Risk Management puzzles, solved step by step

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All topicsCapital and leverage6Compounding and drawdowns8Correlation and diversification8Counterparty exposure and collateral7Credit risk arithmetic10Duration and rates7Liquidity and balance sheet7Logic, estimation and brainteasers7Operational loss and fraud7Options and Greeks7Probability and base rates8Statistics and estimation10VaR and expected shortfall8
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  1. 050An EWMA volatility model with a decay factor of 0.94 had yesterday's daily volatility estimate at 1.2%. Today's return is minus 3%. What is the updated volatility estimate?Statistics and estimationCoreBank market riskQuant risk

    Try it first

    Roughly where does the new daily volatility land?

    Show the worked solution

    About 1.38%. EWMA updates the variance, not the volatility: new variance is 0.94 x 1.2 squared plus 0.06 x 3 squared, which is 1.3536 plus 0.54, or 1.8936. The square root is 1.376%. The shock has only a 6% weight but enters squared, so it supplies 29% of the new variance.

    Why blend variances rather than volatilities?

    A household tracking how much its grocery bill swings would be misled if it averaged the size of the swings in rupees but ignored that one big swing matters far more than several small ones. Variance is the average squared move, so the model updates squared returns, and a move 2.5 times normal size counts 6.25 times as much. An EWMAExponentially weighted moving average: each day the estimate keeps a fixed share of yesterday and adds the rest from today, so older days fade geometrically. of variances is what makes one large shock move the estimate quickly. Blending the volatilities directly gives 1.31%, understating the jump.

    Six per cent of the weight, but more than a quarter of the resultWeights94% on yesterday's variance6%Share ofnew variance0.94 x 1.2 sq = 1.35471.5%0.06 x 3 sq = 0.5428.5%New variance 1.3536 + 0.54 = 1.8936Volatility: square root = 1.38% a day, up from 1.20%
    Today's minus 3% return carries only a 6% weight in the EWMA update, but because it enters squared it supplies 28.5% of the new variance of 1.8936, lifting daily volatility from 1.20% to 1.38%.
    The relationship
    σt2=λ σt−12+(1−λ) rt2=0.94(1.44)+0.06(9)=1.8936,σt=1.376%\sigma_t^2 = \lambda\,\sigma_{t-1}^2 + (1-\lambda)\,r_t^2 = 0.94(1.44) + 0.06(9) = 1.8936, \quad \sigma_t = 1.376\%
    \lambdathe decay factor, 0.94
    \sigma_{t-1}yesterday's volatility estimate, 1.2%
    r_ttoday's return, minus 3%
    What it says in wordsKeep 94% of yesterday's variance, add 6% of today's squared return, then take the square root.

    What happens next, and what are the model's limits?

    If tomorrow is flat, the estimate decays to the square root of 0.94 x 1.8936, about 1.33%. Each day of calm keeps 94% of the variance, so a shock's influence halves in about 11 trading days: ln 0.5 over ln 0.94. That is the design choice behind 0.94: fast enough to react to a new regime within days, slow enough not to swing on every move.

    The limits: EWMA has no pull towards a long-run average, so after a calm spell it can drift very low and understate risk just before volatility returns, which GARCH-type models address with a mean-reversion term. It also treats up and down moves alike, while falling markets often raise volatility more than rising ones. And the choice of 0.94 is a convention for daily data, not a law, so the estimate should be backtested against realised moves.

    Where candidates lose it

    The usual slip is to blend the two volatilities directly, 0.94 x 1.2 plus 0.06 x 3, and answer 1.31%. It looks like the formula but applies it to the wrong quantity and understates the effect of the shock.

    The second is forgetting the square root at the end and quoting 1.89% as a volatility. Say the units at each step: variance in, variance out, then volatility.

    What the interviewer asks next

    • What would the estimate be with lambda of 0.97 instead?
    • How many quiet days until the estimate is back below 1.25%?
    • Why might a risk manager prefer GARCH to EWMA for a ten-day VaR?
  2. 087A desk's daily returns have a volatility of 1%. Over 250 days its average daily return is 0.05%. How precisely is that mean estimated, and can you say the desk has skill?Statistics and estimationCoreAsset manager riskQuant risk

    Try it first

    Is 0.05% a day, measured over one year, clearly different from zero?

    Show the worked solution

    Not precisely enough to claim skill. The standard error of the mean is the daily volatility over the square root of the number of days: 1% over the square root of 250, about 0.063%. The estimate of 0.05% is only 0.79 standard errors from zero, and a two standard error band runs from -0.076% to 0.176%. You would need about 1,600 days, 6.4 years, to clear zero.

    Why is a year of daily data not enough?

    Weigh yourself on a bathroom scale that jumps by two kilos each time you step on it. If you lost 100 grams last month, a week of readings will not show it; the jumps drown the signal. The precision of an average improves only with the square root of the number of observations, so a small edge buried in large daily noise takes years to show. Here the daily noise is 1% and the edge is 0.05%, twenty times smaller.

    The relationship
    SE=σn=1%250=0.063%t=0.050.063=0.79SE = \frac{\sigma}{\sqrt{n}} = \frac{1\%}{\sqrt{250}} = 0.063\% \qquad t = \frac{0.05}{0.063} = 0.79
    sigmadaily return volatility, 1%
    nnumber of daily observations, 250
    thow many standard errors the mean is from zero
    What it says in wordsThe mean's uncertainty is the daily volatility divided by the square root of the number of days, and the edge is less than one of those units from zero.
    One year of data cannot tell a 0.05% edge from zerozero: no skill250 daysone year-0.076%+0.176%1,600 daysabout 6.4 years0.000%+0.100%estimate 0.05%-0.10%0+0.10%+0.20%Average daily return, with a two standard error band
    After 250 days the two standard error band around the 0.05% daily mean runs from -0.076% to 0.176% and straddles zero; after about 1,600 days, 6.4 years, the band narrows to 0.000% to 0.100% and only just clears zero.

    How long would it take, and what does that mean for judging desks?

    Set the t-statistic to 2 and solve for n: n equals (2 times 1% over 0.05%) squared, which is 1,600 days. An annual Sharpe ratio of about 0.79 needs more than six years of data before it is statistically distinguishable from zero. That is longer than most desks keep the same strategy, so a risk team cannot rely on the P&L record alone; it looks at whether the edge has a reason, whether it survives out of sample, and how much the result depends on a few days.

    State the assumptions. The calculation treats daily returns as independent with constant volatility. Fat tails and volatility clustering make the true uncertainty larger, so six years is a floor, not a promise.

    Where candidates lose it

    The common error is annualising the mean to 12.5% and declaring skill, as if a big annual number were proof. The annual volatility grows too, to about 16%, and the ratio of the two is what matters.

    The other slip is dividing by 250 instead of its square root, which makes the mean look fifty times more precise than it is.

    What the interviewer asks next

    • The desk's volatility is 0.5% instead. How many days now?
    • Why is the mean so much harder to estimate than the volatility?
    • How would you judge a new desk that has only six months of history?