Calculating Log-Return Variance for a Two-Outcome Kelly Bet

Summary

The document clarifies an expectation calculation used when studying the Kelly criterion. For a bet whose outcome has two possible values, the log wealth change takes one value after a win and another after a loss. Its squared expectation is found by weighting the square of each possible log change by that outcome’s probability. The variance then follows by subtracting the square of the mean log change from this second moment.

The response explains why the weighted expression is valid: it is the standard expectation of a discrete random variable’s square, with the two log outcomes occurring with probabilities p and q. This is a focused probability clarification rather than a derivation of the Kelly-optimal fraction or a complete betting strategy. It assumes the stated two-outcome setup; cases with more possible outcomes require summing across all outcomes and their probabilities.

Key ideas

• For a discrete random variable, calculate the expectation of its square by weighting each squared outcome by its probability.
• In the two-outcome Kelly setup, the log wealth change has separate values for wins and losses.
• Obtain the variance of log wealth change by subtracting the squared mean from its second moment.
• The calculation applies to the stated two-outcome model and does not determine the optimal Kelly fraction by itself.

Tags

# Kelly Variance - variance of the sum of logs


# Kelly Variance - variance of the sum of logs












I am working through Thorpe's Ch 9 on the Kelly criterion.

On page 9 Thorpe states:

$$Var(ln(1+Y_if)] = p[ln(1+f)]^2 + q[ln(1-f)]^2 - m^2$$

Since $var(X) = E[X^2] - m^2$,

$$p[ln(1+f)]^2 + q[ln(1-f)]^2 = E[X^2]$$

Would it be correct to assume that $E[(ln\sum x_i))^k] = \sum p_i(ln(x_i))^k$ for rvs which can only take on two values? I am assuming this is the case but I am on a steep probability learning curve, so verification from someone with more experience would still be appreciated.

## Answer by nbbo2 (score 0, accepted)

https://quant.stackexchange.com/a/34698

$$ E[X^2] = p[ln(1+f)]^2 + q[ln(1-f)]^2 $$

Is the standard way to compute the expectation of $X^2$, since in this case $X =\ln(1+Y_i f)$ has only two possible values: $\ln(1+f)$ with probability $p$ and $\ln(1-f)$ with probability q.

So it is quite a straightforward calculation.

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

This summary was written by Stratmill's research agent from the original; it is not a copy of the source.