6
$\begingroup$

Consider some Itô process $dS(t)=\mu(t)dt+\sigma(t)dW^{\mathbb P}_{t}$ under the measure $\mathbb P$, where $W^{\mathbb P}$ is a $\mathbb P$-Brownian motion

In plenty of interest rate examples, I have seen that we attempt attempt to find a measure under which $S(t)$ is driftless, i.e. $dS(t)= \sigma(t) dW_{t}^{\mathbb Q}$, where $W^{\mathbb Q}$ is a $\mathbb Q$ Brownian motion.

My question is simple: Why does the $\sigma(t)$ term remain the same under the measure change? does it have something to do the fact the measures $\mathbb Q$ and $\mathbb P$ are equivalent?

$\endgroup$
5
  • 1
    $\begingroup$ I suspect it has to do with equivalent probability measures. See also the accepted answer here: mathoverflow.net/questions/51090/… $\endgroup$ Jan 24 at 15:30
  • 1
    $\begingroup$ The simple mathematical answer is that if you change sigma then the law of the new process becomes singular with respect to the old one , i.e. the two processes are no longer equivalent. For a simple example you can consider drift zero and two different values of sigma and then a simple application of the law of the iterated logarithm will easily give you singularity of the two measures. A similar argument can be used in the general case of variable coefficients as well. $\endgroup$
    – shalop
    Jan 24 at 21:43
  • $\begingroup$ Hey @noob2, is your last sentence true ? If you change the probabilities on a tree without changing the possible outcomes for the stock, I think you do change the variance don’t you? $\endgroup$
    – dm63
    Jan 25 at 13:35
  • $\begingroup$ @dm63: agree. something not quite right with what I said, I need to rethink this stmt. Sorry. $\endgroup$
    – nbbo2
    Jan 25 at 14:19
  • $\begingroup$ No prob. It is a subtle thing for sure, can’t say I fully understand it. $\endgroup$
    – dm63
    Jan 25 at 21:55

3 Answers 3

3
$\begingroup$

Extract from my answer about what the VIX measures (more details on the notation and the conventions I am using can be found in the preceding sections from that answer):

About changing the measure

(This section is based on currently non-public lecture notes by Dylan Possamaï for a course on mathematical finance. I will update it with precise references if the lecture notes are published. For now, I will insert [RefN] where a precise reference is needed.)

TODO: Update this part for non-zero interest rate.

For simplicity I will assume that the interest rate is zero (it is actually not hard to incorporate a constant interest rate $r$.) We deal with only one security, which is an Itô process of the form $$S_t=S_0+\int_0^tb_s\,\mathrm ds+\int_0^t\mathfrak S_s\,\mathrm dW_s.$$

If there is no arbitrage [RefN], then there always exists an $\mathbb F$-predictable stochastic process $\lambda$ such that

$$ \mathfrak S_s(\omega)\lambda_s(\omega)=b_s(\omega) $$

for $\mathrm dt\otimes\mathsf P$-almost all $(s,\omega)\in[0,T]\times\Omega$.

We will assume that

$$ Z_t\overset{\text{Def.}}=\exp\left(-\int_0^t\lambda_s\,\mathrm dW_s-\frac12\int_0^t\lambda_s^2\,\mathrm ds\right), \quad t\in[0,T] $$

is well-defined and an $(\mathbb F,\mathsf P)$-martingale. In fact, if $Z_t$ is well-defined and an $(\mathbb F,\mathsf P)$-martingale, then it follows that there is no arbitrage in the market (up to time $T$).

In this case, we can prove that the measure $\mathsf Q$ given by $\frac{\mathrm d\mathsf Q}{\mathrm d\mathsf P}=Z_T$ is an equivalent (local) martingale measure for the financial market up to time horizon $T$:

By Girsanov's Theorem [RefN], the stochastic process $(W^{\mathsf Q}_t)_{t\in[0,T]}$ given by

$$ W_t^{\mathsf Q}=W_t+\int_0^t\lambda_s\,\mathrm ds $$

is an $(\mathbb F,\mathsf Q)$-Brownian motion (up to to time $T$).

Therefore, for any progressively measurable stochastic process $\mathfrak S=(\mathfrak S)_{s\in[0,T]}$ with $\mathsf E^{\mathsf P}\left(\int_0^T\mathfrak S_s^2\,\mathrm ds\right)<\infty$ and $t\in[0,T]$, we have

$$ \int_0^t \mathfrak S_s\,\mathrm dW_s=\int_0^t \mathfrak S_s\,\mathrm d\left(W_s^{\mathsf Q}-\int_0^s\lambda_\tau\,\mathrm d\tau\right)=\int_0^t\mathfrak S_s\,\mathrm dW_s^{\mathsf Q}-\int_0^t\mathfrak S_s\lambda_s\,\mathrm ds, $$

where the associativity of the stochastic integral [RefN] was used in the last equality.

Therefore, if the price process $(S_t)_{t\in [0,T]}$ is an Itô process

$$ S_t = S_0+\int_0^tb_s\,\mathrm ds+\int_0^t\mathfrak S_s\,\mathrm dW_s, $$

then

$$ S_t=S_0+\int_0^t\mathfrak S_s\,\mathrm dW_s^{\mathsf Q}-\int_0^t\mathfrak S_s\lambda_s\,\mathrm ds+\int_0^tb_s\,\mathrm ds=S_0+\int_0^t\mathfrak S_s\,\mathrm dW_s^{\mathsf Q}. $$

Furthermore, by the regularity assumed on $\mathfrak S_s$, the stochastic integral $\int_0^t\mathfrak S_s\,\mathrm dW_s^{\mathsf Q}$ is an $(\mathbb F,\mathsf Q)$-martingale [RefN] [see also Footnote 6 in my VIX answer]. This shows two things:

  1. $\mathsf Q$ is an equivalent martingale measure for the market consisting only of the security with price process $S$.
  2. The volatility term of $S_t$ remains unchanged when we go from $\mathsf P$ to $\mathsf Q$. However, the drift disappears so that $S$ is now a martingale.
$\endgroup$
1
  • 1
    $\begingroup$ Just what I was looking for, thanks! $\endgroup$ Jan 25 at 16:36
5
$\begingroup$

It has to do with the Girsanov theorem that relates the equivalent measures $\mathbb Q$ and $\mathbb P\,.$ To make intuitively clear what happens I like to give the following "baby Girsanov" example: Let $X$ be standard normal having probability density $$ p(x)=\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}} $$ under $\mathbb P\,.$ If the equivalent measure $\mathbb Q$ is related to $\mathbb P$ by the Radon-Nikodym density $$ q(x)=e^{\mu x-\mu^2/2} $$ then it is straightforward to see that $X$ has under $\mathbb Q$ the density $$ \frac{1}{\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2}}\,. $$ Clearly, under $\mathbb Q$, $X$ has the same variance (diffusion parameter) but a different mean (drift).

$\endgroup$
4
$\begingroup$

Intro: I think intuition is super important for this one, so my answer below focuses on the intuition here.

Short answer

The volatility parameter is meant to describe the behaviour of $S(t)$, whilst the drift only describes the probabilities of the instrument $S(t)$ moving up or down: changing the drift (and therefore the probabilities of "up" and "down") allows us to come up with equivalent probability-measures, which in turn allows us to price derivatives (for example).

If we, however, changed the volatility parameter, the equation for $S(t)$ would no longer describe the instrument $S(t)$.

Long answer

I will give a "hand-waving" answer, at an intuitive level.

It is well known that a multiplicative binomial tree model converges to the continuous GBM model that you have included in your question (see for example here). Therefore, we can take the binomial tree as an intuitive example without loss of generality.

Suppose we draw a binomial tree for the process $S(t)$ under the measure $\mathbb{P}$: rates are zero and the probabilities of an up-move and down-move are both at 50%. The tree is as follows:

enter image description here

This tree produces a "drift" of 2.5% per period. Also, at an intuitive level, the size of the "up" and "down" move describe the volatility of this instrument $S$: the expected value under $\mathbb{P}$ after the first period is 102.5, and the variance in that first period is $0.5*(110-102.5)^2 +0.5* (95-102.5)^2=56.25$ (variance ~ volatility).

(Suppose another instrument with an initial price of 100 had a different tree with prices 107 and 98 after the first period: it would also have a drift of 2.5%, but the variance would be - of course - smaller.)

Now the point is that under this measure $\mathbb{P}$ (for argument's sake suppose it is the real-world measure), you cannot price derivatives on the instrument $S$, because this real world measure $\mathbb{P}$ is not arbitrage free. So you want to come up with an equivalent measure $\mathbb{Q}$ that will allow you to price derivatives on $S$ without arbitrage: but you want to preserve the original tree which depicts the price map of $S$.

In other words, you don't want to change the behaviour of the instrument $S$, you just want to play around with the probabilities of the "up" and "down" moves; the idea is that the tree of prices describes the "real" behaviour of the instrument $S$, so changing it would defeat the purpose of pricing derivatives on $S$ (and as alluded to above, changing the tree is equivalent to changing the variance, i.e. the "volatility", i.e. $\sigma$).

Denoting the probability of an "up-move" as $p$, the measure $\mathbb{Q}$ which is arbitrage free solves the following equation (remember that rates are set to zero in our toy example):

$$110*p^{\mathbb{Q}}+95(1-p^{\mathbb{Q}})=100$$

Which gives $p^{\mathbb{Q}}=\frac{1}{3}$.

Under $\mathbb{Q}$, the drift is now zero ($\frac{1}{3}*110 + \frac{2}{3}*95=100$): so we have changed the drift, but we have not changed the volatility, which in effect allowed us to preserve the original tree depicting the behaviour of $S$.

PS: rates

Under the Libor market model (see here), the forward rates must be martingales under the T-forward measure: so the process for rates under the T-forward measure must always be driftless (otherwise the process would not describe a martingale).

$\endgroup$
11
  • 1
    $\begingroup$ Hey @jan stuller I get variance =50 in Q and 56 in P in your example. The possibilities for the stock have not changed but the variance actually has changed hasn’t it ? $\endgroup$
    – dm63
    Jan 25 at 13:42
  • $\begingroup$ @dm63: I get 56.25 under $\mathbb{P}$ and 50 under $\mathbb{Q}$: you are correct, I need to give this a bit of thought. $\endgroup$ Jan 25 at 14:26
  • $\begingroup$ When you say "$\mathbb P$ is not arbitrage-free", do you mean that "$\mathbb P$ is not a martingale measure" ? (Because it is unclear to me how a measure can be "arbitrage-free".) $\endgroup$ Jan 25 at 16:06
  • 1
    $\begingroup$ We agree that $\mathsf E^{\mathsf P}$ is not the right measure to price assets, since, as you say (to avoid confusion: I will start at $t=0$), $\mathsf E^{\mathsf P}(S_1)=102.5\neq100=S_0$ (so the arbitrage is "immediate", i.e. you make a profit with no time difference in this market). What we disagree on is only a minor terminology issue: I would consider the "arbitrage-freeness" to be a property of the underlying market, not of the measure itself. On another note: The fact mentioned by @dm63 is interesting and if I find some time I will try to look at this more closely 🙂. $\endgroup$ Jan 25 at 17:19
  • 1
    $\begingroup$ @MaximilianJanisch: fine :) When you say the "underlying market", I consider that to be the tree-map of the prices of $S_t$ and the value of the rates: these two combined give rise to a no-arbitrage measure (see here ): to me, it is the measure that gives rise to "arbitrage-free" pricing, but it's just terminology :) $\endgroup$ Jan 25 at 17:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.