Exercise 2: Why an Early Layer's Own Contribution Requires Every Layer Between It and the Output — Possible Solution ==================================================================== THE CHAIN RULE, PER THIS CHAPTER'S OWN EXPLANATION ------------------------------ Per this chapter, "if A affects B, and B affects C, then A's own effect on C is A's effect on B, multiplied by B's effect on C." This is a statement about MULTIPLICATION, not addition or direct measurement — the combined effect of A on C is only knowable by combining BOTH individual local effects (A→B and B→C) together, not by examining A and C in isolation from B. WHY AN EARLY LAYER PLAYS THE ROLE OF "A" IN THIS CHAIN ------------------------------ Per this chapter, "a multi-layer network is a chain of functions — one layer's output feeding the next." An early layer's own weights affect that layer's own output; that output then becomes an input to the next layer, whose own output feeds the layer after that, and so on, all the way to the final loss. The early layer never touches the final loss DIRECTLY at all — its only path of influence runs through every intermediate layer's own transformation of its output, one step at a time, ending at the output layer's own loss computation. WHY SKIPPING ANY INTERMEDIATE LAYER WOULD BREAK THE CALCULATION ------------------------------ Per the chain rule's own multiplicative structure, computing "A's effect on C" when there's an intervening B requires knowing BOTH "A's effect on B" AND "B's effect on C" — omitting either piece leaves an incomplete, incorrect calculation, since the two local effects must be multiplied together to get the true combined effect. Applied to a deep network, an early layer's own true contribution to the final loss can only be computed by multiplying together the local effect at EVERY single layer along the path from that early layer to the output — per this chapter's own backpropagation description, "each layer's own gradient calculation reuses the gradient already computed one layer downstream," precisely because that downstream gradient already encodes the combined effect of every layer between it and the final loss. WHY THIS MEANS THE INFORMATION GENUINELY CAN'T BE COMPUTED LOCALLY ------------------------------ An early layer, considered entirely on its own, has no way of knowing how sensitive the FINAL loss is to its own output — that sensitivity depends entirely on what every subsequent layer does with that output. This is precisely why backpropagation works backward from the output, computing each layer's own gradient only after the layer downstream of it has already been computed — the early layer's own gradient literally cannot be correctly computed until the chain of downstream effects reaching all the way to the loss has been assembled first. WHY THIS WORKS AS AN ANSWER ------------------------------ It applies the chapter's own A-affects-B-affects-C chain rule directly to an early network layer's own relationship to the final loss, explains why the chain rule's own multiplicative structure requires every intervening layer's own local effect to be included, and explains why this is precisely why backpropagation must proceed backward, layer by layer, rather than computing an early layer's own gradient directly and independently.