Today we continue our Mathematics of Infinity series by Victor Argonov with a new part that is driving us closer to the stock exchange topic. This article is devoted to the behaviour of quotes.
The paradoxes that work with pub crawlers random wandering or with casino games are also present in the exchange markets. Their effect often baffles newcomers, and sometimes even devastates financial health of experienced traders.
Most traders are aware of the rule of the thumb in stock exchange tendencies: if you buy a stock and its price suddenly decreases, do not sell it at once. It is likely to go up again in some time, reaching the same level at some moment, or even outgrowing it. The only question is when it will happen. A trader may wait for days, months or even years, and the price will still remain lower than the purchase level. The company is doing rather well, there are no crises around, but the quote still remains below the initial level.
Win-win game with zero profit
Let's assume, there are a lot of stocks listed on the exchange, and they all have different prices. Each day the price of each stock can either become one dollar lower, or one dollar higher. The trader check the prices out only once per day. On the first day he buys 1,000 stocks of one (random) kind. Then he is waiting until the price goes one dollar up, and sells the whole stack getting the profit of $1,000, and repeats the process again.
Let's assume, the companies never go bankrupt, even when the stock price goes below zero. It means, the investor never loses in this game: each deal brings him only profit.
There are two questions:
- What is the average time for the quote to increase by one dollar?
- What is the average income?
This is just the new wording for the same task we solved in the last two parts of the series, so we know what to expect.
So, the probability to sell the stack on the next day is 0.5, on the third day — ⅛, etc. In each 10th case waiting time will be longer than 100 days, 100th case will last 10,000 days; if the time is unlimited, the average waiting time will be eternal, while the average income will be zero.
Although this case is idealized, even such a simplified case demonstrates why such tactics is not perfect for making a fortune on the stock exchange. In trading terms, such orders are called Take profit: sell a stock when its price surpasses the purchase price by a certain value. This tactics is almost lossless, but is hardly useful in its pure state. Maybe there are some ways to make the game more profitable?
No infinity? The zero is still there
The problem remains the same. We are, however, adding one condition: if the stocks have not gone up within X days, the investor should sell them, even if it inclines loss.
- What is the average income?
- What is the most profitable X?
If the time is limited, we finally get rid of the eternal games, but at the same time we impose loss on the investor. Say, the investor decides to set X=100 days. As we know from the pub crawlers problem, the average loss will be aroung $10,000. The probability of this outcome is 1/10. In the other 9 cases the investor will get $1,000 profit. Such strategy is not profitable enough, and it will become even less profitable for bigger Xs. Maybe the solution is to set smaller X?
X=1 won't work, while the probability to be left without money is 1/2. If X=3, 5/8 of all deals will be successful, while 3/8 will be in the red zone. The profit in all profit-making deals will be $1,000, 2/3 of the lossmaking deals will generate loss of $1,000, 1/3 - of $3,000. The average profit is $(5*1000-2*1000-1*3000)/8=$0. With any X.