Saturday 29 August 2026

The one substitution that does the work

Over $[t,t+dt]$ the portfolio value changes by three pieces: the short option $-dV$, the long shares ${\Delta }_{t}dS$, and interest on the cash $r(V-{\Delta }_{t}S)dt$:

$$d\Pi =-dV+{\Delta }_{t}dS+r(V-{\Delta }_{t}S)dt.$$

Apply Itô to $V(S_t,t)$, and here is the point a candidate must not miss: the quadratic variation of $S$ is driven by the realized volatility, since we are computing the actual P&L of the actual path, not a fictitious one. So

$$dV={\partial }_{t}Vdt+{\partial }_{S}VdS+\frac{1}{2}{\partial }_{SS}V{\sigma }_r^2{S}^{2}dt.$$

Substitute, and use ${\Delta }_t={\partial }_{S}V$ so that the ${\partial }_{S}VdS$ term cancels against ${\Delta }_{t}dS$. Every stochastic $dS$ (hence every $dW$) is gone:

$$d\Pi =-{\partial }_{t}Vdt-\frac{1}{2}{\partial }_{SS}V{\sigma }_r^2{S}^{2}dt+rVdt-rS{\partial }_{S}Vdt.$$

Now use that the option is marked with implied vol, so it obeys the Black–Scholes PDE with $\sigma$:

$${\partial }_{t}V=-\frac{1}{2}{\sigma }^2{S}^2{\partial }_{SS}V-rS{\partial }_{S}V+rV.$$

Insert this for ${\partial }_{t}V$. The $rS{\partial }_{S}V$ and $rV$ terms cancel identically, leaving only the two second-order terms:

$$d\Pi =\frac{1}{2}{\partial }_{SS}V{\sigma }^2{S}^{2}dt-\frac{1}{2}{\partial }_{SS}V{\sigma }_r^2{S}^{2}dt.$$

$$\boxed{d\Pi =\frac{1}{2}{\Gamma }_t{S}_t^2({\sigma }^2-{\sigma }_r^2)dt}$$

The drift $\mu$ has vanished entirely, and there is no $dW$: the instantaneous P&L is deterministic given the current state. That is the whole content of the result — the randomness in the option is exactly offset by the randomness in the hedge, and what remains is a clean bet on volatility.

Sign for a vanilla short position

For a call or put $\Gamma >0$, so $d\Pi >0$ precisely when $\sigma >{\sigma }_r$, i.e. realized volatility comes in below the implied volatility you sold at. Being short a positive-gamma option is being "short gamma": you collect the option premium (theta) and profit as long as the stock fails to move as much as the implied vol you charged. If the stock realizes more vol than implied, the gamma cost of rehedging overwhelms the premium and you lose. The magnitude is scaled by the dollar gamma $\frac{1}{2}\Gamma S^2$, which is why traders track "dollar gamma" as the local size of the vol bet.

The one-day number

With $r=0$ the ATM-call gamma is

$$\Gamma =\frac{\varphi (d_1)}{S\sigma \sqrt{T}},d_1=\frac{\sigma \sqrt{T}}{2}.$$

Compute the inputs:

$$\sqrt{T}=\sqrt{30/252}=0.34504,\sigma \sqrt{T}=0.069008,d_1=0.034504.$$

$$\varphi (d_1)=\frac{e^{-d_1^2/2}}{\sqrt{2\pi }}=0.39875,\Gamma =\frac{0.39875}{100\times 0.069008}=0.057785.$$

The realized move is zero, so ${\sigma }_r=0$ and the P&L is pure premium capture:

$$d\Pi =\frac{1}{2}\Gamma S^2{\sigma }^{2}dt=\frac{1}{2}(0.057785)({100}^2)(0.04)\left(\frac{1}{252}\right).$$

$$d\Pi =\frac{1}{2}(0.057785)(10000)(0.04)(0.0039683)=0.0459.$$

The short position gains about $0.046 on the day.

Closing note. Because $r=0$, this equals minus the option's one-day theta: $\Theta =-\frac{S\sigma \varphi (d_1)}{2\sqrt{T}}$ gives $\Theta /252=-0.0459$ per day, and a short position earns $+0.0459$. That is the gamma–theta identity $\Theta +\frac{1}{2}{\sigma }^2{S}^2\Gamma =0$ (the $r=0$ PDE) read off directly. The tempting error is to run Itô with the implied vol $\sigma$ in $dV$; then everything cancels to zero and the whole point — that hedging error is a vol arbitrage — disappears.

1. Forward KL: moment matching

Minimizing $\mathrm{K}\mathrm{L}(p\Vert q)={𝔼}_p[\log p]-{𝔼}_p[\log q]$ over $q$ is the same as maximizing ${𝔼}_p[\log q]$, since ${𝔼}_p[\log p]$ does not involve $q$. For $q=𝒩(m,v)$,

$${𝔼}_p[\log q]=-\frac{1}{2}\log (2\pi v)-\frac{{𝔼}_p[(x-m{)}^2]}{2v}.$$

The general principle: the Gaussian family is an exponential family with sufficient statistics $(x,x^2)$, and minimizing $\mathrm{K}\mathrm{L}(p\Vert q)$ over an exponential family forces the expected sufficient statistics of $q$ to equal those of $p$ (the stationarity conditions ${\nabla }_{\eta }[\log Z(\eta )-{\eta }^{\top }{𝔼}_{p}T]=0$ read ${𝔼}_{q}T={𝔼}_{p}T$). Here that means match the mean and variance.

For the mixture, ${𝔼}_p[x]=0$ by symmetry and ${𝔼}_p[x^2]=1+{\mu }^2$ (each component contributes variance $1$ about mean $\pm \mu$). Hence

$$\boxed{q^{\star }=𝒩(0,1+{\mu }^2)}.$$

Forward KL returns a single broad Gaussian straddling both modes: mass-covering, mean-seeking.

2a. A closed form for the reverse KL

Write the mixture as a single expression:

$$𝒩(x;\mu ,1)+𝒩(x;-\mu ,1)=\frac{1}{\sqrt{2\pi }}{e}^{-(x^2+{\mu }^2)/2}(e^{\mu x}+e^{-\mu x})=\frac{2}{\sqrt{2\pi }}{e}^{-(x^2+{\mu }^2)/2}\cosh (\mu x).$$

Multiplying by $\frac{1}{2}$ and taking logs gives the stated identity

$$\log p(x)=-\frac{1}{2}\log (2\pi )-\frac{x^2+{\mu }^2}{2}+\log \cosh (\mu x),$$

and note the mixing constant $\log 2$ cancelled. Now, with $q=𝒩(m,v)$ having entropy $\frac{1}{2}\log (2\pi ev)$ and ${𝔼}_q[x^2]=m^2+v$,

$$F(m,v)=\underset{{𝔼}_q[\log q]}{\underbrace{-\frac{1}{2}\log (2\pi ev)}}-(-\frac{1}{2}\log (2\pi )-\frac{m^2+v+{\mu }^2}{2}+g(m,v)).$$

Collecting the log terms, $-\frac{1}{2}\log (2\pi ev)+\frac{1}{2}\log (2\pi )=-\frac{1}{2}-\frac{1}{2}\log v$, so

$$F(m,v)=-\frac{1}{2}-\frac{1}{2}\log v+\frac{m^2+v+{\mu }^2}{2}-g(m,v),g(m,v)={𝔼}_{𝒩(m,v)}[\log \cosh (\mu x)].$$

2b. The symmetric configuration is always a local minimum

The tools are two Gaussian differentiation identities for $G(m,v)={𝔼}_{𝒩(m,v)}[f(x)]$: integrating by parts against the Gaussian gives

$${\partial }_{m}G=𝔼[f^{\prime }(x)],{\partial }_{v}G=\frac{1}{2}𝔼[f^{″}(x)],$$

the second because the Gaussian density solves the heat equation ${\partial }_v\varphi =\frac{1}{2}{\partial }_x^2\varphi$. Apply to $f(x)=\log \cosh (\mu x)$, with $f^{\prime }=\mu \tanh (\mu x)$ and $f^{″}={\mu }^2{sech}^2(\mu x)$:

$${\partial }_{m}g=\mu {𝔼}_q[\tanh (\mu x)],{\partial }_{v}g=\frac{1}{2}{\mu }^2{𝔼}_q[{sech}^2(\mu x)].$$

Stationarity in $m$ at $m=0$. Then $q=𝒩(0,v)$ is symmetric and $\tanh$ is odd, so ${𝔼}_q[\tanh (\mu x)]=0$. Since ${\partial }_{m}F=m-{\partial }_{m}g$, we get ${\partial }_{m}F(0,v)=0$ for every $v$.

Curvature in $m$. Differentiate again: ${\partial }_m^{2}F=1-{\partial }_m^{2}g=1-{\mu }^2{𝔼}_q[{sech}^2(\mu x)]$. Evaluated at $m=0$,

$${\partial }_m^{2}F(0,v)=1-{\mu }^2{𝔼}_{𝒩(0,v)}[{sech}^2(\mu x)].$$

This looks like it could go negative for large $\mu$ — the tempting route to a pitchfork. The step a candidate misses is to impose the $v$-stationarity condition simultaneously. From ${\partial }_{v}F=-\frac{1}{2v}+\frac{1}{2}-{\partial }_{v}g$, a symmetric stationary point $(0,v_{\star })$ satisfies

$$-\frac{1}{2v_{\star }}+\frac{1}{2}-\frac{1}{2}{\mu }^2{𝔼}_{𝒩(0,v_{\star })}[{sech}^2(\mu x)]=0⟹{\mu }^2𝔼[{sech}^2(\mu x)]=1-\frac{1}{v_{\star }}.$$

Substituting into the curvature,

$${\partial }_m^{2}F(0,v_{\star })=1-(1-\frac{1}{v_{\star }})=\frac{1}{v_{\star }}>0.$$

Also ${\partial }_m{\partial }_{v}F(0,v)=0$ (an odd integrand in $x$), so the Hessian is diagonal there; combined with ${\partial }_v^{2}F>0$ at a $v$-minimum, $(0,v_{\star })$ is a genuine local minimum of the reverse KL. Such a symmetric stationary point exists for every $\mu$: writing $R(v)=1-{\mu }^2{𝔼}_{𝒩(0,v)}[{sech}^2(\mu x)]$, one has $1/v\to +\infty >R$ as $v\to 0$ and $1/v\to 0<R\to 1$ as $v\to \infty$, so $1/v=R(v)$ has a solution. The mass-covering configuration never destabilizes — there is no bifurcation.

3. The mode-locked branch, and who wins globally

Try $q=𝒩(\mu ,1)$. Using the closed form of $\log p$,

$$\log q-\log p=-\frac{(x-\mu {)}^2}{2}+\frac{x^2+{\mu }^2}{2}-\log \cosh (\mu x)=\mu x-\log \cosh (\mu x).$$

For $x>0$, $\mu x-\log \cosh (\mu x)=\log 2-\log (1+e^{-2\mu x})$. Under $q$, $x$ concentrates at $\mu \gg 0$, so ${𝔼}_q[\log (1+e^{-2\mu x})]\to 0$ and

$${lim}_{\mu \to \infty }\mathrm{K}\mathrm{L}(𝒩(\mu ,1)\Vert p)=\log 2.$$

The interpretation is exact: the mode-locked Gaussian captures one component carrying mass $\frac{1}{2}$, and pays the irreducible $-\log \frac{1}{2}=\log 2$ for ignoring the other half. Checking stationarity confirms this is (asymptotically) a critical point: ${\partial }_{m}F=\mu -\mu {𝔼}_q[\tanh (\mu x)]\to 0$ since $\tanh (\mu x)\to 1$, and ${\partial }_{v}F\to -\frac{1}{2v}+\frac{1}{2}=0$ at $v=1$.

The symmetric branch, by contrast, grows. At $(0,v_{\star })$ with large $\mu$, use $\log \cosh (\mu x)\approx |\mu x|-\log 2$, so $g\approx \mu \sqrt{2v/\pi }-\log 2$ (from $𝔼|x|=\sqrt{2v/\pi }$). Minimizing $F\approx -\frac{1}{2}\log v+\frac{v+{\mu }^2}{2}-\mu \sqrt{2v/\pi }$ over $v$ gives $\sqrt{2\pi v}=2\mu$, i.e. $v_{\star }\approx 2{\mu }^2/\pi$, and

$$F(0,v_{\star })\approx (\frac{1}{2}-\frac{1}{\pi }){\mu }^2\approx 0.18{\mu }^2⟶\infty .$$

So although the mass-covering point is a real local minimum, its cost blows up quadratically, while the mode-locked minimum stays at $\log 2$.

Conclusion. For well-separated modes the global reverse-KL optimum is a single mode, $𝒩(\pm \mu ,1)$ — zero-forcing / mode-seeking — whereas forward KL returns the mass-covering $𝒩(0,1+{\mu }^2)$. This is the asymmetry of KL made quantitative.

Closing notes

• The precise picture is subtler than the usual folklore. Reverse KL does not achieve mode-seeking by turning the symmetric solution into a saddle (part 2b shows it stays a local minimum for all $\mu$); rather the mode-locked minima coexist and overtake it as a first-order switch in the global optimum. A gradient-based variational optimizer initialized broadly can get stuck in the suboptimal mass-covering basin.
• Even the reverse-KL mass-covering local optimum underestimates the spread: $v_{\star }\approx 2{\mu }^2/\pi \approx 0.64{\mu }^2$ versus the forward-KL variance $1+{\mu }^2$. Reverse KL systematically produces variances that are too small — its penalty ${𝔼}_q[\log (q/p)]$ is dominated by regions where $q>0$ but $p\approx 0$, so it refuses to place mass where $p$ is negligible.