Tuesday 25 August 2026

Step 1: Ito on the power map

Take $f(x)=x^n$, so $f^{\prime }(x)=n{x}^{n-1}$ and $f^{″}(x)=n(n-1)x^{n-2}$. Ito's lemma applies because $f$ is $C^2$ on $(0,\infty )$ and $S_t$ stays strictly positive (GBM never hits zero). With $d{S}_t=\mu S_{t}dt+\sigma S_{t}d{W}_t$ the quadratic variation term is $\frac{1}{2}{f}^{″}(S_t)(d{S}_t{)}^2=\frac{1}{2}n(n-1)S_t^{n-2}{\sigma }^2{S}_t^{2}dt$. Hence

$$d{Y}_t=n{S}_t^{n-1}d{S}_t+\frac{1}{2}n(n-1)S_t^{n-2}{\sigma }^2{S}_t^{2}dt.$$

Substituting $d{S}_t$ and collecting terms, each of which carries a factor $S_t^n=Y_t$,

$$d{Y}_t=Y_t[(n\mu +\frac{1}{2}n(n-1){\sigma }^2)dt+n\sigma d{W}_t].$$

The step a candidate most often botches is dropping the $\frac{1}{2}n(n-1){\sigma }^2$ contribution: $Y_t$ is itself a geometric Brownian motion, but with an inflated (or deflated) drift, not simply drift $n\mu$.

Step 2: The expectation

Write the drift coefficient as $a:=n\mu +\frac{1}{2}n(n-1){\sigma }^2$ and the volatility as $b:=n\sigma$. Then $Y_t$ solves a linear SDE with constant coefficients, whose solution is

$$Y_t=Y_0\exp ((a-\frac{1}{2}{b}^2)t+b{W}_t).$$

Taking expectations and using $𝔼\left[e^{b{W}_t}\right]=e^{\frac{1}{2}{b}^{2}t}$ (moment generating function of $W_t~𝒩(0,t)$), the $-\frac{1}{2}{b}^2$ and $+\frac{1}{2}{b}^2$ cancel and only the drift survives:

$$𝔼[S_t^n]=𝔼[Y_t]=S_0^n{e}^{at}.$$

Explicitly,

$$𝔼[S_t^n]=S_0^n\exp \left((n\mu +\frac{1}{2}n(n-1){\sigma }^2)t\right).$$

Equivalently, taking $𝔼[d{Y}_t]$ directly kills the $d{W}_t$ term (it is a true stochastic integral of a square-integrable integrand, hence a martingale with zero mean) and leaves the ODE $m^{\prime }(t)=am(t)$ with $m(0)=S_0^n$, giving the same answer.

Step 3: When is $S_t^n$ a martingale?

A nonnegative Ito process is a martingale exactly when its drift vanishes identically (the $d{W}_t$ part is already a local martingale, and here it is a genuine martingale since $𝔼{\int }_0^t{Y}_s^{2}ds<\infty$). So we need

$$n\mu +\frac{1}{2}n(n-1){\sigma }^2=0.$$

Factor out $n$:

$$n[\mu +\frac{1}{2}(n-1){\sigma }^2]=0.$$

The trivial root $n=0$ gives the constant process $Y_t\equiv 1$. The nonzero root is

$$n=1-\frac{2\mu }{{\sigma }^2}.$$

For $\mu =0.05$, $\sigma =0.20$:

$$n=1-\frac{2(0.05)}{0.04}=1-2.5=-1.5.$$

Answers

1. $d{Y}_t=Y_t[(n\mu +\frac{1}{2}n(n-1){\sigma }^2)dt+n\sigma d{W}_t]$.
2. $𝔼[S_t^n]=S_0^n\exp ((n\mu +\frac{1}{2}n(n-1){\sigma }^2)t)$.
3. $n=0$ or $n=1-2\mu /{\sigma }^2$; the nontrivial value is $n=-1.5$.

Closing note. The tempting wrong answer to part 2 is $S_0^n{e}^{n\mu t}$, which ignores the convexity correction and is only correct when $n\in \{0,1\}$. The martingale exponent $n=1-2\mu /{\sigma }^2$ is the same object that appears in the general solution of the stationary form of barrier/perpetual-option ODEs; its emergence here from a one-line drift condition is worth remembering.

1. The ELBO decomposition and monotonicity

Start from the pointwise identity, valid for any $z$, $$\log p(x;{\theta }^{\prime })=\log p(x,z;{\theta }^{\prime })-\log p(z\mid x;{\theta }^{\prime }).$$ The left side does not depend on $z$, so we may take ${𝔼}_{z~p(·\mid x;\theta )}$ of both sides without changing it: $$\ell ({\theta }^{\prime })=\underset{=Q({\theta }^{\prime }\mid \theta )}{\underbrace{{𝔼}_{z\mid x;\theta }[\log p(x,z;{\theta }^{\prime })]}}-{𝔼}_{z\mid x;\theta }[\log p(z\mid x;{\theta }^{\prime })].$$ Doing the same with ${\theta }^{\prime }=\theta$ gives $\ell (\theta )=Q(\theta \mid \theta )-{𝔼}_{z\mid x;\theta }[\log p(z\mid x;\theta )]$. Subtract: $$\ell ({\theta }^{\prime })-\ell (\theta )=[Q({\theta }^{\prime }\mid \theta )-Q(\theta \mid \theta )]+{𝔼}_{z\mid x;\theta }\left[\log \frac{p(z\mid x;\theta )}{p(z\mid x;{\theta }^{\prime })}\right].$$ The last expectation is exactly $\mathrm{K}\mathrm{L}(p(·\mid x;\theta )\Vert p(·\mid x;{\theta }^{\prime }))$, which proves the identity.

The key facts:

• KL nonnegativity (Gibbs' inequality) makes the second bracket $\ge 0$ for free — this is the term the E-step handles: choosing the averaging distribution to be the posterior at $\theta$ is what makes this a KL and not some signed quantity.
• The M-step controls only the first bracket. By definition $M(\theta )=\mathrm{\backslash arg}{max}_{{\theta }^{\prime }}Q({\theta }^{\prime }\mid \theta )$, so $Q(M(\theta )\mid \theta )\ge Q(\theta \mid \theta )$, hence the first bracket is $\ge 0$ at ${\theta }^{\prime }=M(\theta )$.

Both brackets nonnegative gives $\ell (M(\theta ))-\ell (\theta )\ge 0$. (Note only $Q({\theta }^{\prime }\mid \theta )\ge Q(\theta \mid \theta )$ is needed, not full maximization — this is why generalized EM is also monotone.)

2. Fixed points are stationary — and that is all

The step a candidate misses is Fisher's identity: for every $\theta$, $${\partial }_{{\theta }^{\prime }}Q({\theta }^{\prime }\mid \theta ){|}_{{\theta }^{\prime }=\theta }={\ell }^{\prime }(\theta ).$$ Indeed ${\partial }_{{\theta }^{\prime }}Q({\theta }^{\prime }\mid \theta )={𝔼}_{z\mid x;\theta }[{\partial }_{{\theta }^{\prime }}\log p(x,z;{\theta }^{\prime })]$, and at ${\theta }^{\prime }=\theta$ this is $$\int \frac{{\partial }_{\theta }p(x,z;\theta )}{p(x,z;\theta )}\frac{p(x,z;\theta )}{p(x;\theta )}dz=\frac{{\partial }_{\theta }\int p(x,z;\theta )dz}{p(x;\theta )}=\frac{{\partial }_{\theta }p(x;\theta )}{p(x;\theta )}={\ell }^{\prime }(\theta ),$$ using the assumed interchange of ${\partial }_{\theta }$ and $\int$.

Now if ${\theta }^{\backslash }*=M({\theta }^{\backslash }*)$ is an interior maximizer of $Q(·\mid {\theta }^{\backslash }*)$, its first-order condition is ${\partial }_{{\theta }^{\prime }}Q({\theta }^{\prime }\mid {\theta }^{\backslash }*){|}_{{\theta }^{\prime }={\theta }^{\backslash }*}=0$, which by Fisher's identity equals ${\ell }^{\prime }({\theta }^{\backslash }*)$. Hence ${\ell }^{\prime }({\theta }^{\backslash }*)=0$.

Conclusion. Monotonicity plus this shows EM's limit set consists of stationary points of $\ell$. It buys nothing more: a stationary point may be a local maximum, a saddle, or (with a flat direction) a plateau. Monotone ascent does not distinguish these, and it says nothing about reaching the global maximum. That $\ell$ never decreases is a statement about the sequence of values, not about where they land.

3. The convergence rate is the fraction of missing information

Define $M$ implicitly by the M-step stationarity ${\partial }_{{\theta }^{\prime }}Q(M(\theta )\mid \theta )=0$. Differentiate in $\theta$: $$Q_{20}{M}^{\prime }(\theta )+Q_{11}=0,M^{\prime }(\theta )=-\frac{Q_{11}}{Q_{20}},$$ where $Q_{20}={\partial }_{{\theta }^{\prime }}^{2}Q$ and $Q_{11}={\partial }_{\theta }{\partial }_{{\theta }^{\prime }}Q$, evaluated at $(M(\theta ),\theta )$. At the fixed point $-Q_{20}({\theta }^{\backslash }*\mid {\theta }^{\backslash }*)=I_{\mathrm{c}\mathrm{o}\mathrm{m}}$.

To identify $Q_{11}$, differentiate Fisher's identity ${\ell }^{\prime }(\theta )={\partial }_{{\theta }^{\prime }}Q({\theta }^{\prime }\mid \theta ){|}_{{\theta }^{\prime }=\theta }$ totally in $\theta$ (both the first and second argument move): $${\ell }^{″}(\theta )=Q_{20}(\theta \mid \theta )+Q_{11}(\theta \mid \theta ).$$ At ${\theta }^{\backslash }*$ this reads $-I_{\mathrm{o}\mathrm{b}\mathrm{s}}=-I_{\mathrm{c}\mathrm{o}\mathrm{m}}+Q_{11}$, so $Q_{11}({\theta }^{\backslash }*\mid {\theta }^{\backslash }*)=I_{\mathrm{c}\mathrm{o}\mathrm{m}}-I_{\mathrm{o}\mathrm{b}\mathrm{s}}=I_{\mathrm{m}\mathrm{i}\mathrm{s}}$. Therefore $$M^{\prime }({\theta }^{\backslash }*)=-\frac{Q_{11}}{Q_{20}}=\frac{I_{\mathrm{m}\mathrm{i}\mathrm{s}}}{I_{\mathrm{c}\mathrm{o}\mathrm{m}}}=1-\frac{I_{\mathrm{o}\mathrm{b}\mathrm{s}}}{I_{\mathrm{c}\mathrm{o}\mathrm{m}}}.$$ Since $0\le I_{\mathrm{m}\mathrm{i}\mathrm{s}}\le I_{\mathrm{c}\mathrm{o}\mathrm{m}}$, this lies in $[0,1)$: EM contracts, but at a rate that degrades to $1$ as the latent structure hides more of the information.

The Gaussian instance. With ${\sigma }^2$ known, the complete-data log-likelihood is $-\frac{1}{2{\sigma }^2}{\sum }_{i=1}^n(z_i-\mu {)}^2+\text{const}$. Given the current $\mu$, the posterior over each missing $z_i$ is $N(\mu ,{\sigma }^2)$, so $$Q({\mu }^{\prime }\mid \mu )=-\frac{1}{2{\sigma }^2}[\sum _{i=1}^m(z_i-{\mu }^{\prime }{)}^2+\sum _{i=m+1}^n𝔼[(z_i-{\mu }^{\prime }{)}^2]]+\text{const},$$ and maximizing in ${\mu }^{\prime }$ replaces each missing $z_i$ by its conditional mean $\mu$: $$M(\mu )=\frac{1}{n}(\underset{S}{\underbrace{{\sum }_{i=1}^m{z}_i}}+(n-m)\mu ).$$ This map is affine with slope $(n-m)/n$ everywhere, so from any start EM converges linearly (not just locally) to the fixed point ${\mu }^{\backslash }*=S/m={\overline{z}}_{\text{obs}}$, at exact rate $$\boxed{\rho =\frac{n-m}{n}}$$ Check against the information formula: $I_{\mathrm{c}\mathrm{o}\mathrm{m}}=-{\partial }_{{\mu }^{\prime }}^{2}Q=n/{\sigma }^2$, $I_{\mathrm{o}\mathrm{b}\mathrm{s}}=m/{\sigma }^2$, $I_{\mathrm{m}\mathrm{i}\mathrm{s}}=(n-m)/{\sigma }^2$, so $I_{\mathrm{m}\mathrm{i}\mathrm{s}}/I_{\mathrm{c}\mathrm{o}\mathrm{m}}=(n-m)/n$, matching. If half the data are missing the rate is $1/2$; as $m\to 1$ it approaches $1-1/n$ and EM crawls.

Closing note

The instance is deliberately absurd — the MLE is just the observed-sample mean ${\overline{z}}_{\text{obs}}=S/m$, obtained in closed form. That is the point: even here EM can be made arbitrarily slow purely by increasing the fraction of missing data, and monotonicity is powerless to prevent it. Two tempting errors worth flagging: (i) believing monotone ascent implies convergence to the global maximum (Part 2: only stationarity is guaranteed); and (ii) believing a guaranteed increase at every step implies fast convergence (Part 3: the rate is the fraction of missing information and can sit just below $1$).