If the distribution of returns in symmetric, why not use a coin toss, small risk & high reward?

Question

If the distribution of returns is symmetric then why not

1. use a coin toss to decide whether to buy or sell
2. Calculate the average velocity of the market (ATR - in technical analysis)
3. Place a stop loss on 0.5 ATR away from current price and take a profit 2 ATR away from the current price?

I tried it in FOREX and it doesn't seem to work. Why is this so?

Answer 1

could this strategy be applied for real trading ?

i mean, whatever a trade makes or loses money, trading incurs transaction-costs.

thus, you cannot stand on that (perfect) bell-shaped distribution to trade profitably with any certainty.

Answer 2

There are several reasons:

Expected payoff = 0

I don't know why you selected 0.5 ATR and 2 ATR away from the market but let's go with it for a while. This means that you want to gain 2x while risking only 0.5x. For now let's assume that FX log-returns are normal.

To bring it to a higher level, we can use a piece of Black Scholes formula, namely the probability that an option ends up in the money is N(d$_{2}$) where N(*) is a cumulative normal distribution of a function and d2 is a function defined as in BS. So your expected payoff is E[payoff] = 2*N(d$_{2}$(2)) - 0.5*N(d$_{2}$(0.5)) which equals 0 if you do the calculation. This explains why you don't make money, not why in reality you lose money. Read on.

Returns are not normally distributed

In general, normal distribution is not a good representation of FX (or other) log-returns. The real distribution of returns has fat tails, often skewness, prices have jumps etc. If the assumption of normality is rejected, your model breaks down.

Transaction costs

As edouard mentioned, even if your E[payoff] = 0, you would incur transaction costs (through slippage, fees, bid/ask spread etc.) which would make your E[payoff] < 0.

Hope it helps.