoff function max{$S_T - S_{T/2}, 0$} where $S_T$ is geometric brownian motion. Appreciate any help! Please close this question if this is a duplicated question. Thanks all!

My approach is to take out $S_{T/2}$,

C = $S_{T/2}$ max{$\frac{S_T}{S_{T/2}} - 1, 0$}

then we can define a risk neural measure

E[C] = E[$S_{T/2}$ max{$\frac{S_T}{S_{T/2}} - 1, 0$}] = $\tilde{E}$[max{$\frac{S_T}{S_{T/2}} - 1, 0$}]

where we trick $S_{T/2}$ as the numeraire. Then plug into call option BS formula with $S=1$, $K=1$ and $r=0$. Is this approach correct? My concern is that, can I use $S_{T/2}$ as the numeraire?

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Pandaaaaaaa
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Pricing a call option with pay-off function max{$S_T - S_{T/2}, 0$}

Pricing a call option with payoff function max{$S_T - S_{T/2}, 0$} where $S_T$ is geometric brownian motion. Appreciate any help! Please close this question if this is a duplicated question. Thanks all!