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StatsMaster_J2026-04-06
cfaLevel IQuantitative Methods

How do I use the coefficient of variation to compare investments with different expected returns?

I understand that standard deviation measures absolute risk, but my CFA Level I materials say the coefficient of variation is better for comparing risk when expected returns differ. Can someone show me a concrete example of how CV changes the ranking?

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The coefficient of variation (CV) measures risk per unit of expected return, making it a standardized way to compare investments when their return levels are very different.

Formula

CV = σ / E(R)

Lower CV = more efficient risk-return tradeoff.

Why Standard Deviation Alone Fails

Standard deviation measures total risk in absolute terms. If one investment has an expected return of 25% with σ = 20%, and another has expected return of 5% with σ = 10%, raw standard deviation says the first is riskier. But relative to what you're earning, the second investment is actually riskier.

Worked Example — Whitecliff Investment Committee

Whitecliff is comparing two private equity funds:

FundE(R)σCV
Garrison Ventures22.0%30.0%30/22 = 1.36
Millbrook Capital8.0%14.0%14/8 = 1.75

By standard deviation: Garrison looks riskier (30% > 14%).

By CV: Millbrook is actually riskier per unit of return (1.75 > 1.36). For every 1% of expected return, Millbrook carries 1.75% of risk versus Garrison's 1.36%.

When CV Breaks Down

CV is meaningless when E(R) is zero or negative — you'd be dividing by zero or getting a negative ratio. This is a common exam trap: if one investment has a negative expected return, you cannot use CV to compare it.

Relationship to Sharpe

The Sharpe ratio is conceptually similar but uses excess return (Rp - Rf) instead of total return, and inverts the relationship (higher is better for Sharpe, lower for CV).

If Rf = 0, then Sharpe = 1/CV.

Exam tip: CFA Level I specifically asks you to use CV when comparing investments with "significantly different expected returns." If returns are similar, standard deviation alone is sufficient.

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