> After the third chapter summary I tried to get it to do one of the exercises. > "Could you please complete this exercise from the end of Chapter 2 so I can verify my own solution? > Exercise 2.37. Prove the identities (2.17), (2.18), (2.26), (2.27) and (2.28) in Section 2.2 > It ended up taking two attempting, but seems to have got it on the second try. In a follow up I'm going to verify this myself, and with another LLM. # Attempt 1 Certainly! Let's go through each of these identities one by one. 1. Identity (2.17): Ep[rX] - Eq[rX] = (p - q)(X(1, ↑) - X(1, ↓)) / X(0) Proof: Ep[rX] = p·rX(↑) + (1-p)·rX(↓) = p·(X(1, ↑) - X(0)) / X(0) + (1-p)·(X(1, ↓) - X(0)) / X(0) = (p·X(1, ↑) + (1-p)·X(1, ↓) - X(0)) / X(0) Similarly, Eq[rX] = (q·X(1, ↑) + (1-q)·X(1, ↓) - X(0)) / X(0) Therefore, Ep[rX] - Eq[rX] = (p·X(1, ↑) + (1-p)·X(1, ↓) - q·X(1, ↑) - (1-q)·X(1, ↓)) / X(0) = ((p-q)·X(1, ↑) - (p-q)·X(1, ↓)) / X(0) = (p - q)(X(1, ↑) - X(1, ↓)) / X(0) 2. Identity (2.18): Eπ[rX] = r ≡ R - 1 Proof: Eπ[rX] = π·rX(↑) + (1-π)·rX(↓) = π·(X(1, ↑) - X(0)) / X(0) + (1-π)·(X(1, ↓) - X(0)) / X(0) = (π·X(1, ↑) + (1-π)·X(1, ↓) - X(0)) / X(0) = (R·X(0) - X(0)) / X(0) (by the definition of risk-neutral probability) = (R - 1) = r 3. Identity (2. # Attempt 2 Certainly! Let's prove each of these identities step by step. (2.17) Ep[rX] - Eq[rX] = (p - q)(X(1,↑) - X(1,↓)) / X(0) Proof: Ep[rX] - Eq[rX] = p(X(1,↑)-X(0))/X(0) + (1-p)(X(1,↓)-X(0))/X(0) - [q(X(1,↑)-X(0))/X(0) + (1-q)(X(1,↓)-X(0))/X(0)] = [p(X(1,↑)-X(0)) + (1-p)(X(1,↓)-X(0)) - q(X(1,↑)-X(0)) - (1-q)(X(1,↓)-X(0))] / X(0) = [p(X(1,↑)-X(1,↓)) - q(X(1,↑)-X(1,↓))] / X(0) = (p - q)(X(1,↑) - X(1,↓)) / X(0) (2.18) Eπ[rX] = r ≡ R - 1 Proof: Eπ[rX] = π(X(1,↑)-X(0))/X(0) + (1-π)(X(1,↓)-X(0))/X(0) = [π(X(1,↑)-X(0)) + (1-π)(X(1,↓)-X(0))] / X(0) = [πX(1,↑) + (1-π)X(1,↓) - X(0)] / X(0) = X(0)R/X(0) - X(0)/X(0) (from the definition of π) = R - 1 = r (2.26) Ep[rX] - r = |p - π|σX / √(p(1-p)) Proof: From (2.19), Ep[rX] - r = (p - π)(X(1,↑) - X(1,↓)) / X(0) From (2.25), σX^2 = p(1-p)[(X(1,↑) - X(1,↓)) / X(0)]^2 So, σX = √(p(1-p)) |(X(1,↑) - X(1,↓)) / X(0)| Therefore, Ep[rX] - r = (p - π)(X(1,↑) - X(1,↓)) / X(0) = |p - π| |(X(1,↑) - X(1,↓)) / X(0)| (as Ep[rX] ≥ r) = |p - π| σX / √(p(1-p)) (2.27) X(0) = Ep[X(1)] / (R + Λ(p)σX) Proof: From (2.26), Λ(p) = |p - π| / √(p(1-p)) So, r + Λ(p)σX = Ep[rX] Multiplying both sides by X(0), we get: (r + Λ(p)σX)X(0) = Ep[rXX(0)] = Ep[X(1) - X(0)] = Ep[X(1)] - X(0) Rearranging, X(0)(1 + r + Λ(p)σX) = Ep[X(1)] Therefore, X(0) = Ep[X(1)] / (1 + r + Λ(p)σX) = Ep[X(1)] / (R + Λ(p)σX) (2.28) Ep[rX] - r = βX,Y[Ep[rY] - r] Proof: From (2.19), Ep[rX] - r = (p - π)(X(1,↑) - X(1,↓)) / X(0) and Ep[rY] - r = (p - π)(Y(1,↑) - Y(1,↓)) / Y(0) So, (Ep[rX] - r) / (Ep[rY] - r) = ((X(1,↑) - X(1,↓)) / X(0)) / ((Y(1,↑) - Y(1,↓)) / Y(0)) = (X(1,↑) - X(1,↓))Y(0) / (Y(1,↑) - Y(1,↓))X(0) From (2.24) and (2.25), βX,Y = Covp(rX,rY) / Varp(rY) = (X(1,↑) - X(1,↓))Y(0) / (Y(1,↑) - Y(1,↓))X(0) Therefore, Ep[rX] - r = βX,Y[Ep[rY] - r] Thus, we have proved all the identities mentioned in the exercise. > Here I asked for some clarification on one of the proofs "Can you explain the proof for identity (2.27) in more detail?" Certainly! Let's break down the proof for identity (2.27) step by step. Given: - X(0) is the price of asset X at time 0 - X(1) is the price of asset X at time 1 - R = 1 + r, where r is the risk-free rate - Ep[X(1)] is the expected price of asset X at time 1 under probability measure p - σX is the volatility of asset X - Λ(p) = |p - π| / √(p(1-p)), where π is the risk-neutral probability Step 1: Start with the result from identity (2.26) Ep[rX] - r = |p - π|σX / √(p(1-p)) Step 2: Substitute the definition of Λ(p) Ep[rX] - r = Λ(p)σX Step 3: Add r to both sides Ep[rX] = r + Λ(p)σX Step 4: Multiply both sides by X(0) X(0)Ep[rX] = X(0)(r + Λ(p)σX) Step 5: Expand the left-hand side using the definition of rX X(0)Ep[rX] = Ep[X(0)rX] = Ep[(X(1) - X(0))] = Ep[X(1)] - X(0) Step 6: Substitute the result from Step 5 into the equation from Step 4 Ep[X(1)] - X(0) = X(0)(r + Λ(p)σX) Step 7: Add X(0) to both sides Ep[X(1)] = X(0)(1 + r + Λ(p)σX) Step 8: Divide both sides by (1 + r + Λ(p)σX) X(0) = Ep[X(1)] / (1 + r + Λ(p)σX) Step 9: Substitute R = 1 + r X(0) = Ep[X(1)] / (R + Λ(p)σX) And there you have it! We've derived the identity (2.27): X(0) = Ep[X(1)] / (R + Λ(p)σX) The key steps are: 1. Starting with the result from (2.26) which relates the excess return to volatility and the risk premium Λ(p) 2. Multiplying both sides by X(0) to introduce the asset price 3. Recognizing that X(0)Ep[rX] is equivalent to Ep[X(1)] - X(0) 4. Rearranging the terms to isolate X(0) This identity shows that the current price of an asset X(0) is equal to the expected future price Ep[X(1)] discounted at a rate that includes both the risk-free rate r and a risk premium Λ(p)σX. The risk premium adjusts the discount rate based on the asset's volatility σX and the market price of risk Λ(p).