- katastrofa
**Posts:**10084**Joined:****Location:**Alpha Centauri

Being precise always helps.

- complyorexplain
**Posts:**176**Joined:**

OK. So I asked how is the Girsanov Theorem, which is true, logically connected with this statementBeing precise always helps.

which is false? Logicians say that it is impossible to prove a false statement from a true statement. That is because a proof by definition is a valid argument, and a valid argument by definition is one whose conclusion cannot be false with the premisses true. Do you agree with logicians?

"Why do people think it is generally true?"

Why do you think that people think it is generally true?

Why do you think that people think it is generally true?

- complyorexplain
**Posts:**176**Joined:**

I apologise, I was being insufficiently precise in the OP. I should have distinguished two statements."Why do people think it is generally true?"

Why do you think that people think it is generally true?

(1) Ê[S

(3) E[S

The first (1) is what logicians call necessarily true, because true in virtue of its meaning. It states that in a world without risk premium, the (non-income producing) asset grows on average at the risk free rate. It is true because ‘not having a risk premium’ means ‘grows on average at the risk free rate’.

The second is contingently true, i.e. not necessarily true. Its truth depends on the state of the world. In a a world without risk premium, (3) is true. Otherwise it is false.

To the question “Why do you think that people think it is generally true?”. If ‘it’ refers to proposition (1), then they think it is true because it is true in virtue of its meaning. If ‘it’ refers to proposition (3), well I’m not really sure whether people think it true or not. Here is one reason people might think it true. I think many think the following is true:

(4) E[(S

And everyone thinks this is true

(5) E[(S

But (3) follows logically from (4) and (5).

Hmm. I think I had promised to give up, but taking one more crack. The right hand side of (5) is the forward price of a call option in the BSM model. That can be (and indeed was) derived without any reference to expectations at all, risk neutral or otherwise, by PDE methods. Or, my personal favorite method, as the limiting value of a binomial model. Probabilistic techniques like measure changes are useful for calculating option values and their sensitivities, but usually just that. The undeniable fact that a lot of people get confused about what it all means and forget about the “risk neutral” modifier just generates noise that is best ignored.

- complyorexplain
**Posts:**176**Joined:**

The right hand side of (5) is the forward price of a call option in the BSM model. That can be (and indeed was) derived without any reference to expectations at all, risk neutral or otherwise, by PDE methods.

Yes. Black Scholes: “There is only one formula w(x, t) that satisfies the differential equation (7) subject to the boundary condition (8). This formula must be the option valuation formula” (1973 p643).

So what they did was to find the BSME by no-arbitrage considerations, then solve that equation by tinkering with the formula derived by Sprenkle, which is (4) above, namely this:

(4) E[(S

This was set out in a 1970 paper before they met Merton.

- katastrofa
**Posts:**10084**Joined:****Location:**Alpha Centauri

Thanks to the Girsanov theorem you can replace the physical measure with such an equivalent (risk-neutral) measure r that the expectation can be simply expressed as E[S_t] = S*exp(r*t). Bearish answer to my question completes the general problem formulation.OK. So I asked how is the Girsanov Theorem, which is true, logically connected with this statementBeing precise always helps.

E[S_{T}] = Se^{rt}

which is false? Logicians say that it is impossible to prove a false statement from a true statement. That is because a proof by definition is a valid argument, and a valid argument by definition is one whose conclusion cannot be false with the premisses true. Do you agree with logicians?

You could equally well ask people if MC importance sampling is generally true.

Does that help?

- complyorexplain
**Posts:**176**Joined:**

So you don't agree that it is impossible to derive a false statement from a true statement. Fair enough. Logicians would disagree.Does that help?

Bearish answer was better, in my view.

Last edited by complyorexplain on December 21st, 2020, 3:30 pm, edited 1 time in total.

When they solve the PDE the equation (4) only stand when [$]E[ \ ][$]

means expectation with a probability measure where the spot log normal diffusion as a drift coefficient [$]r S_t dt[$], i.e. what is called risk neutral measure.

means expectation with a probability measure where the spot log normal diffusion as a drift coefficient [$]r S_t dt[$], i.e. what is called risk neutral measure.

- complyorexplain
**Posts:**176**Joined:**

So the BS Formula is true only when there is no risk premium? False.When they solve the PDE the equation (4) only stand when [$]E[ \ ][$]

means expectation with a probability measure where the spot log normal diffusion as a drift coefficient [$]r S_t dt[$], i.e. what is called risk neutral measure.

No, B&S formula is true with the model assumptions (which is not that there is no risk premium), but solution of PDE is expectation only with the SDE compliant with the PDE coefficients (Feynmann-Kac theorem). And under the same risk neutral measure you got [$]E[S_T] = S_t \exp^{r (T-t)}[$], while in the same model on historical measure with drift [$] \mu S_t dt [$] you would have [$]E^{hist}[S_T] = S_t \exp^{\mu (T-t)}[$].So the BS Formula is true only when there is no risk premium? False.When they solve the PDE the equation (4) only stand when [$]E[ \ ][$]

means expectation with a probability measure where the spot log normal diffusion as a drift coefficient [$]r S_t dt[$], i.e. what is called risk neutral measure.

- complyorexplain
**Posts:**176**Joined:**

Mars: "which is not that there is no risk premium"

OK so (3) below is false.

(3) E[S_{T}] = Se^{rT}

We all agree (I hope) that (4) is true

(4) E[(S_{T} – K)+] = E[(S_{T}] N(d1) – KN(d2)

And we agree that (3) and (4) implies

(5) E[(S_{T} – K)+] = Se^{rt} N(d1) – KN(d2)

However as I pointed out a few times, (3) - (5) is an unsound argument, i.e. a valid argument with a false premiss.

OK so (3) below is false.

(3) E[S

We all agree (I hope) that (4) is true

(4) E[(S

And we agree that (3) and (4) implies

(5) E[(S

However as I pointed out a few times, (3) - (5) is an unsound argument, i.e. a valid argument with a false premiss.

(3) is not a premiss. As I said B&S model => PDE => expectation with a specific measure coming from Feynmann-Kac theorem => (3) is true with the expectation using the same measure.

And in that model there are risk premiums.

And in that model there are risk premiums.

- complyorexplain
**Posts:**176**Joined:**

Thanks. Are there any simpler ways to derive the Formula direct from the BS PDE? And how to derive without any appeal to expectations at all?(3) is not a premiss. As I said B&S model => PDE => expectation with a specific measure coming from Feynmann-Kac theorem => (3) is true with the expectation using the same measure.

And in that model there are risk premiums.

- katastrofa
**Posts:**10084**Joined:****Location:**Alpha Centauri

Bearish casts pearls upon this forum - which get trampled sometimes. I just come here because I like Paul quoting Lewis Carroll.So you don't agree that it is impossible to derive a false statement from a true statement. Fair enough. Logicians would disagree.Does that help?

Bearish answer was better, in my view.

Don't drag me into your wikipedia logic drivel, please.