Convexity of Mean Absolute Deviation Portfolio Risk

Summary

The document proves convexity for a portfolio risk measure consisting of expected return plus a positive multiple of the expected absolute deviation from expected return. It assumes portfolio return is linear in portfolio weights, so expected return is also linear. The key step is showing that expected absolute deviation is convex as a function of those weights.

For a convex combination of two portfolios, linearity of expectation expresses the centered return as the same combination of the two centered returns. The triangle inequality bounds the absolute value of that combination by the weighted sum of the individual absolute values. Taking expectations yields the convexity inequality; adding a linear term and scaling by a positive constant preserve convexity. The argument is a mathematical derivation, not an empirical evaluation. It relies on the stated linear-return setup and assumes the expectations in the risk definition exist; it does not compare this measure with other risk criteria or discuss optimization constraints.

Key ideas

• With returns linear in portfolio weights, expected return is a linear function of the portfolio.
• Expected absolute deviation from the mean is convex by the triangle inequality.
• A positive multiple of a convex function plus a linear function remains convex.
• The proof assumes the relevant expectations exist and portfolio returns depend linearly on weights.

Tags

# How can I show convexity of this risk function?


# How can I show convexity of this risk function?












I have the following risk function:

$\mathbf{Risk}(x):=\mathbb{E}[R(x)]+\delta\mathbb{E}[|R(x)-\mathbb{E}[R(x)]|]$

where $R(x)$ is the portfolio return and $\delta$ is any positive scalar. My textbook assumes this function is convex without proving it. Any ideas how it could be shown it is convex?

## Answer by MainCom (score 1, accepted)

https://quant.stackexchange.com/a/69920

Since $R(x)$ is linear in $x$, we just need to prove $\mathbb{E}(x) + \delta \mathbb{E}(\mathbb{|x-\mathbb{E}(x)|})$ is convex. Now $\mathbb{E}(x)$ is linear and $\delta$ is a positive constant, so we only need to prove $f(x):=\mathbb{E}(|x-\mathbb{E}(x)|)$ is convex in $x$. Then it is straightforward to prove by definition: for $0\leq t <1$,

$f(tx+(1-t)y) = \mathbb{E}(|tx + (1-t)y -\mathbb{E}(tx + (1-t)y)|) = \mathbb{E}(|t(x-\mathbb{E}(x)) + (1-t)(y-\mathbb{E}(y))|) \leq \mathbb{E}(|t(x-\mathbb{E}(x))| + |((1-t)(y-\mathbb{E}(y))|) = tf(x) + (1-t)f(y)$,

as desired. The inequality used is simply the triangle inequality.

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