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Question on an exercice of Joshi's Book

lukaluke

lukaluke

Joined
12/11/13
Messages
17
Points
11
Hi everyone,

The purpose of the exercise is to compute
\[max(S_T(S_T - K), 0)\]
(Question 2.7 page 17)

Using a change of numeraire (using the stock measure), we have:
\[E [\max(S_T - K, 0)] = E [S_T 1_{S_T > K}] - E [K 1_{S_T > K}]\]
Then he says that we need to know how the final stock price is distributed in the stock measure. Until there I agree. Then he says, that we get:
\[S_T = S_0 \exp\left\{ \large( r + \dfrac{\sigma^2}{2} \right)T + \sigma \sqrt{T} N(0,1)\right\}\]
And I do not find the same thing, so I was wondering if there was a typo or if my approach was the wrong one.

To determine how the stock price is distributed in the stock measure, I applied Ito Lemma to
\[dY_t = \displaystyle \large( \dfrac{1}{S_t} \right)\]
then I applied Ito Lemma to \(d \ln(Y_t).\)
And eventually I got:
\[S_T = S_0 \exp\left\{ \large( r - \dfrac{\sigma^2}{2} \right)T + \sigma \sqrt{T} N(0,1)\right\}\]
I doubled checked the maths, so if you could tell me what I did wrong I would really appreciate.
Thank you ;)
 
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He said he used the stock measure.
I guess B/S is a martingale instead?
 
Hi.
Thank you for your answer.
You mean that I should try instead to compute:
\[dY_t=d\large(\dfrac{B(t,T)}{S_t}\right)?\]
 
Thank you but I have already looked at this forum and I couldn't find anything.
 
Not yet. The problem is that I have an interview tomorrow and I would have liked understanding that question (just in case).
 
lukaluke said:
Hi.
Thank you for your answer.
You mean that I should try instead to compute:
\[dY_t=d\large(\dfrac{B(t,T)}{S_t}\right)?\]
Click to expand...

Your initial post was very vague. You said you used Ito's lemma and got your incorrect answer. But you never explained where you used the martingale condition or how you found the drift of your stock under numeraire measure.

If you use Ito on something you know is a martingale, then you can set the resulting expression's drift equal to 0. That at least gives you something. See what you get if you follow the above suggestion by quotes and use this methodology (set drift term = 0, find stock drift mu). You should get \(\mu = r + \sigma^2\).
 
it's a straightforward change of measure:
define \(Q^S\) the stock measure and \(Q\) the risk neutral one, then we have \(\frac{dQ^S}{dQ} = \frac{\frac{S_T}{S_0}}{\frac{B_T}{B_0}} = \exp(-\frac{\sigma^2}{2}T+\sigma W_T ):= L_T\) where \(B_T = \exp(rT)\) bank account numeraire. you recognize that \(L_T\) is the exponential martingale and under \(Q\) its SDE is given by \(\frac{dL_t}{L_t}=\sigma .dW_t\) this means that using Girsanov theorem \(dW_t^S = dW^Q_t - \sigma dt\) , if you replace \(dW^Q\) by its value \(dW_t^S + \sigma dt\) in \(S_T\) you get the desired solution with \(r+\frac{\sigma^2}{2}\)
 
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Hi,

Thank you for your answers. Under \(Q\) I know that \(S_t/S_0\) is a martingale, but what I do not really understand is which quantity is a martingale under \(Q^S\)... :S
 
any payoff divided by \(S_t\) is a martingale under \(Q^S\)
so we have \( \frac{Price_t}{S_t} = \mathbb{E}^{Q^S}_t\left[\frac{Payoff(T)}{S_T}\right]\)

and

under \(Q\) it's \(\frac{S_t}{B_t} \) that is a martingale NOT \(\frac{S_t}{S_0} \)
 
Hi.
Sorry for the mistake. Thank you for correcting me.
Considering your suggestion C S, I eventually used Ito on \(d\large( \dfrac{B(t,T)}{S_t} \right) \) and get the right result, but what I still do not understand is where this quantity comes from.

To start from scratch, we have:
\(Payoff_t = E^{Q}[\max(S_T(S_T - K), 0)] = S_t E^{Q_S}[max(S_T - K, 0)]\)
using a change of numeraire.

Then how do you found out that you had to apply Ito on the above quantity. Sorry if this is a silly question, but it is not clear in my course how to proceed...
Thanks for your time and help. ;)
 
The basic concept is to understand what should be a martingale under the numeraire measure. Under the standard BS assumptions, when you pick an asset as numeraire, that determines a new measure under which any asset price process divided by the numeraire asset price process is a martingale. So if you pick stock as numeraire, now any asset in your universe when "discounted" by the numeraire (stock price) will be a martingale under the stock measure. The usefulness of this trick is that you often don't actually need to figure out what the new measure looks like. You just know any discounted price process is a martingale and that will enable many useful computations.
 
Hi,

Thank you for your answer. I think I start to understand. Let \(A = \{ S_T > K\}\). The following steps are:
\[Payoff_t = S_t E^{Q^S}[(S_T - K) 1_{A}] = s_t E^{Q^S}[1_{A}] - K E^{Q^S}[B(T,T) 1_{A}]\]
Where we used that \(\dfrac{S_T}{S_T} = 1\) is a martingale under \(Q^S\) and now we need to compute:
\[d\large( \dfrac{B(t,T)}{S_t} \right)\]
to compute the second term (using the fact that under \(Q^S\), it is a martingale).

Is it correct?
 
throll gave the right answer, and I think if you read and understand Girsanov, things will become clear to you.
 
The formula for \(S_T\) has nothing to do with the payoff function at all.
 
If you are reading Concepts and Practice of Mathematical Finance, reread 6.108 & 6.109 on page 169
 
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