# How to find the minimum variance portfolio?

I am doing some revision questions on my Portfolio Theory module, and have come across the following question:

Consider an investor who has constructed a risky portfolio from N securities. The investment opportunity set is described by the equation:

$$\sigma^2 = 10 - 5{\times}E(r) + 0.5\times(E(r))^2$$

Find the minimum variance portfolio.

I can't find any info in my notes, but my intuition says differentiate, set to zero and rearrange for E(r)?

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## migrated from economics.stackexchange.comMay 1 '12 at 19:55

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Does anything strike you as odd about the resulting value of sigma squared, when you follow your intuition? Could there be a mistake in the transcription of the question? (I could be wrong, I don't do MPT, it just strikes me as a bit peculiar, that's all) – EnergyNumbers Jan 7 '12 at 7:08

If you are trying to minimize sigma-squared, then the points where the derivative is zero will be at least local minima or maxima. If this is not intuitive, imagine a parabola and calculate the derivative at various points.

Another step would be to prove that the function is globally concave so that the local minima/maxima are in fact global, but your prof probably won't require that. In comparison with the parabola example, finding where the dy/dx is zero in y = x ^ 3 won't find the global.

I'm not sure what you mean by rearrange for E(r).

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We want to minimize $\sigma^2$ as a function of $E(r)$. We also likely want $E(r) \ge 0$, but this won't be necessary.

$$0 =\frac{d \sigma^2}{dE(r)}=-5+E(r).$$

Now, looking at the second derivative

$$\frac{d^2 \sigma^2}{dE(r)^2} = 1>0$$

It is convex at every point, so we only need to check minima. And $E(r)=5$ is the minimum variance portfolio.

However, the variance at that point, $\sigma^2=-2.5$, so perhaps there is a mistake in the question. Can you clarify?

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