Risk Management puzzles, solved step by step

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  1. 023You estimate a correlation of 0.5 between two assets from 36 monthly observations. Roughly what is its standard error, and is a later estimate of 0.3 from the next 36 months evidence that the relationship has changed?Statistics and estimationHardQuant riskAsset manager risk

    Try it first

    Is the drop from 0.5 to 0.3 statistically meaningful?

    Show the worked solution

    The standard error is about 0.125, and 0.3 is not evidence of a change. A rough rule is one minus the correlation squared, over the square root of n: 0.75 over 6. Using Fisher's transformation, the 95% interval for 0.5 runs from about 0.21 to 0.71, and 0.3 sits inside it. Testing the difference between two 36 month estimates gives z of about 0.97, well below 1.96.

    Why is a correlation from 36 months so imprecise?

    Think of judging whether two friends like the same films after watching three dozen films together. They might agree on 20 by chance even if their tastes barely overlap. A correlation is estimated from how often two series move together, and with 36 points random coincidences move the estimate a lot: the standard error of 0.5 is about 0.125, a quarter of the estimate itself. The rough rule is one minus the correlation squared over the square root of the sample size, 0.75 over 6.

    The relationship
    SE(ρ^)≈1−ρ2n=0.756=0.125z=tanh⁡−1(ρ^),    SE(z)=1n−3\text{SE}(\hat\rho) \approx \frac{1 - \rho^2}{\sqrt{n}} = \frac{0.75}{6} = 0.125 \qquad z = \tanh^{-1}(\hat\rho), \;\; \text{SE}(z) = \frac{1}{\sqrt{n - 3}}
    rhothe estimated correlation, 0.5
    nthe number of observations, 36
    tanh^{-1}Fisher's transformation, which makes the sampling error close to normal
    What it says in wordsThe error of a correlation shrinks only with the square root of the sample size, and Fisher's transformation gives a fair interval around it.
    Two correlations from 36 months each: the intervals overlap-0.20.00.20.40.60.81.0Correlation0.210.71First 36 months: 0.50-0.030.57Next 36 months: 0.30overlap 0.21 to 0.57Rough standard error: (1 - 0.5 squared) / 6 = 0.125. Test of the difference: z = 0.97, well under 1.96.
    The 95% interval for a correlation of 0.5 from 36 months runs from 0.21 to 0.71, and the interval for 0.3 from another 36 months runs from -0.03 to 0.57, so the two overlap heavily and the drop is within sampling noise.

    How do you test whether the relationship changed?

    Transform both estimates with Fisher's z, which turns 0.5 into 0.549 and 0.3 into 0.310. The difference, 0.240, divided by its standard error of the square root of 2 over 33, about 0.246, gives a z of 0.97: well short of the 1.96 needed for 95% confidence. A gap that size would appear by chance about 33% of the time even if nothing had changed. The honest statement is that the data cannot tell 0.5 from 0.3.

    Then say why a risk manager cares. Hedge ratios, portfolio volatility and diversification benefits are all built on correlations like this one, and treating 0.5 as exact makes a model look more certain than its data allow. The practical responses are longer or higher-frequency data, shrinking noisy estimates towards a sensible prior, and stress testing the portfolio at correlations across the whole interval rather than at the point estimate.

    Where candidates lose it

    The trap is reading the drop from 0.5 to 0.3 as a regime change. It is a 40% fall in the number but well inside the noise that three years of monthly data carry.

    The second miss is quoting a standard error without saying it is rough. The simple formula is an approximation; Fisher's transformation is what you use for an interval or a test.

    What the interviewer asks next

    • How many monthly observations would you need to bring the standard error down to 0.05?
    • Would weekly data over the same three years help, and what could go wrong?
    • How would you stress test a portfolio whose risk depends on this correlation?
  2. 073A rating grade carries a predicted default rate of 1%. It has 500 obligors, and 9 defaulted last year. At 95% confidence, is the grade miscalibrated?Statistics and estimationHardRating agencyModel validation

    Try it first

    Nine defaults against five expected. What does an exact one-sided test say?

    Show the worked solution

    No, not at 95% confidence, though it is close. With 500 obligors at 1%, you expect 5 defaults. The exact binomial chance of 9 or more is about 6.7%, above the 5% cut-off; the smallest count that would reject is 10. A normal approximation without a continuity correction gives z = 1.80 and wrongly rejects. One year of data rarely proves a grade wrong.

    How surprising are nine defaults if the 1% is right?

    If a coin lands heads 7 times in 10, you would not call it biased; 70 in 100 you would. Small samples swing. With 500 obligors at 1%, the expected count is only 5, and counts that small are lumpy. The question is not whether 9 is bigger than 5, but how often a correct 1% grade produces 9 or more by chance, and the exact answer is about one year in fifteen.

    Nine defaults against five expected is borderline, not proof0%5%10%15%01234567891011121314Defaults in the grade in one year (500 obligors, PD 1%)reject from 10P(10 or more) = 3.1%observed 9:P(9 or more) = 6.7%expected 5
    At a true 1% default rate across 500 obligors the count centres on 5, and 9 or more defaults happens 6.7% of the time. The one-sided 95% critical value is 10 defaults, so the observed 9 sits just inside the acceptance region.

    Why does the normal approximation get it wrong here?

    The standard deviation of the count is the square root of 500 x 0.01 x 0.99, about 2.22. Nine is 1.80 standard deviations above five, past 1.645, so a quick z-test rejects. But the binomial is skewed and discrete at small counts; with a continuity correction z falls to 1.57 and the conclusion flips back. For counts under about ten, use the exact binomial.

    The relationship
    P(X≥9)=∑k=9500(500k)(0.01)k(0.99)500−k≈0.067>0.05P(X \ge 9) = \sum_{k=9}^{500} \binom{500}{k}(0.01)^k(0.99)^{500-k} \approx 0.067 > 0.05
    Xthe number of defaults in the grade
    500obligors in the grade
    0.01the predicted default rate under test
    What it says in wordsAdd up the chance of every outcome at least as bad as nine; if that total is above 5%, you do not reject.

    Two further limitations make the case for caution stronger. Defaults are correlated through the economy, so a single bad year moves many obligors together and the true spread of counts is wider than the binomial says. And a validation team would look at several years and at neighbouring grades before recalibrating. The right answer is to flag the grade for watching, not to declare it broken.

    Where candidates lose it

    The trap is the quick z-test: (9 - 5) / 2.22 = 1.8, above 1.645, reject. It is the answer most candidates give, and it is wrong because a normal curve is a poor stand-in for a skewed count with a mean of five.

    The second miss is failing to mention correlation. Independent defaults are the most favourable assumption for rejecting; real defaults cluster, which makes nine even less conclusive.

    What the interviewer asks next

    • How many defaults would reject at 99% confidence?
    • The grade shows 8, 9 and 7 defaults in three consecutive years. What now?
    • How does default correlation change the test?