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I am reading Buehler et al. (2022) "Learning to Trade II: Deep Hedging" and the slide on p. 44 states

Fun fact: in discrete time, we can change also the volatility of a process by changing measure.

I am familiar with continuous-time stochastic processes and the change of measure there, which (through Girsanov's theorem) would only change the drift part but not the diffusion part of the process.

Now, according to the quote above, I am assuming that a discrete time version of Girsanov's theorem exists with the additional property of changing the diffusion part in a difference equation. However, I haven't been able to find papers or textbooks describing this. I would be grateful if somebody could point me to such an article (preferably with examples) or could outline how this could be derived.

p.vitzliputzli
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2 Answers2

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The authors of those slides don't mention the Girsanov theorem and I don't think we need it to see the fun fact that we can change the volatility of a discrete process by changing the measure. Take $X_i\sim N(0,1)$ i.i.d. and consider the martingale $$ M_n=\sum_{i=1}^nX_i\,. $$ Its volatility is one. To simplify things I assume $n\le N<\infty\,.$ The probability space on which $M_n$ is defined is $\Omega=\mathbb R^N$ equipped with the $N$-fold product of the standard normal PDF $$ p(x)=\frac{1}{\sqrt{2\pi}}e^{-x^2/2}\,. $$ An obvious equivalent measure that changes $M$'s volatlity to $\sigma$ is the $N$-fold product of $$ q(x)=\frac{1}{\sqrt{2\pi\sigma}}e^{-x^2/(2\sigma)}\,. $$

Kurt G.
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  • The context of the slides is quantitative finance and the quote there is certainly a reference to the change of measure with Girsanov from a statistical or real-world measure to a risk-neutral one in continuous-time. I wouldn't call it a "fun fact", but within this context it might be indeed surprising that in discrete-time also the diffusion part could be affected. – p.vitzliputzli Jun 11 '23 at 09:12
  • @p.vitzliputzli . Why Girsanov ? I looked at Föllmer & Schied, Stochastic Finance, p. 361 and don't see how their Girsanov theorem changes the volatility of the martingale. The reason why in my answer we need a discrete martingale is that Brownian motion $B_t$ is not $B_t=\int_0^tX_s,ds$ unless we introduce the artefact of the Dirac measure valued white noise process $X_t,.$ I have no appetite to work this out in detail. – Kurt G. Jun 11 '23 at 13:43
  • In your example, doesn't the discrete case boil down to the following observation: Here you take a sum of i.i.d. random variables as a process, since the law of the sum of i.i.d. normal distributed random variables is equal to the product of the laws, one can apply a "Girsanov like Measure change" by applying Radon-Nikodym derivative to each individual law and then again using independence to have a measure change for the overall process? – a.s. graduate student Jun 12 '23 at 10:56
  • @a.s.graduatestudent What in the transition $p\to q$ is Girsanov? That would rather be the transition from $p(x)$ to $e^{\mu x-\mu^2/2}p(x),.$ Can you work out that density? Hint: complete the square. (The rest of your comment is acceptable.) – Kurt G. Jun 12 '23 at 11:08
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A discrete-time version of Girsanov's theorem is described in Föllmer, Hans and Schied, Alexander. Stochastic Finance: An Introduction in Discrete Time, Berlin, Boston: De Gruyter, 2002. In particular, relevant is the section 10.2."Minimal Martingale Measures" on pp. 358. The discrete-time version of Girsanov's theorem can be found on p. 361.

p.vitzliputzli
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