0
$\begingroup$

I am trying to find a way to link Risk-neutral moment of simple return to risk-neutral moment of log-returns.

Specifically, by making the same standard assumptions of the Black-Scholes model with the stock having the SDE

$$\frac{d S_{t}}{S_{t}}=\mu d t+\sigma d W(t)$$

with $W(t)$ being a standard Brownian motion, $\mu$ and $\sigma$ being constant, I apply Ito's lemma and obtain

$$S_{T}=S_{t} e^{\mu(T-t)+\sigma(W(T)-W(t))}$$

Then, by knowing the VIX, which is the risk-neutral (integrated) variance over the next 30days

$$V I X^{2}=E_{t}^{Q}\left[\int_{t}^{t+30 d} \sigma^{2} d t\right]$$

I obtain the risk-neutral 2nd moment of the simple gross return as:

\begin{aligned} E_{t}^{Q}\left[\left(\frac{S_{T}}{S_{t}}\right)^{2}\right] &=E_{t}^{Q}\left[e^{2\left[\left(\mu-\frac{1}{2} \sigma^{2}\right)(T-t)+\sigma(W(T)-W(t))\right]}\right.\\ &=e^{2\left(r_{f}-\frac{1}{2} \sigma^{2}\right)(T-t)} E_{t}^{Q}\left[e^{Z}\right], \quad Z \sim N\left(0,4 \sigma^{2}(T-t)\right) \\ &=e^{2\left(r_{f}-\frac{1}{2} \sigma^{2}\right)(T-t)} e^{0+\frac{4}{2} \sigma^{2}(T-t)} \\ &=e^{2 r_{f}(T-t)+\sigma^{2}(T-t)} \end{aligned}

by using the property of expectation of the exponential of a normal random variable and with $\sigma^2(T-t)$ being my VIX estimate.

I tried using the same methodolody on the 3rd moment, which should yield:

\begin{aligned} E_{t}^{Q}\left[\left(\frac{S_{T}}{S_{t}}\right)^{3}\right] &=E_{t}^{Q}\left[e^{3\left[\left(\mu-\frac{1}{2} \sigma^{2}\right)(T-t)+\sigma(W(T)-W(t))\right]}\right.\\ &=e^{3\left(r_{f}-\frac{1}{2} \sigma^{2}\right)(T-t)} E_{t}^{Q}\left[e^{Z}\right], \quad Z \sim N\left(0,9 \sigma^{2}(T-t)\right) \\ &=e^{3\left(r_{f}-\frac{1}{2} \sigma^{2}\right)(T-t)} e^{0+\frac{9}{2} \sigma^{2}(T-t)} \\ &=e^{3 r_{f}(T-t)+3 \sigma^{2}(T-t)} \end{aligned}

I tried testing whether this approach would give the same result as the Breeden and Litzenberger formula of writing the contract $g(S_T) = (\frac{S_T}{S_t})^n$ in terms of options, and while for the 2nd moment it gives very similar result for the 3rd it gives totally different ones.

I was wondering why it's wrong.

Thanks a lot!

$\endgroup$
2
  • $\begingroup$ Your methodology for computing the third moment is correct. I think that it's your procedure to write this moment in term of options which is wong. so can you please provide us with the Breeden and Litzenberger formula you used for to compute third moment? $\endgroup$ Feb 4 '20 at 18:09
  • $\begingroup$ Hi, thanks for the comment. I think I solved it. The idea is that estimating the 3rd moment using Breeden and Litzenberger and therefore option prices yields a measure of market skewness, while if we use the above assumption of normality it clearly yields 0. They will never match. They do match for the 2nd moment since a normality assumption for volatility is acceptable but not for skew. Do you think this makes sense? $\endgroup$
    – Caidong
    Feb 6 '20 at 11:26

Your Answer

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

Browse other questions tagged or ask your own question.