Multiplying all trades by the same constant just scales down the final "pile of shares" by that constant, so if it earned $\epsilon \times magnitude$ value before with its pile-o-shares, it'll earn $\epsilon \times magnitude$ value after scaling with its scaled pile-o-shares. If it takes $poly\left(k\right)$ time to write down the algorithm for ${\overline{T}}^k$, it'll take $poly\left(k\right)$ time to write down the algorithm for the screened version since it only takes $poly\left(k\right)$ time to generate ${\xi }_k$, and not much more to generate the scaling factor. If a trader doesn't trade on certain days, multiplying its trades by a constant will still result in not trading on those days. If it took $poly\left(n\right)$ time or less for ${\overline{T}}^k$ to write down a trade before, it'll still take (a larger) $poly\left(n\right)$ time or less to write down the trade afterwards, the trader is just writing down the cached scaling factor/expressible feature $max(1,{\xi }_k{)}^{-1}$ at the end of every expressible feature.

Thus, scaling the "unsafe" trader down doesn't affect its ability to be efficiently emulated, and it doesn't affect the $ϵ$-ROI property.

The basics of "guaranteed net value" is that both of the two following conditions are fulfilled. "almost all trades have been made already" and "the total trades that have been made so far have some guaranteed value according to all the $PC$ worlds, enough to cover the losses from the remaining trades".

"Almost all trades have been made already" is equivalent to $\sum _{i\le m}||T_i^k\left(\overline{P}\right)|{|}_{mg}\ge (1-\frac{\epsilon }{3}){\alpha }_k$.

Pretty much, only an $\frac{\epsilon }{3}$ fraction of the total lifetime amount of trading is left.

"The total trades that have been made so far have some guaranteed value according to all the $PC$ worlds" is equivalent to $\forall W\in PC\left(D_n\right),W\left(\sum _{i\le m}{T}_i^k\left(\overline{P}\right)\right)\ge \left(\frac{2\epsilon }{3}\right){\alpha }_k$

Pretty much, a $\frac{2\epsilon }{3}$ fraction of the total lifetime amount of trading has been earned in the form of guaranteed value. Note that this is only guaranteed assuming that the trader has $\epsilon$-ROI.

Notice that even if everything goes to hell from this point on, only $\frac{\epsilon }{3}{\alpha }_k$ value can possibly be lost, since there are so few trades left, and the magnitude of remaining trades provides a hard bound on how much could be lost. Also, since there's $\frac{2\epsilon }{3}{\alpha }_k$ guaranteed value that has been earned, once the employee has hit this point, it can't possibly lose, it'll have $\frac{\epsilon }{3}{\alpha }_k$ net worth guaranteed, in the limit.

By the earlier discussion about how the "trading firm" can just directly emulate the market (at a massive time cost), both of these are indeed computable, if infeasible. There will be some $n\gg m$ where it's verifiable in $n$ steps that yes, on day m, the k'th employee has "guaranteed net value", or, using the terminology from the paper, "its holdings have matured"

Ok, we'll want a "budget on day $k$" variable for how much money would be left over if all the employees with uncertain net value lost everything. And this will be used as a scaling factor for the $k$'th employee. If the budget on day 1,000 is 50 cents, then the trading firm will have a scaling factor of 0.5 on all future trades of the 1,000th employee. After all, even if their magnitude is 1, and all their trades are disastrous, only 50 cents at most could be lost.

So, $T_n:=\sum _{k\le n}{\beta }_k^{†}{T}_n^k$

The day-$n$ trade of the trading firm as a whole is the sum of the trades of all the employees so far on day n, with each employee individually scaled down by their personal "budget" feature.

And, to define the "budget" feature... it starts at 1 and subtracts the sum of the total money risked by all employees with uncertain net value.

$open(i,k):=0$ if it's verifiable in $k$ computation steps that ${\overline{T}}^i$ has guaranteed net value, otherwise it's 1. We'll use this to only select the employees with uncertain net value.

${\beta }_k^{†}:=1-\sum _{i<k}open(i,k){\beta }_i^{†}{\alpha }_i^{†}$

Remember, $†$ means it's an expressible feature. ${\beta }_i^{†}$ is the expressible feature that returns the budget for day $i$, ${\alpha }_i^{†}$ is the expressible feature that returns the magnitude of the $i$'th employee. So, since $open(i,k)=0$ for every employee that has uncertain net value on day k, yes, this takes 1 dollar and subtracts the total money being risked by copying all previous employees. (The $i$'th employee would risk ${\alpha }_i$ dollars, and their trades get scaled down by a factor of ${\beta }_i$ too.)

What's our "trading firm" going to have to compute on day $n$? Well, it's going to have to compute all the ${\overline{T}}^k$, and compute all the ${\beta }_k^{†}$, and run all the ${\overline{T}}^k$ to return an actual trade when they're given $n$.

Well, we already assumed that $({\overline{T}}^k{)}_k$ was efficiently emulatable, so that does imply that it takes poly(k) time to generate ${\overline{T}}^k$, and they all have the same poly(n) runtime bound.

That takes care of part of it. What is our "trading firm" going to have to do to compute all the ${\beta }_k^{†}$? Well, it'll have to compute all the ${\alpha }_i^{†}$, and all the $open(i,k)$.

Ok. To compute a particular ${\alpha }_i^{†}$, it takes poly(i) time (because $\overline{\alpha }$ is a $\overline{P}$-generable sequence). We're assembling a list of k of them, so it'll take $O(k\ast poly(k\left)\right)$ time to generate that list. To compute a particular $open(i,k)$ value, it takes k steps, and we're doing that for k-1 different values of i, so it'll take $O\left(k^2\right)$ time to generate that list of values. And previously computed values of ${\beta }_i^{†}$ can be cached and reused.

Thus, since it takes polynomial time to generate all the $\alpha$'s and all the $open$'s and generate and cache the $\beta$'s as you go and add all the stuff up, ${\beta }_k^{†}$ is computable in poly(k) time.

Since it takes poly(k) time to generate ${\beta }_k^{†}$ and poly(k) time to generate ${\overline{T}}^k$ and we're doing this $n$ times over and adding them up and $k\le n$, it takes poly(n) time to write down the entire collection of scaled traders. And since there are n of them and they all run in poly(n) time (and we're also summing them up), it takes poly(n) time for the whole computation to complete. Whew!

Instead of constantly writing out ${\alpha }_k^{†}\left(\overline{P}\right)$ and ${\beta }_k^{†}\left(\overline{P}\right)$, they will be abbreviated as ${\alpha }_k$ and ${\beta }_k$. Please remember, these are numbers, while ${\alpha }_k^{†}$ and ${\beta }_k^{†}$ are functions.

By how we defined ${\beta }_k$ $${\beta }_k:=1-\sum _{i<k}open(i,k){\beta }_i{\alpha }_i$$ Since ${\alpha }_k\le 1$ because they were all scaled to be like that, $${\beta }_k{\alpha }_k\le 1-\sum _{i<k}open(i,k){\beta }_i{\alpha }_i$$ By relabling variables, $${\beta }_n{\alpha }_n+\sum _{k<n}open(k,n){\beta }_k{\alpha }_k\le 1$$ Because ${\beta }_n{\alpha }_n\ge open(n,n){\beta }_n{\alpha }_n$, $$\sum _{k\le n}open(k,n){\beta }_k{\alpha }_k\le 1$$ By the definition of ${\beta }_k$ and the above line and induction, ${\beta }_k\ge 0$. The value of $\overline{T}$'s holdings at time n is $$W\left(\sum _{i\le n}{T}_i\right(\overline{P}\left)\right)=\sum _{i\le n}W\left(T_i\right(\overline{P}\left)\right)$$ And by the definition of the trader $T_n$, $$=\sum _{k\le n}W\left(\sum _{i\le n}{\beta }_k{T}_i^k\right(\overline{P}\left)\right)$$ and this can be split into two pieces, the value of the traders with uncertain value, and the value of the traders with guaranteed value. $$=\sum _{uncertain,k\le n}W\left(\sum _{i\le n}{\beta }_k{T}_i^k\right(\overline{P}\left)\right)+\sum _{guaranteed,k\le n}W\left(\sum _{i\le n}{\beta }_k{T}_i^k\right(\overline{P}\left)\right)$$ Looking at the first term with the traders of uncertain value, the magnitude serves as a lower bound on the value of the trader according to all worlds. Also, ${\beta }_k$ is the same for every $i$ in the inside sum, so it can be factored out. $$\sum _{uncertain,k\le n}W\left(\sum _{i\le n}{\beta }_k{T}_i^k\right(\overline{P}\left)\right)\ge -\sum _{uncertain,k\le n}{\beta }_k\sum _{i\le n}||T_i^k\left(\overline{P}\right)|{|}_{mg}$$ By adding in the rest of the sequence, it increases the magnitude so far, and thus the potential losses. Also, the sum of the magnitudes of all trades made by the k'th employee is just ${\alpha }_k$. Also, since we're summing up over the employees of uncertain value, $open(i,k)$ always equals 1, and equals 0 on every trader outside of that set. $$\ge -\sum _{uncertain,k\le n}{\beta }_k\sum _{i\in N^{+}}||T_i^k\left(\overline{P}\right)|{|}_{mg}=-\sum _{k\le n}open(k,n){\beta }_k{\alpha }_k$$ This is $\ge -1$ by multiplying the 4th equation in this section by -1. Thus, the magnitude of the "uncertain" traders up to day n is -1 at worst.

Now to analyze the value of the second term, assessing the wealth of the employees of guaranteed value. By the earlier discussion on the precise conditions for declaring a employee to have "guaranteed value", $$\sum _{guaranteed,k\le n}W\left(\sum _{i\le n}{\beta }_k{T}_i^k\right(\overline{P}\left)\right)\ge \sum _{guaranteed,k\le n}\frac{\epsilon }{3}{\beta }_k{\alpha }_k$$ Combining these two lower bounds on the employees of guaranteed and uncertain value to assess the value of the trading firm on day n, $$\sum _{k\le n}W\left(\sum _{i\le n}{\beta }_k{T}_i^k\right(\overline{P}\left)\right)\ge -1+\sum _{guaranteed,k\le n}\frac{\epsilon }{3}{\beta }_k{\alpha }_k$$ Since we have previously established that ${\beta }_k\ge 0$ and ${\alpha }_k\ge 0$, the value of the trader as a whole is bounded below by -1, so it has bounded losses. Now, unbounded value must be demonstrated. This can be done by demonstrating that the second term diverges as $n\to \infty$.

For any employee $k$, there is an (extremely large) day $n$ at which it will be verified to have guaranteed value. Therefore, $$\underset{n\to \infty }{lim}\sum _{guaranteed,k\le n}\frac{\epsilon }{3}{\beta }_k{\alpha }_k=\sum _k\frac{\epsilon }{3}{\beta }_k{\alpha }_k$$ Since $\frac{\epsilon }{3}$ is a constant, showing that $\sum _k{\beta }_k{\alpha }_k=\infty$ is sufficient to show divergence. Assume that $\underset{k\to \infty }{lim}{\alpha }_k\ne 0$. Then there is some $\delta$ such that ${\alpha }_k>\delta$ infinitely often. Also assume that $$\sum _k{\beta }_k{\alpha }_k$$ is finite, for a proof by contradiction. Then there is some day $n$ such that $$\sum _{i>n}{\beta }_i{\alpha }_i\le \frac{1}{2}$$ Let $N$ be a day by which all employees with $k\le n$ have been verified to have guaranteed value. Then $$\sum _{i<N}open(i,N){\beta }_i{\alpha }_i=\sum _{i\le n}0\ast {\beta }_i{\alpha }_i+\sum _{n<i<N}open(i,N){\beta }_i{\alpha }_i$$ And by the fact that $n<N$ and $\sum _{i>n}{\beta }_i{\alpha }_i\le \frac{1}{2}$ $$\le 0+\sum _{n<i<N}{\beta }_i{\alpha }_i\le \frac{1}{2}$$ Now for a little bit of symbol manipulation. $${\alpha }_k{\beta }_k={\alpha }_k(1-\sum _{i<k}open(i,k\left){\beta }_i{\alpha }_i\right)$$ $\sum _{i>N}open(i,N){\beta }_i{\alpha }_i\le \frac{1}{2}$ as previously demonstrated, and the same applies to any $k>N$. Since there are infinitely many days where $k>N$ and ${\alpha }_k>\delta$, all those infinitely many occasions will have $${\alpha }_k(1-\sum _{i<k}open(i,k\left){\beta }_i{\alpha }_i\right)\ge {\alpha }_k(1-\frac{1}{2})\ge \frac{\delta }{2}$$ And this infinite recurrence of guaranteed value shows that $$\sum _k{\alpha }_k{\beta }_k=\infty$$, disproving the assumption it is finite.

Since this sum diverges, by the discussion above, $\underset{n\to \infty }{lim}\sum _{guaranteed,k\le n}\frac{\epsilon }{3}{\beta }_k{\alpha }_k$ diverges so the "trading firm" has unbounded earnings.

To summarize, this trading firm has bounded losses, and assuming that $\underset{k\to \infty }{lim}{\alpha }_k\ne 0$, this trading firm has unbounded earnings. This is impossible for a logical inductor. Therefore, one of the following conditions must fail.

• $\epsilon$-ROI on all "employees"
• The sequence of "employees" is efficiently emulatable so they can be combined into a "trading firm"
• There is a $\overline{P}$-generable sequence $\overline{\alpha }$ that describes the magnitude of the sequence of "employees"
• $\underset{k\to \infty }{lim}{\alpha }_k\ne 0$

Proving all four forces a contradiction, and future proofs will proceed by proving these four properties.