# What is the strategy for this piece of information

Heavy Math background, very light finance background:

Suppose I have a stock $S$ whose price is measured by the market once on times $t_0$ $t_1$ $t_2$.

Now the market has some opinion for how the stock behaves and it has priced the stock and options derived on it accordingly: (say S(t+1) - S(t)) is normally distributed with mean $0$ and standard deviation $\sigma$.

Now suppose a oracle (or insider?) approaches you and says that $a_0 \sigma > |S(t+1) - S(t) | + |S(t+2) - S(t+1)| > a_1 \sigma$.

For some constants $a_0, a_1, a_0 - a_1 < \sigma$ which is a much tighter bound than what the market can have any reasonable opinion about.

What sort of portfolio can you construct to profit off of of this? Using just going long and short, calls and puts, as well as long and short the underlying asset I can't seem to cook up any portfolio and am wondering if there is an algorithmic way to make a portfolio, or a systematic way to prove its not possible.

Say I posed the question as the oracle lets you know the $S(t+1) > S(t) + \sigma$. Then a portfolio would be to buy call options with strike price equal to the current price.

Say I posed the question as knowing that $|S(t+1) - S(t)| > \sigma$ then a portfolio, would be to go and buy a call and a put option with strike price equal to the current price. (Since either way the stock is high enough or low enough that one of the options can be executed to cover the cost of the initial and yield some profit, independent of which direction the stock moves).

In that sense, here I have another inequality, and I want to construct a corresponding portfolio for this inequality.

• What is the implied volatility relative to $\sigma$? – amdopt Aug 15 '18 at 12:36
• implied volatility should be sigma after reading online (i also forgot to look at log prices, so i guess adjust accordingly) – frogeyedpeas Aug 15 '18 at 15:04
• Yes, commonly IV is denoted as $\sigma$ however, you use it as standard deviation. Regardless of what it 'should be', did you mean to use $\sigma$ as IV or standard deviation in this equation: |S(t+1)−S(t)|> σ? – amdopt Aug 15 '18 at 15:12
• I meant to use it as standard deviation (over the course of one day). If I read correctly IV is the same as standard deviation but with a generally unknown time range attached to it (though commonly 1 year). EDIT: I see IV is a percentage so ignore my previous sentence and just assume standard deviation the whole way through – frogeyedpeas Aug 15 '18 at 15:16
• Consider the latter case, i didn’t rigorously state it but what I meant to convey is that the Oracle has given information which the other market participants don’t have. – frogeyedpeas Aug 15 '18 at 15:25

Say I posed the question as the oracle lets you know the S(t+1)>S(t)+σ. Then a portfolio would be to buy call options with strike price equal to the current price.

Well actually, no, I wouldn't do this. In this case the precise information possessed is that the stock price at expiry will be greater than the current price by $\sigma$, but you do not know by how much. If you buy a call option with a strike price equal to the current price you have unnecessarily introduced market risk into your profit, i.e. the higher the price goes then the more money you make, but your profit is undetermined.

However, if you go long the market (+1 delta) and sell a call option with a strike at $S(t)+\sigma$ then with certainty you will accrue a profit of the option premium plus $\sigma$. The value of the information is then a fixed deterministic amount. This is a covered call, but equivalently you could also just sell a put option with strike $S(t)+\sigma$, (put-call parity).

Say I posed the question as knowing that |S(t+1)−S(t)|>σ then a portfolio, would be to go and buy a call and a put option with strike price equal to the current price.

Well for the same reason the information gives certainty about a range so, instead of doing a straddle as your comment suggests, I would do a reverse iron butterfly. The value of that information can also be determined as a fixed amount, in that case, with the same profit for all outcomes conditional upon your insider information. # With regards to the question

You will observe that as at time, $t+1$, $|S(t+1)-S(t)|=k$ is a known parameter. Therefore as a trader with flexibility of timing of execution, one particular choice (I am not saying I can prove this is optimal) would be to wait until time $t+1$ and then you have the information about the price at time $S(t+2)$: i.e.

$$a_0 \sigma - k > |S(t+2)−S(t+1)| > a_1 \sigma - k$$

Now you are reduced to some specific cases:

• Case 1: $a_1 \sigma - k > 0$ then the reverse iron butterfly can be applied as above (since you have a lower bound of market movement). You also have an upper bound of market movements to further refine your trading window from which you profit by selling a strangle. (executed at time $t+1$ dependent on $k$).

• Case 2: $a_1 \sigma - k < 0$ and $a_0 \sigma - k > 0$ then you only have an upper bound, the lower bound is replaced by 0 (so the reverse iron butterfly no longer works) but the upper bound strategy in case 1 will still generate fixed profit by selling a strangle dependent upon $k$.

• Case 3: $a_0 \sigma - k < 0$. Well this actually contradicts your piece of information since by definition of modulus $|S(t+2)−S(t+1)| \geq 0$ hence you can conclude that $k = |S(t+1) - S(t)| \leq a_0 \sigma$, and therefore this opens a trade option at time $t$ with expiry $t+1$ again playing on the fact that you know the price range. So again you sell a strangle here with strikes dependent upon $a_0 \sigma$.

If I was assessing the value of the piece of information I would assess the value of the strangle in case 3 as a fixed amount, and plus the values of cases 1 and 2 integrated over all legitimate values of $k$, which represents the expectation of the informational value from time $t+1$ to $t+2$.

Have only pondered on this for a brief period so I welcome peer review and criticism of these ideas.

• Thank you, this gave a lot of food for thought and I didn't realize the subtely between strategies that are guaranteed profitable (but unknown value) vs. profitable strategies with an exact known value. I do have one criticism, the strategy you give requires active portfolio management, how would you make a portfolio at time $t$ that is guaranteed to acrue a profit, that doesn't require you to actively re-balance at later times? – frogeyedpeas Aug 20 '18 at 0:44

What sort of portfolio can you construct to profit off of this? Using just going long and short, calls and puts, as well as long and short the underlying asset I can't seem to cook up any portfolio

This is by no means a comprehensive list, but three strategies come to mind when looking for a long vega position.

2. Short butterfly/Short Iron Condor--both function the same way. You profit from a move beyond of predetermined limits that depend on the strikes that you choose.

3. Calendar/Reverse Calendar spreads--You could do a lot with these including be vega AND theta positive at the same time. You may also find these called Time spreads or Horizontal spreads.

In that sense, here I have another inequality, and I want to construct a corresponding portfolio for this inequality.

Here you seem to be looking for an option on one of the options strategies (or combination of them) I mentioned above. I'm not sure why this would be necessary when (according to the scenario you describe) you already have a stock that can be bought or shorted and vanilla options derived from that stock which you can do pretty much anything with. That being said I'm sure you could find a counterparty to trade with but, why reinvent the wheel with a more complex wheel that functions the same way and yields the same result?

I'm happy to elaborate on anything above but a basic Google search for those options strategy names will produce tons of results. Good luck!

• Yea so my observation is that none of the strategies listed above go to the level of abstraction I need for my portfolio: Both long straddle and short butterfly correspond the the case of $|S(t+1) - S(t)| > \sigma$ I.e. the stock is going to make a BIG move. The calendar spread gets close where its focusing on: either the stock will move big above its t_0 price and then down below its t_0 price (or the reverse). The inequality I listed in the question basically states: "The stock will make a large move and then stabilize" OR "The stock will be stable and then make a large move". – frogeyedpeas Aug 15 '18 at 21:11
• A sort of game theory way of describing it: the price can go either (up = up_large, down = down_large, stable). This means over 2 days, there are 9 possible behaviors. Of these i'm interested in 4 of them (up, stable), (down, stable), (stable, up), (stable, down) since this is what the Oracle has told us will happen (which the market doesn't know). The calendar lets you take a position on: (up, down) and reverse calendar is (down, up). The straddle and butterflies correspond to a single time step, and profit off of the following set { (up), (down) }. What i'm looking for is a bit subtler – frogeyedpeas Aug 15 '18 at 21:17
• I'm not convinced but suspicious that much like its impossible to use just stocks to replicate the behavior of a long straddle, that it might be impossible to use just options and stocks to build the portfolio I want, and I need to use compound options. I'm not sure though, and I don't have any rigorous reason to believe this except for "I thought about and its hard", but that could just as well be my incompetence in finance – frogeyedpeas Aug 15 '18 at 21:19
• Does the Oracle give a time frame? Or at least a time frame that we can be 95% confident of? If so, laddering or tranching one (or a combination) of the strategies I mentioned above would work. I have only listed the broad strategies, not how one could profit if one were creative... – amdopt Aug 16 '18 at 18:41