Geometric growth rate and Kelly's Criterion question In the Wikipedia page about Kelly Criterion, the author calculated the expected wealth after N bets as $$W * (1+g)^N$$ where $W$ is the initial wealth, and $g$ is the expected geometric growth rate. For example, with \$25 starting wealth, a 60% chance of winning/losing the whatever you wager, if our strategy is to bet 20% of current wealth, then the article says that  $1+g = (1+0.2*1)^{0.6}(1-0.2*1)^{0.4} = 1.02034$ and expected wealth at round N is $W_N = 25*(1.02034)^N$
But if I look at it from another perspective: the expected wealth after 1 round is $0.6(1.2W) + 0.4(0.8W) = 1.04W$. So shouldn't $W_N$ be $25*(1.04)^N$ instead?
That is, there should only be one answer for expected wealth at round N, given our paramters p,q,f (here f is known so Kelly Criterion doesn't really come into play). However, the Wiki page has a completely different computation than what I have.
 A: The point of the Kelly Criterion is to maximize the expected geometric growth rate -- not the expected arithmetic return and not the expected terminal wealth. Your reasoning about computing expected values is not incorrect, but it is not what is used in deriving the Kelly Criterion.
With fractional betting, the wealth $W_n$ after $n$ rounds is
$$W_n = W_{n-1}(1+fX_n),$$
where $f$ is the fixed deterministic fraction and $X_n$ is a binary random variable such that $p =\mathbb{P}(X_n = +1)$ and $q= 1-p = \mathbb{P}(X_n = -1)$.
The expected wealth after $n$ rounds conditional on the wealth after $n-1$ rounds, $W_{n-1}$, is
$$\mathbb{E}(W_n\,| W_{n-1}) = W_{n-1}\left[p(1+f)-q(1-f) \right],$$
and for $f = 20\%$, $p = 60\%$ and $q = 40\%$ we obtain
$$\mathbb{E}(W_n\,| W_{n-1}) = W_{n-1}[0.6(1+0.2) +0.4(1-0.2)]= 1.04W_{n-1}$$
This is similar to what you derivived as far as it goes.
Derivation of the Kelly Criterion
The Kelly Criterion is derived by maximizing the expected geometric growth rate
$$\tag{*}\mathbb{E}\left[\frac{1}{n}\log \frac{W_n}{W_0}\right],$$
which is not the same as
$$\frac{1}{n} \log  \frac{\mathbb{E}(W_n)}{W_0}$$
The terminal wealth after $n$ rounds is
$$W_n = W_0\prod_{k=1}^n (1+fX_k)$$
Taking the logarithm of both sides we get
$$\log W_n = \log W_0 + \sum_{k=1}^n \log(1+fX_k)$$
Whence, for independent and identically distributed $X_k$,
$$\mathbb{E} \left[\frac{1}{n}\log  \frac{W_n}{W_0}\right] = \frac{1}{n} \sum_{k=1}^n \mathbb{E}[\log(1+fX_1)]= \mathbb{E}[\log(1+fX_1)]\\ = p\log (1+f) +q\log(1-f)$$
To find the maximum, take the derivative with respect to $f$ and equate to $0$.  The Kelly optimal fraction will be $f^* = p-q$.
Resolution
With a fractional betting strategy we always have
$$\mathbb{E} \left(\frac{W_n}{W_0}\right) = \mathbb{E}\left[\prod_{k=1}^n (1+f X_k)\right]= \prod_{k=1}^n \mathbb{E}(1+fX_1) = (1 + (p-q)f)^n$$
and, hence,
$$\tag{1}\left[\mathbb{E} \left(\frac{W_n}{W_0}\right)\right]^{1/n} = 1 + (p-q)f$$
To maximize (1) we would choose $f=1$ (betting all accumulated wealth on each round) and we would get $\mathbb{E}(W_n) = W_0(1+p-q))^n$. However, there would be a minuscule probability of attaining this wealth as the probability of ruin is $P_{\text{ruin}} = 1 - p^n$ which approaches $1$ rapidly as $n$ is increased.
However, (1) is not the expected growth rate that is maximized by Kelly Criterion. That would be
$$g(f) = \mathbb{E}\left[\log \left(\frac{W_n}{W_0}\right)^{1/n}\right], $$
which is maximized with $\text{argmax }\, g(f)=f^*=p-q$ and
$$\mathbb{E}\left[\log \left(\frac{W_n}{W_0}\right)^{1/n}\right] = g(f^*) = p\log(1+f^*) + q\log(1- f^*)$$
It is true that with the optimal betting fraction,
$$\exp\left(\mathbb{E}\left[\log \left(\frac{W_n}{W_0}\right)^{1/n}\right] \right)= (1+f^*)^p(1-f^*)^q,$$
but this does not mean $\mathbb{E}(W_n) = W_0[1+g(f^*)]^n$.
