Deriving the Second Moment from Mean and Variance
Summary
The explanation addresses a step in a derivation used in Kelly-style growth analysis: why the expected square of an outcome can be approximated by the variance when the mean is small. It writes each outcome as its mean plus a zero-mean deviation, then expands the square into the squared mean, squared deviation, and cross term.
Taking expectations removes the cross term because the deviation has mean zero. The squared deviation contributes the variance, while the squared mean is neglected under the small-mean approximation. The result is that the expected squared outcome is approximated by the variance. This is an explanatory approximation, not an exact identity in general: the expected square equals variance plus squared mean, so omitting the latter requires the stated small-mean assumption.
Key ideas
- Express each outcome as its mean plus a zero-mean deviation.
- Expanding the square produces a squared mean, a squared deviation, and a cross term.
- The cross term vanishes in expectation because the deviation has mean zero.
- The expected square is approximately the variance when the squared mean is negligible.
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# Can you clarify Kelly's derivation in Paul Wilmott Introduces Quantitative Finance?
# Can you clarify Kelly's derivation in Paul Wilmott Introduces Quantitative Finance?
Can anybody explain how the $\phi_i$ became $\mu$ and how $\phi_i^2$ became $\sigma^2$. Am I correct to assume that since $\phi_i$ is the outcome, $\mu$ is the average of the outcome? But I don't understand how $\phi_i^2$ is the variance $\sigma^2$.
Extract from Paul Wilmott Introduces Quantitative Finance:
## Answer by Phil H (score 1)
https://quant.stackexchange.com/a/40473
Suppose we decompose each $\phi_i$ into the mean $\mu$ plus some extra bit $\kappa_i$. Then what would the squared term look like?
$$\phi_i^2 = (\mu + \kappa_i)^2 = \mu^2 + \kappa_i^2 + 2\mu\kappa_i$$
Now we've been told that the mean is small, which is generally code for "the square can be neglected", leaving us with $\kappa^2$ and $2\mu\kappa$.
We are not, however, looking for the value of $\phi_i^2$, but for $\mathbb{E}(\phi_i^2)$. Expectations are linear, i.e. $\mathbb{E}(a+b) = \mathbb{E}(a)+\mathbb{E}(b)$, so
$$\mathbb{E}(\phi_i^2) = \mathbb{E}(\kappa_i^2) + \mathbb{E}(2\mu\kappa_i)$$
$\mathbb{E}(\kappa_i^2)$ is by definition the variance (it is the expectation of the square of the deviations), thus $=\sigma^2$.
$\mathbb{E}(2\mu\kappa_i) = 2\mu\mathbb{E}(\kappa_i)$. Remembering our decomposition, $\kappa_i$ is the extra bit, so again by definition, it has a mean 0, so $\mathbb{E}(\kappa_i) = 0$.
$$\mathbb{E}(\phi_i^2) = \sigma^2$$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.