Exercises — Score-based generative models

Take the log first — that is the whole trick of a "score": $\log p(x)=-\frac{1}{2}{x}^2-\frac{1}{2}\log (2\pi ).$ The second term is a constant (no $x$), so its derivative is $0$. Differentiate: $s(x)=\frac{d}{dx}\log p(x)=-x.$ WHY the log: logging turns the product/exponential into a sum, and the annoying constant $\frac{1}{\sqrt{2\pi }}$ becomes an additive constant that dies under the derivative. That is the "$Z$ disappears" magic in miniature.

At $x=+3$: $s=-3$ — the arrow points left, back toward the mean $0$ (higher density). At $x=-3$: $s=+3$ — points right, again toward $0$. The score always aims at the peak. ✅

False. $\log p=\log \tilde{p}-\log Z$, and $\log Z$ is a constant in $x$, so ${\nabla }_x\log Z=0$. Hence ${\nabla }_x\log p={\nabla }_x\log \tilde{p}$. You never touch $Z$. ✅

(a) $\tilde{x}=5+2(1.5)=8$.

(b) From the parent note, for the Gaussian $q_{\sigma }(\tilde{x}\mid x)=\mathcal{N}(\tilde{x};x,{\sigma }^2)$, ${\nabla }_{\tilde{x}}\log q_{\sigma }(\tilde{x}\mid x)=-\frac{\tilde{x}-x}{{\sigma }^2}=-\frac{ϵ}{\sigma }.$ Plug in: $-\frac{1.5}{2}=-0.75$. Check via the other form: $-\frac{\tilde{x}-x}{{\sigma }^2}=-\frac{8-5}{2^2}=-\frac{3}{4}=-0.75$. ✅ Both agree — the two formulas are the same thing since $\tilde{x}-x=\sigma ϵ$, so $-\frac{\tilde{x}-x}{{\sigma }^2}=-\frac{\sigma ϵ}{{\sigma }^2}=-\frac{ϵ}{\sigma }$.

What it means: the score $-0.75$ points back down the noise you added, toward the clean $x=5$. That is exactly "denoising."

Recall $s_{\theta }(\cdot )$ is the network's output; we want it to approximate the true score. The training target is $-ϵ/\sigma =-1.5/2=-0.75$ (same value as the true conditional score in 2.1b — the network learns exactly that score).

The denoising loss is $\frac{1}{2}\Vert s_{\theta }+ϵ/\sigma {\Vert }^2=\frac{1}{2}(s_{\theta }-(-ϵ/\sigma ){)}^2$. With $s_{\theta }=-0.5$ and target $-0.75$: $\frac{1}{2}(-0.5-(-0.75){)}^2=\frac{1}{2}(0.25{)}^2=\frac{1}{2}(0.0625)=\mathrm{0.03125.}$ ✅

One term of the objective is $\lambda (\sigma )\mathbb{E}[\frac{1}{2}\Vert s_{\theta }(\tilde{x},\sigma )+\frac{ϵ}{\sigma }{\Vert }^2].$ (Recall $s_{\theta }$ is the learned score — the network's output.) Substitute $\lambda (\sigma )={\sigma }^2$ and pull the scalar inside the norm. Why is that legal? For any vector $u$ and scalar $\sigma$, ${\sigma }^2\Vert u{\Vert }^2=\Vert \sigma u{\Vert }^2$ — because $\Vert \sigma u{\Vert }^2={\sum }_i(\sigma u_i{)}^2={\sigma }^2{\sum }_i{u}_i^2={\sigma }^2\Vert u{\Vert }^2$. So the constant ${\sigma }^2$ outside a squared norm is the same as one factor $\sigma$ multiplying the vector inside it. Applying this with $u=s_{\theta }+ϵ/\sigma$: ${\sigma }^2\cdot \frac{1}{2}\Vert s_{\theta }+\frac{ϵ}{\sigma }{\Vert }^2=\frac{1}{2}\Vert \sigma (s_{\theta }+\frac{ϵ}{\sigma }){\Vert }^2=\frac{1}{2}\Vert \sigma s_{\theta }+ϵ{\Vert }^2.$ WHY this matters: define the reparameterised network output ${\widehat{ϵ}}_{\theta }\equiv -\sigma s_{\theta }$. Then the term is $\frac{1}{2}\Vert ϵ-{\widehat{ϵ}}_{\theta }{\Vert }^2$ — a plain "predict the unit-variance noise $ϵ$" loss. The typical magnitude of $ϵ$ is $O(1)$ at every noise level, so gradients across scales have comparable size. Without the ${\sigma }^2$ weight, the raw target $ϵ/\sigma$ blows up as $\sigma \to 0$ and the smallest-noise level would dominate training. ✅

At $x_t=0$: the drift $\alpha s(0)=\alpha \cdot 0=0$. So the expected displacement is $\mathbb{E}[x_{t+1}-x_t]=0.$ But the noise term $\sqrt{2\alpha }{z}_t$ with $z_t\sim \mathcal{N}(0,1)$ has variance $\mathrm{Var}(\sqrt{2\alpha }{z}_t)=2\alpha \cdot 1=2\alpha .$ What stops collapse: exactly this injected noise. If we used only the drift, the chain would settle at the mode and stay — giving one point, not a distribution. The $\sqrt{2\alpha }{z}_t$ term keeps kicking $x$ off the peak; the drift then pulls it back. The tug-of-war between "pull uphill" and "random kick" is what makes the stationary spread of the chain match the true distribution's spread (its detailed-balance property). ✅

Distribution: conditioned on $x_0$, $x_t$ is Gaussian with mean ${\alpha }_t{x}_0$ and variance ${\sigma }_t^2$: $q(x_t\mid x_0)=\mathcal{N}(x_t;{\alpha }_t{x}_0,{\sigma }_t^2).$ Score: using ${\nabla }_y\log \mathcal{N}(y;\mu ,\Sigma )=-{\Sigma }^{-1}(y-\mu )$ with $y=x_t,\mu ={\alpha }_t{x}_0,\Sigma ={\sigma }_t^2$: ${\nabla }_{x_t}\log q(x_t\mid x_0)=-\frac{x_t-{\alpha }_t{x}_0}{{\sigma }_t^2}.$ Substitute $x_t-{\alpha }_t{x}_0={\sigma }_tϵ$: $=-\frac{{\sigma }_tϵ}{{\sigma }_t^2}=-\frac{ϵ}{{\sigma }_t}.$ Conversion: if the net outputs ${\widehat{ϵ}}_{\theta }$, then the score estimate is $s_{\theta }=-{\widehat{ϵ}}_{\theta }/{\sigma }_t$ — one division by the (known) scalar ${\sigma }_t$. ✅

Numeric check: $x_t=0.8(2)+0.6(0.5)=1.6+0.3=1.9$. Score $=-\frac{x_t-{\alpha }_t{x}_0}{{\sigma }_t^2}=-\frac{1.9-1.6}{0.36}=-\frac{0.3}{0.36}=-0.8\overline{3}$. Cross-check $-ϵ/{\sigma }_t=-0.5/0.6=-0.8\overline{3}$. ✅ Match.

Geometric means each level is the previous times $r=1/5$: ${\sigma }_1=10,{\sigma }_2=10\cdot \frac{1}{5}=2,{\sigma }_3=2\cdot \frac{1}{5}=\mathrm{0.4.}$ So ${\sigma }_L={\sigma }_3=0.4$. Step sizes ${\alpha }_i=0.1({\sigma }_i/0.4{)}^2$:

• ${\alpha }_1=0.1(10/0.4{)}^2=0.1(25{)}^2=0.1\cdot 625=\mathrm{62.5.}$
• ${\alpha }_2=0.1(2/0.4{)}^2=0.1(5{)}^2=0.1\cdot 25=\mathrm{2.5.}$
• ${\alpha }_3=0.1(0.4/0.4{)}^2=0.1\cdot 1=\mathrm{0.1.}$

Reading it: big $\sigma$ ⇒ big steps (coarse exploration of all of space); small $\sigma$ ⇒ tiny steps (careful refinement near the data). Exactly the annealing intent. ✅

Notation note: we are in one dimension, so the gradient ${\nabla }_x$ is just the ordinary derivative. Below we write $p^{\prime }(x)$ as shorthand for $\frac{d}{dx}p(x)={\nabla }_{x}p(x)$ — it is the same $\nabla$ used everywhere else on this page, only spelled with a prime to keep the fraction readable.

(a) Let $g_{\mu }(x)=\mathcal{N}(x;\mu ,1)=\frac{1}{\sqrt{2\pi }}{e}^{-(x-\mu {)}^2/2}$. Then $p=\frac{1}{2}{g}_{-3}+\frac{1}{2}{g}_3$. By the chain rule, ${\nabla }_x\log p=\frac{{\nabla }_{x}p}{p}=\frac{p^{\prime }(x)}{p(x)}$, so the score of a mixture is a weighted average of component scores, weighted by posterior responsibility: $s(x)=\frac{p^{\prime }(x)}{p(x)}=\frac{\frac{1}{2}{g}_{-3}^{\prime }(x)+\frac{1}{2}{g}_3^{\prime }(x)}{p(x)}.$ Since $g_{\mu }^{\prime }(x)=-(x-\mu )g_{\mu }(x)$, $s(x)=\frac{\frac{1}{2}(-(x+3))g_{-3}(x)+\frac{1}{2}(-(x-3))g_3(x)}{\frac{1}{2}{g}_{-3}(x)+\frac{1}{2}{g}_3(x)}.$

(b) At $x=0$: by symmetry $g_{-3}(0)=g_3(0)$, so they cancel in the numerator's structure. Numerator $\propto -(0+3)+-(0-3)=-3+3=0$. Hence $s(0)=0.$ Interpretation: $x=0$ is a local minimum (a valley) of $\log p$ — the low-density point between the two modes. In one dimension there are no saddles, so this is a genuine minimum: the slope vanishes there, and on either side $\log p$ rises toward a mode. The score gives no directional guidance at the very bottom of the valley (equal pull from both sides).

(c) Single-scale Langevin from $x_0=0$: the drift is $\approx 0$, so only the small noise $\sqrt{2\alpha }z$ moves you. To cross from the $x=0$ valley to either peak at $\pm 3$ you must random-walk across a low-density gap — exponentially unlikely per step, so the chain gets stuck near one mode (poor mode coverage).

Fix via annealing: at a large ${\sigma }_1$, the target is convolved with wide Gaussian noise, which fills in the valley — the two blurred modes merge into one broad hump whose score smoothly points toward the middle mass. Sampling there mixes freely across the whole region. As we anneal $\sigma$ down, the single hump splits back into two, and by then the chain is already distributed across both basins. This is precisely why multi-scale score matching exists. ✅

Compute the two responsibilities' unnormalised weights at $x=2$:

• $g_{-3}(2)\propto e^{-(2+3{)}^2/2}=e^{-25/2}=e^{-12.5}$.
• $g_3(2)\propto e^{-(2-3{)}^2/2}=e^{-1/2}=e^{-0.5}$.

Numerator of $s$: $-(2+3)e^{-12.5}+-(2-3)e^{-0.5}=-5e^{-12.5}+1\cdot e^{-0.5}$. Denominator: $e^{-12.5}+e^{-0.5}$. $s(2)=\frac{-5e^{-12.5}+e^{-0.5}}{e^{-12.5}+e^{-0.5}}.$ Numerically $e^{-12.5}\approx 3.72665\times 10^{-6}$, $e^{-0.5}\approx 0.6065307$: $s(2)\approx \frac{-5(3.72665\text{e-}6)+0.6065307}{3.72665\text{e-}6+0.6065307}=\frac{0.6065121}{0.6065344}\approx \mathrm{0.999963.}$ Rounding to 4 decimal places: the fifth decimal is $6(\ge 5)$, so $0.99996\dots$ rounds up at the fourth place, giving $s(2)\approx 1.0000\text{(to 4 d.p.)}.$ Positive ⇒ the arrow points right, toward the mode at $x=3$. Makes sense: at $x=2$ you're on the inner slope of the right mode, so uphill is rightward (toward $3$), and the far-left mode contributes almost nothing (its weight $e^{-12.5}$ is negligible). ✅

The normalizer $Z$ vanishes from the score because ${\nabla }_x\log Z=$ ::: $0$ (it is constant in $x$). The denoising target given noise $ϵ$ at level $\sigma$ is ::: $-ϵ/\sigma$. The Langevin noise term coefficient is ::: $\sqrt{2\alpha }$. The ${\sigma }^2$ loss weighting reduces the objective to predicting ::: the unit noise $ϵ$. A mixture's score is a ::: responsibility-weighted average of component scores.