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Discussion in 'Strategy Development' started by kut2k2, Nov 28, 2011.

Not open for further replies.
1. kut2k2

A trader is using a system where, after each trade, he wins W units with a probability of p (0 < p < 1) or he loses L units with a probability of q = 1 - p.

The trader has a starting bankroll of B0 units and he has in mind a surrender bankroll of X units (B0 > X). If the trader's bankroll drops to or below X, he admits defeat (i.e., he is RUINED).

What is his risk of ruin?

The bankroll after n trades is Bn = B0 + sum[t_i]i=1...n

The expectation of each trade is E[t] = pW-q|L|

The variance of each trade is V[t] = pq(W + |L|)Â²

The expected value of the trader's bankroll is

E[Bn] = E[ B0 + sum[t_i]i=1...n ]

E[Bn] = B0 + nE[t] = B0 + n(pW-q|L|)

The variance of the trader's bankroll is

V[Bn] = nV[t]

V[Bn] = npq(W + |L|)Â²

Central Limit Theorem:

Bn ~ N(E[Bn], V[Bn])

Bn ~ N(B0 + nE[t], nV[t])

The z-score of ruin is

Zr = (X - E[Bn])/sqrt[V[Bn]]

Zr = (X - B0 - nE[t])/sqrt[nV[t]]

Zr = (-(B0-X)/sqrt[n] - sqrt[n]*E[t])/sqrt[V[t]]

If E[t] <=0, then Zr will increase as n increases, making ruin more and more likely.

BUT

If E[t] > 0, then Zr becomes more negative as n increases, making ruin less and less likely.

So the long-term risk of ruin is certain (100%) unless the trading system has positive expectation.

2. kut2k2

Unsurprisingly this result is fully consistent with the RoR (risk of ruin) formulae derived elsewhere via the Gambler's Ruin Theorem and summarized below:

Let W = |L| = one (1) betting unit.

Let R{T|X|C} be the probability of failure (aka risk of ruin) for a gambler with current bankroll C attempting to reach a target bankroll T before falling to a surrender bankroll X. T > X+1.

R{T|X|X} = 1

R{T|X|T} = 0

If p = q:

R{T|X|C} = (T-C)/(T-X)

If p <> q:

R{T|X|C} = ((q/p)^(C-X) - (q/p)^(T-X))/(1 - (q/p)^(T-X))

R{infinity|X|C} = 1 if q > p

R{infinity|X|C} = 1 if q = p

R{infinity|X|C} = (q/p)^(C-X) if q < p

R{infinity|X|C} < 1 if q < p AND C > X

3. kut2k2

Naturally the most interesting formula is

R{infinity|X|C} = (q/p)^(C-X) if q < p

Mathematics professor Peter A. Griffin wanted to use this formula to calculate the risk of ruin for his expert blackjack systems but his real-world payoffs were not all equal (just like for trading systems).

So Griffin came up with the brilliant idea of an "even-payoff equivalent" system, a fictional system which has even-money payoffs as well as first and second moments (expectation and variance) identical to the real-world system.

Let U = bet size in the even-payoff equivalent system.

Let z = win rate in the even-payoff equivalent system.

Now the expectation (E) and the variance (V) must be the same for both systems:

E = pW - q|L| = zU - (1-z)U

V + EÂ² = pWÂ² + qLÂ² = zUÂ² + (1-z)UÂ²

Solving for z and U gives:

U = (pWÂ² + qLÂ²)^Â½

z = 0.5(1 + (pW - q|L|)/U)

So (q/p)^(C-X) becomes ((1 - z)/z)^((C-X)/U)

If we define the trader's edge as

edge = (pW - q|L|)/U

Then

RoR = ((1 - edge)/(1 + edge))^((C-X)/U)

4. virtualmoney

Say I am interested in q/p ratio to be profitable,
Set (q/p)^(C-X)<1,
=> q/p < InverseLog(Log1/(C-X))
Is this requirement correct?

5. kut2k2

You want a positive-expectation trading system. So get away from the Gambler's Ruin formula and use the Trader's Ruin formula:

RoR = ((1-edge)/(1+edge))^((C-X)/U)

where edge = (pW - q|L|)/U

and U = (pWÂ² + qLÂ²)^Â½

RoR < 1 if edge > 0 and C > X

6. virtualmoney

Setting RoR = ((1-edge)/(1+edge))^((C-X)/U)<1,
What is the edge requirement expression?

edge > 0

8. zedDoubleNaught

If there was only one trade, could you rephrase T, X as:

T = target profit for the trade
X = stop loss for the trade

or is that oversimplifying too much?
Thanks for writing out the math step by step, it is very helpful.

9. kut2k2

You could say

T = C + W = X + |L| + W

In that case the RoR is just q.
You're welcome.