Discrepancy between two expected profit formulas

Question

I have two formulas for calculating the expected ratio of the account value after a trade to before it (we'll call this expected account ratio or EAR) that both make intuitive sense to me, yet they produce different results. The first formula uses the definition of expected value. The second uses a geometric mean. I'm unsure which accurately describes financial behavior and I'd like to understand which one is correct and why.

Terms: p is the probability of a trade winning with some profit before losing with probability 1-p with some loss. AB = available balance, or the amount in the account before the trade.

$$EAR1 = p \cdot \frac{AB + \text{profit}}{AB} + (1-p) \cdot \frac{AB + \text{loss}}{AB}$$ $$EAR2 = \left(\frac{AB + \text{profit}}{AB}\right)^p \left(\frac{AB + \text{loss}}{AB}\right)^{1-p}$$

The rationale for the first approach is that we calculate the actual EAR after a successful trade and weight it by p, and compute the actual EAR after a failed trade and weight it by 1-p. This in theory should give us the expected EAR after one trade.

The rationale for the second approach is as follows. Imagine that we make 2% per successful trade and lose 2% per failed trade. Since multiplication is commutative, the order in which wins and losses happen doesn't matter - just the totals of each. Assume p=0.6. Then if we take 10 trials, our expected EAR at the end of all 10 is 1.02^6 * 0.98^4. To find the per trade EAR, we take the 10th root of it. By the laws of exponents, this is equivalent to taking 1.02^0.6 * 0.98^0.4.

I can't figure out why the two approaches give different results other than putting them side by side and seeing that indeed they are not equal quantities since they cannot be made to cancel to zero. But both seem to describe a reasonable way of describing the behavior of an account after a trade. What am I missing?

Answer 1

$EAR2$ is the correct formula because taking the expectation of a binomial random variable assumes independence of trials, which is not the case in any context that involves compounding, such as this question's context of trading. Binomial expectation relies on each outcome contributing equally and independently to the overall outcome. Since a 2% win following a 2% produces a greater than 2% win on top of the original available balance, the binomial model fails and we must use the geometric mean instead.

Answer 2

If the increasing rate and decreasing rate are $2\%$ then your factors $1.02$ and $0.98$ are right. If you have $n$ trials, then you have $\binom{n}{x}$ ways of making $x$ times profit and $n-x$ times making loss. The probability of making profit is $p$ and therefore the probability of making loss is $1-p$

Then the probability of making profit $x$ times and making $n-x$ times loss (in exactly that order) is $p^x\cdot (1-p)^{n-x}$.

The outcome after making profit $x$ and making $n-x$ times loss (in exactly that order) is $1.02^x\cdot 0.98^{n-x}$

The number of ways making profit $x$ times and making $n-x$ times loss is $\binom{n}x$. Let $EAR$ be the expected account ratio. If $EAR=1$, the winning/loosing is $0$

Then $(EAR)^n$ after $n$ trials is

$$(EAR)^n=\sum\limits_{x=0}^n \binom{n}x 1.02^x\cdot 0.98^{n-x}\cdot p^x\cdot (1-p)^{n-x}$$

$$=\sum\limits_{x=0}^n \binom{n}x (1.02p)^x\cdot (0.98\cdot (1-p))^{n-x}$$

The binomial theorem is $ (a+b)^n=\sum\limits_{x=0}^{n}{{n \choose x}\cdot a^{x}\cdot b^{n-x}} $. If we apply that we get

$$(EAR)^n=(1.02p+0.98\cdot (1-p))^n$$

Now we calculate $EAR$ by using the geometric mean

$$EAR=1.02p+0.98\cdot (1-p)$$

If $p=0.5$, then $EAR=1$. This is not surprisingly, since the possible amounts of winning and loosing are equal at a given level.