Catastrophic forgetting: Exercises — AI-ML deep dive

Formula The one loss we keep reusing (EWC)

$L_{EWC} (θ) = L_{new} (θ) + \frac{λ}{2} \sum_{i} F_{i} (θ_{i} - θ_{i}^{*})^{2}$ Reading it as English: "be good at the new task, AND for every weight $i$ , pay a fine of $\frac{λ}{2} F_{i}$ times the squared distance you drag that weight away from its old-task home $θ_{i}^{*}$ ."

$θ_{i}$ — the current value of weight number $i$ (a knob you are turning).
$θ_{i}^{*}$ — where that knob sat after the old task ("home position").
$F_{i}$ — the Fisher importance of that knob: how much the old task cared about it. Big $F_{i}$ = stiff spring = do not move me. See Fisher Information.
$λ$ — a single global dial for "how loyal to the past."

Domain constraints (important!): $λ \geq 0$ and every $F_{i} \geq 0$ . The Fisher information is an average of squared gradients, so it can never be negative; $λ$ is a strength dial we choose, and we never pick it negative. A negative $λ$ or $F_{i}$ would flip the spring into a push-away force — it would actively encourage forgetting, the opposite of the goal. So both live in $[0, \infty)$ .

Level 1 — Recognition

Recall Solution — L1.1

(b) abrupt. The word catastrophic is the whole point: the drop is sudden, not a gentle fade. In the toy example of the parent note, Task-A accuracy plunges from 98% to ~40% after just training on Task B.

Recall Solution — L1.2

$F_{i}$ , the Fisher information of weight $i$ . Large $F_{i}$ ⇒ stiff spring ⇒ moving that weight hurts the old task a lot. $λ$ is a global strength; $θ^{*}$ is the anchor point; only $F_{i}$ is per-weight importance.

Recall Solution — L1.3

1→B, 2→A, 3→C. Mnemonic from the parent: "REPLAY the FIsher, FREEZE the base." Replay puts old loss back in the objective; LoRA freezes the base so it literally cannot be overwritten; low LR just drifts less per step.

Level 2 — Application

Recall Solution — L2.1

Penalty $= \frac{λ}{2} F (θ - θ^{*})^{2} = \frac{1}{2} \cdot 8 \cdot (3.0 - 2.0)^{2} = 4 \cdot 1 = 4.0$ . What this means: dragging this important weight one unit away costs 4 units of loss — a strong pull back toward home.

Recall Solution — L2.2

Move weight 1: $\frac{1}{2} (100) (1)^{2} = 50$ . Move weight 2: $\frac{1}{2} (0.01) (1)^{2} = 0.005$ . EWC strongly prefers moving weight 2 (cost $0.005 ≪ 50$ ). This is the whole idea: the new task is pushed to learn in the spare capacity (low- $F$ directions) and leave the important old weight alone.

Recall Solution — L2.3

Penalty $= \frac{λ}{2} \cdot 8 \cdot 1 = 4 λ$ .

$λ = 0 \Rightarrow 0$ — no protection at all, i.e. plain fine-tuning (pure forgetting).
$λ = 1 \Rightarrow 4$ .
$λ = 1000 \Rightarrow 4000$ — huge, weight nearly frozen at $θ^{*}$ . So $λ$ interpolates from "forget freely" ( $0$ ) to "freeze the past" ( $\infty$ ). (Recall $λ \geq 0$ — negative is not allowed.)

Level 3 — Analysis

Recall Solution — L3.1

Follow the amber dashed curve: gradient descent on $L_{new}$ only "sees" the amber valley, so it rolls $θ$ from $θ^{*}$ rightward to $θ_{new}$ . But look at the cyan solid curve at that landing point — it is now high up the wall of the old valley, so old-task loss is large ⇒ accuracy collapses. EWC adds a spring pulling toward $θ^{*}$ , so its minimum — the white dashed vertical marker — sits at a compromise point between the two valleys: decent on new (amber), still low on old (cyan). See Continual Learning for the general picture.

Recall Solution — L3.2

Objective: $J (θ) = \frac{1}{2} (θ - 10)^{2} + \frac{1}{2} \cdot 4 \cdot θ^{2}$ . The calculus, step by step. To find the bottom of a bowl we set its slope to zero, so we differentiate. For a term $\frac{G}{2} (θ - c)^{2}$ : bring the power down and multiply by the inside-derivative (chain rule) — the $2$ from the square cancels the $\frac{1}{2}$ , and the inside $(θ - c)$ differentiates to $1$ , leaving $\frac{G}{2} \cdot 2 (θ - c) \cdot 1 = G (θ - c)$ . With $G = 1$ that is $(θ - 10)$ . The penalty term $\frac{F}{2} (θ - 0)^{2}$ differentiates the same way to $F θ = 4 θ$ . Add and set to zero: $J^{'} (θ) = (θ - 10) + 4 θ = 5 θ - 10 = 0 \Rightarrow θ = 2$ . Interpretation: without the penalty, new task alone would want $θ = 10$ . The old-task spring ( $F = 4$ ) is 4× stiffer than the new-task spring ( $G = 1$ ), so the compromise sits much closer to home: $θ = \frac{G c + λ F θ ^{*}}{G + λ F} = \frac{1 \cdot 10}{1 + 4} = 2$ .

Recall Solution — L3.3

$θ_{2}$ (with $F_{2} = 0$ ) has zero penalty — the EWC spring is slack, so it is completely free. $θ_{1}$ is stiff. The "spare capacity" argument is the same idea LoRA exploits, but LoRA is stronger: it freezes the base entirely ( $θ_{1}, θ_{2}$ both fixed) and adds brand-new trainable parameters, so old weights are untouchable by construction rather than merely discouraged.

Level 4 — Synthesis

Recall Solution — L4.1

A defensible recipe:

LoRA on attention projections (parameter isolation) — base weights frozen ⇒ general skill cannot be overwritten. Mechanism: LoRA and Adapters.
Replay 5–10% synthetic general instructions (rehearsal with pseudo-data, since real pretraining data is unavailable) — puts general loss back into the gradient. Mechanism: rehearsal.
Low learning rate (LR) ( $\sim 2 \times 1 0^{- 5}$ ), 1–2 epochs — minimizes drift from $θ^{*}$ . Here LR is exactly the step size $η$ from the gradient-descent bullet: a smaller $η$ moves $θ$ less per step, so it strays less far from the old home. Mechanism: Learning Rate & Optimization. Each part attacks a different lever (frozen base, gradient content, step size), so they stack.

Recall Solution — L4.2

EWC wins when the old data is private/unavailable (medical, legal) — you keep only $θ^{*}$ and $F_{i}$ , no raw examples, so no privacy leak. Rehearsal wins when the old data is freely available and tasks are very different, because replaying real examples defends the actual old loss rather than a quadratic Laplace approximation of it (see Laplace Approximation) — the approximation can be poor far from $θ^{*}$ . This connects to Multi-task Learning, the joint-training ideal both methods approximate.

Recall Solution — L4.3

$L_{EWC} (θ) = L_{C} (θ) + \frac{λ _{A}}{2} \sum_{i} F_{i}^{A} (θ_{i} - θ_{i}^{* A})^{2} + \frac{λ _{B}}{2} \sum_{i} F_{i}^{B} (θ_{i} - θ_{i}^{* B})^{2} .$ Each past task contributes its own spring set anchored at its own home. This is the multi-anchor extension of EWC used in Continual Learning across a task sequence.

Level 5 — Mastery

Recall Solution — L5.1

Naive: $Δ_{naive} = 98 - 40 = 58$ points. EWC: $Δ_{EWC} = 98 - 90 = 8$ points. Forgetting prevented $= \frac{Δ _{naive} - Δ _{EWC}}{Δ _{naive}} = \frac{58 - 8}{58} = \frac{50}{58} \approx 86.2%$ . EWC eliminated about 86% of the forgetting the naive run suffered.

Recall Solution — L5.2

$J (θ) = \frac{G}{2} (θ - c)^{2} + \frac{λ F}{2} (θ - θ^{*})^{2}$ . The calculus, step by step. Same rule as L3.2: differentiating $\frac{G}{2} (θ - c)^{2}$ gives $\frac{G}{2} \cdot 2 (θ - c) = G (θ - c)$ (the $2$ cancels the $\frac{1}{2}$ , inside-derivative is $1$ ), and differentiating $\frac{λ F}{2} (θ - θ^{*})^{2}$ gives $λ F (θ - θ^{*})$ . Set the total slope to zero: $J^{'} (θ) = G (θ - c) + λ F (θ - θ^{*}) = 0$ . Expand: $Gθ - G c + λ F θ - λ F θ^{*} = 0 \Rightarrow (G + λ F) θ = G c + λ F θ^{*}$ . Divide by $(G + λ F)$ (positive, since $G > 0$ and $λ, F \geq 0$ , so never zero): $θ_{EWC} = \frac{G c + λ F θ ^{*}}{G + λ F}$ It is a stiffness-weighted average of the two homes $c$ and $θ^{*}$ .

$λ \to 0$ : numerator $\to G c$ , denominator $\to G$ , so $θ_{EWC} \to c$ — pure fine-tuning, lands in the new-task minimum (full forgetting).
$λ \to \infty$ : the $λ F$ terms dominate top and bottom, so $θ_{EWC} \to \frac{λ F θ ^{*}}{λ F} = θ^{*}$ — frozen at the old home (no new learning). Check with L3.2 numbers ( $G = 1, c = 10, λ = 1, F = 4, θ^{*} = 0$ ): $\frac{10 + 0}{1 + 4} = 2$ . ✓

Recall Solution — L5.3

$F \approx \frac{1}{4} (2^{2} + (- 2)^{2} + 1^{2} + (- 3)^{2}) = \frac{1}{4} (4 + 4 + 1 + 9) = \frac{18}{4} = 4.5$ . A larger $F$ means this parameter's log-likelihood is steep/sensitive ⇒ the old task cared about it ⇒ EWC gives it a stiff spring. Note $F \geq 0$ automatically — it is an average of squares. See Fisher Information.

Connections

Catastrophic forgetting — the parent note these exercises drill.
Fisher Information — the $F_{i}$ importance weights (L5.3).
Laplace Approximation — why the quadratic penalty is valid near $θ^{*}$ .
LoRA and Adapters — the freeze-the-base fix (L3.3, L4.1).
Continual Learning — multi-task-sequence setting (L4.3).
Multi-task Learning — the joint-training ideal (L4.2).
Learning Rate & Optimization — low-LR drift control (L4.1).