In this question, the risk-neutral probability distribution $q(S_T=s)$ for the underlying at time $t = T$ is given by the Breeden-Litzenberger identity as:

$$ \frac{1}{P(0,T)} \frac{ \partial^2 C }{\partial K^2} (K=s,T) $$

In practice, this can be computed numerically for a given $K = s$ with a centered, second-order finite-difference approximation:

$$ \frac{ \partial^2 C }{\partial K^2} (K=s,T) \approx \frac{ C(K=s-\Delta K,T) - 2 C(K=s,T) + C(K=s+\Delta K,T)}{ (\Delta K)^2 } $$

i.e., the payoff for a butterfly spread around $K = s$ normalized for its width.

If instead we are interested in the CDF $Q(S_T \geq s)$, can we look at the first order finite-difference approximation:

$$ \frac{ \partial C }{\partial K} (K=s,T) \approx \frac{ C(K=s-\Delta K,T) - C(K=s+\Delta K,T)}{ (2 \Delta K) } $$

to determine the implied risk-neutral CDF? And similar to the above, is it appropriate to think of this as a normalized payer spread option (buying a call at $K = s + \Delta K$ and selling a call at $K = s - \Delta K$)?

In particular, if the underlying is a swap rate and the market data are swaption prices, what sort of issues (and remedies?) arise from this interpretation (i.e., potential liquidity constraints, or discretization complications with $\Delta K$)? Finally, is there a standard method for constructing a meaningful error bound on the cumulative probability implied by a certain strike?

  • 1
    $\begingroup$ Yes absolutely, the approximation of the cdf through a call spread (ie the price of a digital call) holds. The size of the wedge $\Delta K $ matters indeed: too small = run numerical precision/stability issues, too large = the approximation of the derivative using a finite difference approach does not hold anymore. You could also have chosen a forward vs backward first order difference (precision).. This all depends on the available list of instruments as you mention (liquidity). For the error bound, just use Taylor (cf. expression of the remainder), to get the truncation error. $\endgroup$ – Quantuple Jul 30 '16 at 7:24
  • $\begingroup$ From the first steps of deriving the Dupire equation p.3 here, $$\frac{\partial C}{\partial K} = P(t, T) \int_K^{\infty} \frac{\partial C}{\partial K} \pi (S,T) dS = -P(t, T) \int_K^{\infty} \pi (S,T) dS = -P(t,T) + \Pi (S \leq K)$$ how would I take into account the $P(t,T)$ term? To get the CDF $\Pi$ do I have to add on this PV factor to the finite difference approximation? $\endgroup$ – jake_r Aug 4 '16 at 15:37
  • 1
    $\begingroup$ I don't understand your equation nor does it correspond to anything I can find on page 3 of your link? of course there is a PV term that will appear, because a call price is a discounted expectation $\endgroup$ – Quantuple Aug 4 '16 at 15:45
  • $\begingroup$ I restated eq. 9 (mistakenly: left out the $(S-K)$ term) on p. 3 to list the equation explicitly since I didn't address the PV term in the simple finite-difference approximation in my original question. Is there a meaningful interpretation for the finite difference term $ \frac{ \partial C }{\partial K} (K=s,T) \approx \frac{ C(K=s-\Delta K,T) - C(K=s+\Delta K,T)}{ (2 \Delta K) } $ without adjusting for the PV to get to the CDF? $\endgroup$ – jake_r Aug 4 '16 at 17:24
  • $\begingroup$ First this ratio is $-\partial C/\partial K $ and yes it is the probability you are looking for, divided by the discount factor. $\endgroup$ – Quantuple Aug 4 '16 at 17:28

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.