In a few papers I have seen the following notation: $$ \langle X_t \rangle $$ Also, in Bergomi's book, at page 8, we have the following equality: $$ \biggr\langle \int_0^T e^{-rt}s^2 \frac{d^2P_{\hat{\sigma}}}{dS^2}\sigma^2_t dt \biggr\rangle=\biggr\langle\int_o^Te^{-rt}S^2\frac{d^2P_{\hat{\sigma}}}{dS^2}\hat{\sigma}dt\biggr\rangle $$ What does the angular bracket symbol mean? Is it the expectation under the risk neutral measure? Does its meaning vary as a function of the contect, or does it always has the same precisely-defined meaning?

Edit: It seems the notation originates from physics. I'd be very grateful if someone can explain what this notation means in physics.

Edit: To give more context, I quote from Bergomi's book : Indeed, to a good approximation, the variance of returns scales linearly with their time scale, thus $\langle \delta S^2 \rangle$ is of order $\delta t$ and $\delta S$ is of order $\sqrt{\delta t}$. The book had not specified any dynamic for $(S_t)_{t\geq 0}$.

Edit: Related question What do angle brackets (⟨⟩ ) mean in mathematics/statistics (autocorrelations)?

Edit: Also on Bergomi's book, on p. 9, we are told: Let is write the daily return $r_i$ as: $$ r_i=\sigma_i\sqrt{\delta t}z_i $$ where $\sigma_i$ is the realized volatility for day $i$, and $z_i$ is centered and has unit variance: $\langle z_i\rangle=0$, $\langle z_i^2\rangle=1$.

It seems this notation has dual meaning: both quadratic variation and expectation.

  • 2
    $\begingroup$ It is the quadratic variation of an Ito's process: $<X>_t:=\int_0^t\sigma_s^2ds$ $\endgroup$ Oct 29, 2020 at 14:39
  • $\begingroup$ @FrancescoTotti This is one possible interpretation. However, I don't think that this is it. The notation seems to come from physics. I edited the question to make this more clear. $\endgroup$
    – fwd_T
    Oct 29, 2020 at 15:28
  • 4
    $\begingroup$ The angle-bracket process is known as the predictable quadratic variation, whereas the straight-bracket process is the quadratic variation. For continuous processes, e.g. driven by a Brownian Motion, these 2 are equal but they might differ for jump-diffusions. This question and answer from Math Stack Exchange should be helpful. $\endgroup$ Oct 29, 2020 at 15:45
  • 2
    $\begingroup$ Pls also see here: quant.stackexchange.com/questions/46098/… $\endgroup$ Oct 29, 2020 at 17:24
  • $\begingroup$ @Magicisinthechain This is useful! I hadn't seen that answer. So $\langle \delta S^2 \rangle $ means $\mathbb{E}_{\mathbb{Q}}[d\langle S \rangle_t]$ (where inside the expectation I have the quadratic variation) ? $\endgroup$
    – fwd_T
    Oct 29, 2020 at 17:38


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.