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For simplicity let's assume that the discount rate $r = 0$, then a price of a call option with strike $K$ and maturity $T$ on an asset with positive price can be computed as $C(K) = \Bbb E_Q(S_T-K)^+$. A known fact that this leads to $C''(K)$ being the density of the risk-neutral distribution $(S_T)_*Q$ of the asset price at maturity. I was playing with this fact to compute first two moments of this distribution for the price itself and its logarithm.

What I got is $\Bbb E_QS_T = C(0)$, that is the forward price, which makes perfect sense. Computations are pretty easy thanks to integration by parts: $$ \Bbb E_QS_T = \int_0^\infty KC''(K)\mathrm dK = \int_0^\infty K\mathrm dC'(K) = \int_0^\infty C'(K) \mathrm dK =C(0) $$ Then I got $\Bbb E_QS^2_T = 2\int_0^\infty C(K) dK$ using the same technique, which does not look natural to me, and in particular I don't see why would that always be greater than $C^2(0)$. Perhaps, monotonicity and concavity guarantees that.

Nevertheless, what I am having problem with is computing $\Bbb E_Q\log S_T$ and $\Bbb E_Q\log^2 S_T$ since if I do integration by parts there, on the very first step I get two terms that are infinite. I do however see $\frac1{K^2}$ starting to appear there, which afaik should be a part of the final result, however I was unable to get the explicit formulas. Can someone help here?

SBF
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  • Have you tried writing $$E[f(S_T)] = \int_0^\alpha f(K) P''(K) dK + \int_\alpha^\infty f(K) C''(K) dK$$ where $\alpha$ is some cut-off point, eg ATM. That usually takes care of 'infinities'. – Frido Jun 30 '23 at 17:15
  • @Frido thanks, but I don’t see how would that help taking care of infinity at zero: both second derivatives are the same, but the problem is in $f$ – SBF Jun 30 '23 at 18:58
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    Using both second derivatives of calls and puts means that after integration by parts the boundary terms are more well behaved since low strike puts and high call strikes tend to zero as do their dual deltas. If you only use calls then the low strike calls and dual delta tend to a nonzero value which can cause the infinities when multiplied by f' or f. Does this make more sense? – Frido Jun 30 '23 at 19:28
  • I see you've given this question a bounty. Honestly I think the answer is given already by my previous comment, and if you'd like more detail take a look at the Carr-Madan formula: https://quant.stackexchange.com/q/27626/65759 – Frido Jul 05 '23 at 12:53
  • @Frido your comment indeed helped, so I have split the integral into two. It did not help much however in the computation of moments, hence the bounty. – SBF Jul 05 '23 at 17:43

1 Answers1

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With calculating moments I'll assume you mean calculating $$ E[S_T^n] \enspace\text{for} \enspace n \in \mathbb N $$ Now in general \begin{align*} E[f(S_T)] &= \int_0^\infty f(K) C''(K) dK \\ &= \int_0^\alpha f(K) C''(K) dK + \int_\alpha^\infty f(K) C''(K) dK \\ &= \int_0^\alpha f(K) P''(K) dK + \int_\alpha^\infty f(K) C''(K) dK \\ \end{align*} because $P''(K) = C''(K)$ as you've already pointed out, and $\alpha$ is some cut-off point (eg ATM).

Integrating by parts twice gives \begin{align} E[f(S_T)] &= [f(K)P'(K)]_0^\alpha + [f(K)C'(K)]_\alpha^\infty \\ &\quad - [f'(K)P(K)]_0^\alpha - [f'(K)C(K)]_\alpha^\infty \\ &\quad + \int_0^\alpha f''(K) P(K) dK + \int_\alpha^\infty f''(K) C(K) dK \end{align}

In general, for example if $f(S_T) = S_T^2$ or $f(S_T) = \log S_T$ the terms $P'(K), P(K)$ decay faster than $f(K), f'(K)$ for $K \to 0$ and $C'(K), C(K)$ decay faster than $f(K), f'(K)$ for $K \to \infty$. Thus the integrals and the boundary terms will remain finite.

Let me know if this is answers your question, and if not pls let me know which part has not been clarified.

Frido
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  • Thanks. Is that the final form one can hope to achieve for $f(x) = \log^n x$ and there is no simplification to in? Also, is there a natural choice of $\alpha$? – SBF Jul 06 '23 at 04:58
  • @SBF Yes that is the final form one can hope for. I see what you mean that issues can still be present for the choice $\log^n x$, you'll need to check under what circumstances you can still get infinities. Also, not all higher moments of a stock need be finite. As for the cut-off point there is no canonical choice, althought ATM(F) appears to be the most common choice. – Frido Jul 06 '23 at 07:14
  • Thanks for the answers – SBF Jul 07 '23 at 05:00
  • @SBF You're welcome, I hope it has helped somewhat. – Frido Jul 07 '23 at 06:45