Deterministic sampling with DDIM

Toggle between Deterministic (DDIM) and Stochastic (DDPM-style), then step through the same 4-step schedule with the same starting noise and the same (perfect) predictor. The deterministic path retraces a single fixed trajectory back to $x_0=5$ every time. The stochastic path, built from the exact same ingredients plus one injected random draw per step, wanders off that trajectory and never quite gets back.

Every reverse step first estimates $x_0$ directly from the current $x_t$, then rebuilds $x_{t_{\text{prev}}}$ from that estimate using the forward process's own closed form:

• $\bar\alpha_t$ — the cumulative signal-retained fraction at timestep $t$: how much of the original signal survives, all the way from $t=0$.
• $\hat\epsilon$ — the model's predicted noise direction at $x_t$, assumed here to be exactly correct.
• $\hat{x}_0$ — the estimate of the clean data, computed directly from $x_t$ and $\hat\epsilon$, regardless of which path reached $(x_t, t)$.
• $\eta$ — the noise-injection knob. $\eta=0$ is DDIM: fully deterministic. $\eta=1$ replaces the predicted direction with an injected draw $z$, a DDPM-style stochastic step.
• $t_{\text{prev}}$ — the next timestep in the schedule — any earlier timestep, not necessarily $t-1$.

1. x0 is estimated the same way no matter how x_t was reached

   Because $\hat{x}_0$ depends only on $(x_t, t, \hat\epsilon)$ — never on the sequence of steps that produced $x_t$ — a perfect predictor recovers the exact same $\hat{x}_0$ whether $t_{\text{prev}}$ is one step back or ten.

2. That's what makes skipping timesteps free

   Since $t_{\text{prev}}$ can be any earlier timestep in the formula above, a schedule can skip straight from $t=4$ to $t=2$ to $t=0$, using the model only twice instead of four times, and land on the identical trajectory a full step-by-step schedule would trace.

3. Injected noise breaks that guarantee

   Setting $\eta > 0$ swaps part of the deterministic direction for a random draw $z$. Once $x_t$ is knocked off the clean marginal it was estimated from, the next step's $\hat{x}_0$ estimate inherits that error — which is exactly why a stochastic path drifts even from identical starting noise and an identical predictor.

Drag $\eta$ from $0$ up to $1$. At $\eta=0$ (pure DDIM) the reconstruction is exact. As $\eta$ rises, more of each step's direction comes from the fixed random draws instead of the predictor, and the final reconstructed $x_0$ drifts further from the true value — even though the predictor itself never changes.

With $x_0=5$, $\hat\epsilon=1$ always correct, and $\bar\alpha = [1, 0.8, 0.6, 0.4, 0.2]$ for $t=0,1,2,3,4$:

1. Full schedule: every step, t = 4 -> 3 -> 2 -> 1 -> 0

   Each hop estimates $\hat x_0=(x_t-\sqrt{1-\bar\alpha_t})/\sqrt{\bar\alpha_t}$ (since $\hat\epsilon=1$), then rebuilds $x_{t_{\text{prev}}}=\sqrt{\bar\alpha_{t_{\text{prev}}}}\hat x_0+\sqrt{1-\bar\alpha_{t_{\text{prev}}}}$:

   • $t=4\to3$: $\hat x_0=(3.1305-\sqrt{0.8})/\sqrt{0.2}=(3.1305-0.8944)/0.4472=5.0000$; $x_3=\sqrt{0.4}(5)+\sqrt{0.6}(1)=3.1623+0.7746=3.9369$
   • $t=3\to2$: $\hat x_0=(3.9369-\sqrt{0.6})/\sqrt{0.4}=(3.9369-0.7746)/0.6325=5.0000$; $x_2=\sqrt{0.6}(5)+\sqrt{0.4}(1)=3.8730+0.6325=4.5054$
   • $t=2\to1$: $\hat x_0=(4.5054-\sqrt{0.4})/\sqrt{0.6}=(4.5054-0.6325)/0.7746=5.0000$; $x_1=\sqrt{0.8}(5)+\sqrt{0.2}(1)=4.4721+0.4472=4.9193$
   • $t=1\to0$: $\hat x_0=(4.9193-\sqrt{0.2})/\sqrt{0.8}=(4.9193-0.4472)/0.8944=5.0000$; $x_0=\sqrt{1}(5)+\sqrt{0}(1)=5$

   Every hop's $\hat x_0$ lands exactly on $5$ — the perfect predictor recovers the true $x_0$ regardless of which timestep it's called from.

2. Skip schedule: t = 4 -> 2 -> 0, two model calls instead of four

   Starting from the identical $x_4 \approx 3.1305$:

   • $t=4\to2$: $\hat x_0=(3.1305-\sqrt{0.8})/\sqrt{0.2}=5.0000$ (same computation as the full schedule's first hop, since it depends only on $x_4$); $x_2=\sqrt{0.6}(5)+\sqrt{0.4}(1)=4.5054$ — the exact same value the full schedule passed through
   • $t=2\to0$: $\hat x_0=(4.5054-\sqrt{0.4})/\sqrt{0.6}=5.0000$; $x_0=\sqrt{1}(5)+\sqrt{0}(1)=5$ exactly
3. Single jump: t = 4 -> 0

   $\hat x_0=(3.1305-\sqrt{0.8})/\sqrt{0.2}=5.0000$ (the same estimate every schedule above computed from $x_4$); $x_0=\sqrt{1}(5)+\sqrt{0}(1)=5$ exactly. With a perfect predictor, the number of steps only changes the compute cost — never the destination.

4. Now inject noise: eta = 1, fixed draws

   Running the same full schedule with $\eta=1$ (direction $=z$ instead of $\hat\epsilon$) and fixed draws $z=[0.5,-0.5,0.5,0.3]$:

   • $t=4\to3$: $\hat x_0=5.0000$ (unchanged, from $x_4$); direction $=z=0.5$; $x_3=\sqrt{0.4}(5.0000)+\sqrt{0.6}(0.5)=3.1623+0.3873=3.5496$
   • $t=3\to2$: $\hat x_0=(3.5496-\sqrt{0.6})/\sqrt{0.4}=(3.5496-0.7746)/0.6325\approx4.3876$ — already off the true $5$, since $x_3$ was knocked off the clean marginal; direction $=z=-0.5$; $x_2=\sqrt{0.6}(4.3876)+\sqrt{0.4}(-0.5)\approx3.3990-0.3162=3.0828$
   • $t=2\to1$: $\hat x_0=(3.0828-\sqrt{0.4})/\sqrt{0.6}\approx3.1629$; direction $=z=0.5$; $x_1=\sqrt{0.8}(3.1629)+\sqrt{0.2}(0.5)\approx2.8295+0.2236=3.0526$
   • $t=1\to0$: $\hat x_0=(3.0526-\sqrt{0.2})/\sqrt{0.8}\approx2.9129$; direction $=z=0.3$; $x_0=\sqrt{1}(2.9129)+\sqrt{0}(0.3)\approx2.913$

   Final $x_0\approx2.913$ — an error of $5-2.913\approx2.09$, because each step's injected noise knocks the next step's estimate off the clean marginal it assumes.