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Let $\tilde{W}_t := (1+R)^{-t}W_t$ and $\tilde{S}_t := (1+R)^{-t}S_t$ be respectively discounted wealth process and discounted asset price. Then, show that $$\tilde{W}_t = w_0 + \sum_{i=1}^{t}\Delta_i(\tilde{S}_{i+1} - \tilde{S}_i)$$ For simplicity, we fix the notation $(\delta\cdot S)_t$ for $\sum_{i=1}^{t}(S_{i+1} - S_i)$.

Background information: Consider a discrete-time market with time horizon $T$ in which trading occurs at only at time $t = 0,\ldots,T$. We consider a risky asset for which the price at time $t$ is given by a random variable $S_t$ and a risk-free asset (bond) with interest rate $R$.

A self-financing portfolio strategy consists of an initial wealth $w_0$ and a vector $\Delta = (\Delta_0,\ldots,\Delta_{T-1})$ such that $\Delta_t$ shows the units of risky asset held in the portfolio at time $t$. At time $t = 0$, the initial wealth $W_0 := w_0$ is split in $\Delta_0 S_0$ dollars in risky asset and $W_0 -\Delta_0 S_0$ is risk-free asset. Then, at time $t = 1$, the dollar value of risky investment will change to $\Delta_0 S_1$ when the market price of risky asset changes to $S_1$, and the dollar value of risk-free investment will be affected by the interest rate and changes to $(1 + R)(W_0 - \Delta_0 S_0)$ . At this, point we want to follow the portfolio strategy and adjust the risky portion of the portfolio to $\Delta_1 S_1$ . To do so, we change the position of the portfolio by transferring dollar between risky and risk-free portions, i.e. to make the risky portion of portfolio $\Delta_1 S_1$, we have to withdraw/deposit $\Delta_0 S_1 - \Delta_1 S_1$ dollars from/into the risk-fee portion to buy/sell the required amount of risky assets. To generalize this process, let $Y_t$ be the risk-free portion of the portfolio at time $t$. The risky portion is always rebalanced to be $\Delta_t$ number of risky assets, equivalently dollar value of $\Delta_t S_t$, at time $t$; i.e. $W_t = \Delta_t S_t + Y_t$.

It is easy to see that $$Y_{t+1} = (1+R)Y_t + \Delta_t S_{t+1} - \Delta_{t+1}S_{t+1}$$ where the term $\Delta_t S_{t+1} - \Delta_{t+1}S_{t+1}$ is the amount required to rebalance the portfolio for the strategt $\Delta$. Then, the total wealth $W_t$ satifies

$$W_t = w_0 + R\sum_{i=0}^{t-1}Y_i + \sum_{i=0}^{t-1}\Delta_i(S_{i+1} - S_i)$$ If desired, I can provide a paragraph that led to this recursive formula.

Attempted solution: $$\tilde{W}_t = \frac{1}{(1+R)^t}\left[w_0 + R\sum_{i=0}^{t-1}Y_i + \sum_{i=0}^{t-1}\Delta_i(S_{i+1} - S_i)\right]$$ $$=\frac{w_0}{(1+R)^t} + \frac{R\sum_{i=0}^{t-1}Y_i}{(1+R)^t} + \frac{\sum_{i=0}^{t-1}\Delta_i(S_{i+1} - S_i)}{(1+R)^t}$$ We know that $\tilde{S}_t := (1+R)^{-t}S_t$ so then $S_t = \tilde{S}_t(1+R)^t$. Changing the last sum from $i = 0$ to $i = 1$ and substituting in for $S_t$ we get $$\frac{w_0}{(1+R)^t} + \frac{R\sum_{i=0}^{t-1}Y_i}{(1+R)^t} + \sum_{i=1}^{t}\Delta_i(\tilde{S}_{i+1} - \tilde{S}_i)$$ I feel like I may be doing something wrong here and/or I am not sure where to proceed from here any suggestions is greatly appreciated.

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Let $B_t$ be the value of the risk-free asset at time $t$. Then $B_0=1$ and $B_{t+1} = (1+R) B_t$. Moreover, let $\beta_t$ be units invested in the risk-free asset at time $t$. It is clear that $\beta_0 = w_0 - \Delta_0 S_0$. Since the strategy is self-financing, \begin{align*} \Delta_{t-1} S_{t-1} + \beta_{t-1} B_{t-1} = \Delta_t S_{t-1} + \beta_t B_{t-1},\tag{1} \end{align*} that is, the value will not change after the re-balance at time $t-1$, which will be held until time $t$.

From (1), \begin{align*} \beta_t &= \beta_{t-1} -(\Delta_t - \Delta_{t-1}) \frac{S_{t-1}}{B_{t-1}}\\ &=\beta_{t-1} -(\Delta_t - \Delta_{t-1}) \widetilde{S}_{t-1}. \end{align*} Then, for the normalized wealth process $\widetilde{W}_t=W_t/B_t$, \begin{align*} \widetilde{W}_t &= \Delta_t \widetilde{S}_t + \beta_t\\ &= \widetilde{W}_{t-1} + \Delta_t \widetilde{S}_t - \Delta_{t-1} \widetilde{S}_{t-1}+ \beta_t-\beta_{t-1}\\ &=\widetilde{W}_{t-1} + \Delta_t (\widetilde{S}_t -\widetilde{S}_{t-1})\\ & \qquad \cdots\cdots \\ &=\widetilde{W}_0 + \sum_{i=1}^t \Delta_i (\widetilde{S}_i -\widetilde{S}_{i-1})\\ &= w_0 + \sum_{i=1}^t \Delta_i (\widetilde{S}_i -\widetilde{S}_{i-1}). \end{align*}

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