Expected Values and Distributions of Black–Scholes Call Prices
Summary
The document explores the distribution and moments of a European call’s value when the underlying asset follows a lognormal process in a Black–Scholes setting. It asks how to calculate expected option prices and higher moments, including the variance of the option payoff, and sketches conditioning on whether the asset finishes above the strike. It also notes that squaring a lognormal variable changes its log parameters, a possible route toward the needed truncated moments.
Two further approaches are outlined. Before expiry, the call price is monotonic in the underlying, so a change of variables could derive its density if the price function can be inverted. At expiry, the document gives a density for positive call payoffs and a point mass at zero for out-of-the-money outcomes, then suggests using the backward diffusion equation to obtain earlier distributions. These are questions and proposed directions, not a completed derivation: no general inverse or pre-expiry density is established, and the assumptions and measure used for each expectation need to be specified carefully.
Key ideas
- A call price is a nonlinear function of the underlying, so its distribution depends on the underlying’s distribution.
- The variance of a call payoff requires its second moment as well as its expected value.
- Truncated lognormal moments can help evaluate payoff moments conditional on finishing in the money.
- Monotonicity allows a change-of-variables approach to the pre-expiry call-price density if the inverse price map is available.
- At expiry, the call payoff distribution includes both a continuous positive-payoff part and a mass at zero.
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# Expectation of option value
# Expectation of option value
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.
Add 1
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.
Add 2
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 $t<T$.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.