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Say we are in a BS world where the (conditional on t) price of a call is given by the usual

$$V(S_t)=V(S_t;K,r,\sigma,T|F_t) = \Phi(d_1)S_t - \Phi(d_2)Ke^{-r(T-t)}$$

Now, what about the unconditional (or actually conditional on s < t, say t=0) expectation of this price? That is, what does the following equal to

$$E[V(S_t)|F_0] = \int_{0}^{\infty}V(S)f_{S}dS = ?$$ where $f_{S}$ is the distribution of a log-normal rv

$$S=S_0e^{(\mu - 0.5\sigma)t+\sigma\sqrt{t}Z}$$

And what about $$E[S_tV(S_t)|F_0] = ?$$ and $$E[S_t^2V(S_t)|F_0] = ?$$

Also, the price is computed as the expectation, say

$$V(t)=e^{-r(T-t)}E[V(T)|F_t]$$,

but what about other moments? What is, for example the variance $$Var[V(T)|F_t] = E[V^2(T)|F_t] - (E[V(T)|F_t])^2 = ?$$

For a call I get to this

$$E[V^2(T)|F_t]=E[(S(T)-K)^2|F_t,S(T)>K]P(S(T)>K|F_t)=\left(E[S^2(T)|...]- 2KE[S(T)|...]+ K^2 \right)P(S(T)>K|F_t) = \left(E[S^2(T)|...] - KE[S(T)|...] \right)P(S(T)>K|F_t) - K\left(E[S(T)|...] - K \right)P(S(T)>K|F_t) = \left(E[S^2(T)|...] - KE[S(T)|...] \right)P(S(T)>K|F_t) - KV_C(t)$$

but then it needs a conditional expectation of a square of log-normal RV $E[S^2(T)|F_t,S(T)>K]$ which I haven't been able to work out so far. I think it could be solved by writing it as

$$S^2(T) = S^2(t)e^{2(\mu - 0.5\sigma)(T-t)+2\sigma\sqrt{T-t}Z}$$

and so it seems to be also log-normal with $\times 2$ the location and scale parameters.

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Since for European calls under BS we have

$$\frac{\partial V_C(t)}{\partial S(t)} = e^{-q\tau}\Phi(d_1)$$

we have that the density function of the value of the European call option is

$$f_C(v) = \frac{e^{q\tau}}{\Phi(d_1(s))}f_S(s)$$

where

$$ f_{S}(s; \mu, \sigma, t) = \frac{1}{\sqrt{2 \pi}}\, \frac{1}{s \sigma \sqrt{t}}\, \exp \left( -\frac{ \left( \ln s - \ln S_0 - \left( r - q - \frac{1}{2} \sigma^2 \right) t \right)^2}{2\sigma^2 t} \right).$$

Since the price of a European call is monotonic in $S(t)$, "all we have to do" is to find the inverse $s = V^{-1}_C(v)$ and then we get complete information on the distribution of $V_C$, not just its expectation as we do at the moment. Unfortunately, I have not been able to find that inverse and perhaps there is no expression for it in terms of the "common" functions. However, it seems to me that the distribution of $V_C$ should have been already derived by someone somewhere, but I haven't been able to find such literature.

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If we consider the distribution at the expiry, then we have for a European call under the BS framework

$$f_C(v;T|v>0) = \frac{1}{\sqrt{2 \pi \sigma^2 T}}\, \frac{1}{v+K}\, \exp \left( -\frac{ \left( \ln(v+K) - \ln S_0 - \left( r - q - \frac{1}{2} \sigma^2 \right) T \right)^2}{2\sigma^2 T} \right)$$

and

$$f_C(v;T|v=0) = \delta(v)P(S_T<K).$$

Thus, since we know the terminal density, I would have thought that it is possible to apply the backwards diffusion equation to derive it for times $t<T$.

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    $\begingroup$ Why do you need these? Note that $E\big(V(S_t) \mid \mathcal{F}_0\big) = e^{-r(T-t)}E\Big(E\big((S_T-K)^+\mid \mathcal{F}_t\big) \mid \mathcal{F}_0\Big) = e^{-r(T-t)}E\big((S_T-K)^+ \mid \mathcal{F}_0\big)$. $\endgroup$ – Gordon Oct 25 '18 at 17:02
  • $\begingroup$ @Gordon Thank you for your comment. Yes, we can even assume that $r=0$ for ease of exposition. I am trying to assess performance of a so-called American Monte Carlo method which uses least-square regression of option values on risk factors and these terms come up in the numerator of the OLS estimate (well, not exactly these but they converge to these in probability). $\endgroup$ – Confounded Oct 25 '18 at 17:10
  • $\begingroup$ @Gordon 'why do you need these?' That's a rather strange comment to a very reasonable question. Why should someone not consider the intertemporal price distribution of a claim to be of potential interest? $\endgroup$ – James Spencer-Lavan Oct 25 '18 at 23:06
  • $\begingroup$ @JamesSpencer-Lavan: that is for my own curiosity, as I haven't yet encountered such payoffs. $\endgroup$ – Gordon Oct 26 '18 at 1:41

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