Spent the first couple minutes just making sure I understood the position accumulation mechanic correctly, which I think was the right call because it's easy to misread.
Start by formalizing the random walk and the trading strategy, then express the cumulative profit as a sum of products of past and current steps. Compute the variance by analyzing the covariance structure of these products, and finally discuss the expected profit and continuous-time extension.
Pro tip: Emphasize that the profit is a quadratic function of the random walk, and use the fact that the steps are independent to simplify the variance calculation. This shows you can handle complex stochastic processes efficiently.
Let S_t be the price at time t, with S_0=0 and S_t = sum_{i=1}^t X_i, where X_i are i.i.d. with P(X_i=1)=P(X_i=-1)=1/2. At each time t, you buy 1 share if X_t=1 and short 1 share if X_t=-1, so your position after time t is X_t.
The profit from the position taken at time t is X_t * (S_T - S_t) for t < T, and the position at time T yields no profit yet. So total profit at time T is P_T = sum_{t=1}^{T-1} X_t (S_T - S_t) = sum_{t=1}^{T-1} X_t sum_{i=t+1}^T X_i.
Swap sums to get P_T = sum_{i=2}^T X_i sum_{t=1}^{i-1} X_t = sum_{i=2}^T X_i S_{i-1}. Alternatively, note that P_T = (S_T^2 - T)/2, since S_T^2 = sum X_i^2 + 2 sum_{i<j} X_i X_j = T + 2 sum_{i<j} X_i X_j, and P_T = sum_{i<j} X_i X_j.
Using P_T = (S_T^2 - T)/2, and S_T^2 ~ chi-squared with 1 degree of freedom scaled? Actually S_T is sum of T i.i.d. ±1, so S_T^2 has mean T and variance 2T? Wait, Var(S_T^2) = E[S_T^4] - (E[S_T^2])^2. For simple random walk, E[S_T^4] = 3T^2 - 2T? Let's derive: S_T = 2B - T where B~Bin(T,1/2). Then S_T^2 = 4B^2 -4BT + T^2. E[B]=T/2, E[B^2]=T/4 + T^2/4, E[B^3]= T/8 + 3T^2/8 + T^3/8? Actually use known: E[S_T^4] = 3T^2 - 2T for T>=1? Check T=1: E[S_1^4]=1, 3-2=1 ok. T=2: S_2^4: values 0,4,0,4? Actually S_2 can be -2,0,0,2 with probs 1/4,1/4,1/4,1/4. E[S_2^4]= (16+0+0+16)/4=8. 3*4-2*2=12-4=8 ok. So Var(S_T^2)= (3T^2-2T) - T^2 = 2T^2 - 2T = 2T(T-1). Then Var(P_T) = Var((S_T^2 - T)/2) = (1/4) Var(S_T^2) = (1/4)*2T(T-1) = T(T-1)/2.
Expected profit: E[P_T] = E[(S_T^2 - T)/2] = (T - T)/2 = 0. This makes sense because each trade has zero expected profit due to martingale property. Continuous-time: replace random walk with Brownian motion B_t, strategy: buy if dB_t>0? Actually in continuous time, the strategy is to hold sign of dB_t? But dB_t is infinitesimal, so more appropriate: hold position equal to sign of previous increment? The discrete analogue leads to P_T = (B_T^2 - T)/2, and variance T^2/2? Actually for Brownian motion, if we define profit as integral of sign of past increments? The continuous limit of the discrete strategy yields P_T = (B_T^2 - T)/2, which has variance T^2/2? Let's check: Var(B_T^2) = 2T^2, so Var((B_T^2 - T)/2) = T^2/2. Expected profit 0.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.