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Correlated returns interval

StatisticsHard

You have $n = 400$ daily returns with standard deviation $σ = 10$ . They follow a stationary series with lag $1$ autocorrelation $ρ = 0.6$ and no dependence beyond what an AR model implies. Assume the variance of the sample mean is $(\frac{σ ^{2}}{n}) \times \frac{1 + ρ}{1 - ρ}$ . The interval half-width is $2$ times the square root of this variance. What is the half-width?

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Hint

Find the inflation factor on the variance, then take its root.

Solution

The inflation factor is $\frac{1 + 0.6}{1 - 0.6} = \frac{1.6}{0.4} = 4$ .
The independent-case variance is $\frac{σ ^{2}}{n} = \frac{100}{400} = 0.25$ .
The actual variance is $0.25 \times 4 = 1$ , so the standard error is $1$ .
The half-width is $2 \times 1 = 2$ .
So the answer is $2$ .

Answer: $2$

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