Are currency markets random?

[question]

In the simplest case where the market is a random walk and that the ratio of winning and losing is 50:50 you can still make money if you limit losses and let profit runs. One cannot control let alone predict the market. The only variable one can control is YOURSELF.

That logic is flawed. You will get stopped out more often than not (with tight stops) and thus your win ratio will not be 50-50%

[awcapital]

Originally Posted by question

That logic is flawed. You will get stopped out more often than not (with tight stops) and thus your win ratio will not be 50-50%

What is a tight stop ??? 1 pip 30 pips 100 pips ? Its all relative. A system is deemed to have a 50:50 win loss ratio (based on certain backtesting criteria) means that it will get stopped out just as often as it wins. That's why it's 50:50. Whether one thinks the stop is tight or loose is irrelevant.

Alternatively, in a 50:50 system, to breakeven one has an equal chance of getting stopped out as be in a winning trade. The $ breakeven point is when the winnings ($) = stop loss size ($)

I hope it makes sense

[TraderABC]

Originally Posted by awcapital

What is a tight stop ??? 1 pip 30 pips 100 pips ? Its all relative. A system is deemed to have a 50:50 win loss ratio (based on certain backtesting criteria) means that it will get stopped out just as often as it wins. That's why it's 50:50. Whether one thinks the stop is tight or loose is irrelevant.

Alternatively, in a 50:50 system, to breakeven one has an equal chance of getting stopped out as be in a winning trade. The $ breakeven point is when the winnings ($) = stop loss size ($)

I hope it makes sense

If price is random, then it has 50:50 chance of going either way. However where do you think it will touch first, 30 pips away or 300? Thus you cannot say "Oh, I'll set my stop loss at 30 and take profit at 300" . While it is true that your max win to max loss ratio will be 10-1. Your frequency of such wins will be MINISCULE.

It is true that if your profit/loss ratio is 3/1 your wins are much larger. But with 3/1 ratio you will be stopped out 3x as much, thus your mathematical expectency is 0 at best.

[awcapital]

Originally Posted by TraderABC

If price is random, then it has 50:50 chance of going either way. However where do you think it will touch first, 30 pips away or 300? Thus you cannot say "Oh, I'll set my stop loss at 30 and take profit at 300" . While it is true that your max win to max loss ratio will be 10-1. Your frequency of such wins will be MINISCULE.

It is true that if your profit/loss ratio is 3/1 your wins are much larger. But with 3/1 ratio you will be stopped out 3x as much, thus your mathematical expectency is 0 at best.

Agree. The lower the win loss ratio, the higher profits per win needed. That was all I was trying to illustrate. With regards to what profit targets is needed to be profitable, I guess in a 50:50 world just 1 pip more than stop loss size.

[TraderABC]

Originally Posted by awcapital

Agree. The lower the win loss ratio, the higher profits per win needed. That was all I was trying to illustrate. With regards to what profit targets is needed to be profitable, I guess in a 50:50 world just 1 pip more than stop loss size.

First, with slippage, price shifting, etc that 1 pip is not enough.

2nd, I probably didn't accent enough risk/reward ratio and random walk enough.


IF risk/reward ratio is 1/1, then in best case you will break even.
if risk/reward 2/1, then your losses are 2x as big and you'll probably get margin call much quicker than big gains. On other hand you'll be winning 2x as often so if you DO get lucky you will break even at best...

If 1/2, then while gaining 2x as much, you will be loosing 2x as much... Thus you will be lucky to break even AGAIN..

With spread, slippage, price shifting, and other events no matter which r/r ratio you choose the result is the same according to random walk.

[awcapital]

Originally Posted by TraderABC

First, with slippage, price shifting, etc that 1 pip is not enough.

2nd, I probably didn't accent enough risk/reward ratio and random walk enough.


IF risk/reward ratio is 1/1, then in best case you will break even.
if risk/reward 2/1, then your losses are 2x as big and you'll probably get margin call much quicker than big gains. On other hand you'll be winning 2x as often so if you DO get lucky you will break even at best...

If 1/2, then while gaining 2x as much, you will be loosing 2x as much... Thus you will be lucky to break even AGAIN..

With spread, slippage, price shifting, and other events no matter which r/r ratio you choose the result is the same according to random walk.

I agree with that. But I think we are talking about 2 different things !

Risk/Reward ratio relates to magnitudes of losses and winnings.
Win:Loss ratio simply is the probability between winning and losing.

If the market follows a random walk, then there is even chances of winning or losing. Not talking about magnitudes.

Spread, slippage etc can be incorporated into the breakeven calculation (at least for spread, we can estimate slippage). Also, unless during major announcements there's hardly any slippages when trading the majors.

[question]

TraderABC, here's an exercise in statistics if the markets are truly random:

1. Let's accept that news events and new information can be considered random to the outside viewer in the fact that it can either be good or bad.

2. Let's assume that the random security we're talking about is EUR/USD which is currently trading at 1 EUR to 1 USD.

3. If the market movements are totally random, after a sufficiently large number of iterations, the EUR/USD will be trading at or very near to 1 EUR to 1 USD.(why?)

4. The reason for this is that the standard deviation of an event decreases after each successive iteration.

5. Toss a coin. Assume heads equal 1 and tails equal 0. After one toss, it is impossible to have an average value of 0.5. But after 2 tosses, you have a 50% chance of having a value of 0.5. Toss the coin more and more, and you will see that more and more of your average value possibilities will hover around 0.5.

6. This whole exercise proves there is some non-random component to the markets.

[awcapital]

Originally Posted by question

TraderABC, here's an exercise in statistics if the markets are truly random:

1. Let's accept that news events and new information can be considered random to the outside viewer in the fact that it can either be good or bad.

2. Let's assume that the random security we're talking about is EUR/USD which is currently trading at 1 EUR to 1 USD.

3. If the market movements are totally random, after a sufficiently large number of iterations, the EUR/USD will be trading at or very near to 1 EUR to 1 USD.(why?)

4. The reason for this is that the standard deviation of an event decreases after each successive iteration.

5. Toss a coin. Assume heads equal 1 and tails equal 0. After one toss, it is impossible to have an average value of 0.5. But after 2 tosses, you have a 50% chance of having a value of 0.5. Toss the coin more and more, and you will see that more and more of your average value possibilities will hover around 0.5.

6. This whole exercise proves there is some non-random component to the markets.

I don't quite agree with that.

Firstly, in the coin example, you need at least 100 samples to form the basic normal distribution to have statistically significant probability of 50:50 between heads or tail.

Secondly, the random walk theory simply states that the market follow an autoregressive order (1). Which means that the best prediction for what the market will do at t+1 is t-1. In otherwords, only the most recent time period matters.

Thirdly, the standard deviation of regression errors (i.e. difference between the mean regression line and actual prices) actually remains constant. Because in a random walk situation, the only thing that matters is the most recent price data. So regardless of the number of previous iteration, the standard deviation of errors that counts is only the most recent one. That's why prices don't converge to 1 EURUSD.

I hope it makes sense