Timeline for Find a formula for the price of a derivative paying $\max(S_T(S_T-K),0)$
Current License: CC BY-SA 4.0
38 events
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| Nov 10, 2020 at 14:27 | comment | added | Kevin | @PontusHultkrantz great question. The problem is that $e^{-rt}S_t^2$ is not a martingale (Jensen’s inequality). The price (value) process, $N_{t,T}^\alpha$, is however (by construction). Remember that a numéraire is defined to be the price of an asset (which is the unit for the price of all other assets). | |
| Nov 10, 2020 at 14:17 | comment | added | Pontus Hultkrantz | @Kevin: why is it that we chose $N^α_{t,T}$ as the time-t price of an asset (claim) paying $S^α_T$ at time $T$? Why not chose $N_t \equiv S_t^2$, such that $\frac{dQ^2}{dQ^1} = \frac{S_T}{S_0}$ and $E^{Q^1}[S_T 1_{(.)}] = S_0 E^{Q^2}[ 1_{(.)}]$? | |
| Jul 12, 2020 at 10:33 | history | edited | Kevin | CC BY-SA 4.0 |
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| Jul 12, 2020 at 10:24 | history | bounty ended | Trajan | ||
| Jul 12, 2020 at 10:24 | vote | accept | Trajan | ||
| Jul 12, 2020 at 10:22 | history | edited | Kevin | CC BY-SA 4.0 |
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| Jul 12, 2020 at 10:16 | history | edited | Kevin | CC BY-SA 4.0 |
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| Jul 12, 2020 at 10:09 | comment | added | Trajan | i think you forget how much you know at times | |
| Jul 12, 2020 at 10:08 | comment | added | Trajan | that needs to go in the answer, for everyone else | |
| Jul 12, 2020 at 9:56 | comment | added | Trajan | $\mathcal{P}[\{S_T\geq K\}]=N\left(\frac{\ln\left(\frac{S_0}{K}\right)+\mu T-\frac{1}{2}\sigma^2 T}{\sigma \sqrt{T}}\right)$. Last thing. How has this been derived? | |
| Jul 11, 2020 at 23:30 | history | edited | Kevin | CC BY-SA 4.0 |
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| Jul 11, 2020 at 20:37 | comment | added | Kevin | @Permian That's great to hear. Please have a look at the information I added. I started from the statement of the theorem and outlined how it impacts the drift of the stock price. Please ask further questions if I couldn't answer everything | |
| Jul 11, 2020 at 20:36 | history | edited | Kevin | CC BY-SA 4.0 |
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| Jul 11, 2020 at 18:38 | comment | added | Trajan | I get the points above now thanks, its only the bit around girsanov that I dont get now | |
| Jul 11, 2020 at 18:20 | comment | added | Kevin | @Permian Let me know whether you are okay with the two points above. They were really only little maths manipulations. No tricks or such. I can provide a thorough edit to my answer and add more details to Girsanov's theorem. Just let me know whether there's anything else that you don't understand in my answer (other than this one paragraph on Girsanov). If you tell me how I can help you, I'll edit the answer accordingly :) | |
| Jul 11, 2020 at 18:14 | comment | added | Kevin | @Permian In the BS world, $S_t=S_0e^{(r-0.5\sigma^2)t+\sigma W_t}$ and $S_t^\alpha=S_0^\alpha e^{(r-0.5\sigma^2)\alpha t+\alpha\sigma W_t}$. So, you only need to collect these terms to obtain $S_t^\alpha$ in the equations in my answer and that's it. The $\frac{1}{2}\alpha^2\sigma^2(T-t)$ remains unaffected and the rest is summarised as $S_t^\alpha$. Does this make sense? | |
| Jul 11, 2020 at 18:11 | comment | added | Kevin | @Permian If $X_t=e^{\sigma W_t-\frac{1}{2}\sigma^2t}$ is a martingale, then $\mathbb{E}[e^{\sigma W_t-\frac{1}{2}\sigma^2t}|\mathcal{F}_s]=e^{\sigma W_s-\frac{1}{2}\sigma^2s}$. Now, you can take the exponential $e^{-\frac{1}{2}\sigma^2t}$ out of the conditional expectation and bring it on the right side, right? | |
| Jul 11, 2020 at 17:22 | comment | added | Trajan | $= e^{-r(T-t)}S_t^\alpha\exp\left(\alpha\left(r-\frac{1}{2}\sigma^2\right)(T-t)+\frac{1}{2}\alpha^2\sigma^2(T-t)\right)$. I cant see how you have pulled the $S_t$ here either | |
| Jul 11, 2020 at 17:20 | comment | added | Trajan | I cant see the exact way Girsanov has been applied either. To be honest, the whole paragraph from "To conclude...Bjork" is hard to understand | |
| Jul 11, 2020 at 17:08 | comment | added | Trajan | "he stock price in the Black-Scholes model has drift $r+\sigma^2$". How can you spot this? I dont understand the linked answers at all. After this point you have lost me | |
| Jul 11, 2020 at 17:07 | comment | added | Trajan | $\mathbb{E}[e^{\sigma W_t}|\mathcal{F}_s]=e^{\frac{1}{2}\sigma^2(t-s)+\sigma W_s}$ no sorry I still cannot see how this has been derived | |
| Jul 11, 2020 at 11:05 | comment | added | Kevin | The point of my answer is to give you a more elegant approach for which you don't need to complete many tedious integrals. In particular, the formulae are easy to generalise. What I like the most about this approach is that it applies the change of numeraire twice. So, this may help you better understand how to apply it. Let me know if you have further questions! | |
| Jul 11, 2020 at 11:03 | comment | added | Kevin | @Permian It comes from the fact that $X_t=e^{\sigma W_t-\frac{1}{2}\sigma^2t}$ is a martingale, i.e. $\mathbb{E}[X_t|\mathcal{F}_s]=X_s$ | |
| Jul 11, 2020 at 10:52 | comment | added | Trajan | This method is attractive though | |
| Jul 11, 2020 at 10:51 | comment | added | Trajan | $\mathbb{E}[e^{\sigma W_t}|\mathcal{F}_s]=e^{\frac{1}{2}\sigma^2(t-s)+\sigma W_s}$ where does this come from? | |
| Jul 11, 2020 at 10:50 | comment | added | Trajan | Im struggling a bit with this. Joshi's papers has too many unexplained equations, with the detail glossed over. | |
| Jul 8, 2020 at 17:36 | history | edited | Kevin | CC BY-SA 4.0 |
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| Jul 6, 2020 at 18:59 | history | undeleted | Kevin | ||
| Jul 6, 2020 at 18:59 | history | edited | Kevin | CC BY-SA 4.0 |
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| Jul 6, 2020 at 18:15 | history | deleted | Kevin | via Vote | |
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| Jul 6, 2020 at 18:08 | history | edited | Kevin | CC BY-SA 4.0 |
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| Jul 6, 2020 at 17:59 | history | edited | Kevin | CC BY-SA 4.0 |
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| Jul 6, 2020 at 17:53 | history | answered | Kevin | CC BY-SA 4.0 |