Risk Management puzzles, solved step by step

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  1. 027A stock has 2% daily volatility and no drift. What is the probability that it touches a level 10% below today's price at some point in the next 20 trading days, and how does that compare with the probability of ending below that level?Compounding and drawdownsHardBank market riskQuant risk

    Try it first

    Compared with the chance of finishing below the level, the chance of touching it at some point is:

    Show the worked solution

    About 24% to touch, against about 12% to finish below: roughly double. Twenty days at 2% a day is 8.94% of volatility. A 10% fall is ln 0.9, or 10.5% in log terms, which is 1.18 standard deviations, so the chance of ending below is 11.9%. The reflection principle doubles it for a touch: 23.9%.

    Why is touching twice as likely as finishing below?

    Picture a drunk walker on a pavement with a kerb one step to the left. By the end of the evening he may be well to the right, yet at some point he almost certainly stepped off the kerb. Paths wander. Once a driftless path has touched the level, it is equally likely to finish above it or below it, so for every path that finishes below there is a twin that touched and came back. This is the reflection principleFor a random walk with no drift, the path after it first hits a level is equally likely to be its mirror image about that level.: flip the path after the first touch and you swap an above-finisher for a below-finisher.

    Every path that ends below the line has a twin that touched it and came backtoday-10%stop level05101520Trading daynever touchestouch, day 9ends -1.8%mirror ends -19.2%Ends below the line11.9%Touches at any time23.9%twice the first
    A path that touches the 10% stop level on day 9 and recovers has a mirror twin, drawn dashed, that finishes below the line; because the two are equally likely, the chance of touching is twice the chance of finishing below, 23.9% against 11.9%.

    How do you get the numbers in the room?

    Scale volatility by the square root of time: 2% times the square root of 20 is 8.94%. Work in log returns, because a 10% fall is a log move of 10.5%, not 10%. That is 1.18 standard deviations, and the normal table gives 11.9% below it. Double for the touch. If you skip the log step and use 10% flat you get 26.4%, close enough to earn credit if you say it is an approximation.

    The relationship
    P(touch)=2 Φ ⁣(ln⁡0.90.0220)=2 Φ(−1.18)≈2×0.119=0.239P(\text{touch}) = 2\,\Phi\!\left(\frac{\ln 0.9}{0.02\sqrt{20}}\right) = 2\,\Phi(-1.18) \approx 2 \times 0.119 = 0.239
    \Phithe standard normal cumulative probability
    0.02\sqrt{20}20-day volatility, 8.94%
    \ln 0.9the log return of a 10% fall, minus 0.105
    What it says in wordsFind the chance of ending beyond the level, then double it because every finisher beyond has a twin that touched and came back.

    Where does the doubling rule stop being exact?

    Two places. With drift the mirror twins are no longer equally likely, and a positive drift pulls the touch probability below double. And if the stop only triggers on daily closes, a path can dip through during the day and recover by the close, so fewer touches count; a standard correction for daily monitoring gives about 19%. The practical reading is that a stop-loss level is hit far more often than a finishing-price distribution suggests, which is why stops set from end-of-period VaR fire more than the desk expects.

    Where candidates lose it

    Most candidates compute the finishing probability, 11.9%, and offer that as the answer to the touching question. The interviewer asked about touching precisely to see whether you know the two differ, and by how much.

    The other loss is knowing the doubling rule but not why. Draw the mirror path out loud; it takes ten seconds and proves you are reasoning, not reciting.

    What the interviewer asks next

    • What is the probability of touching a level 10% above instead?
    • How does a positive drift change the answer?
    • A barrier option knocks out at this level. Why is it cheaper than the vanilla option by more than the finishing probability suggests?
  2. 065You can bet repeatedly at even odds with a 60% chance of winning each bet. Compare the long-run growth of your capital if you bet 20% of it each time with betting 50% each time.Compounding and drawdownsHardQuant riskAsset manager risk

    Try it first

    Betting 50% a time on a 60% edge: what happens over many bets?

    Show the worked solution

    Betting 20% grows capital about 2.0% a bet; betting 50% shrinks it about 3.3% a bet. Growth per bet is the average log return: 0.6 ln(1 + f) + 0.4 ln(1 - f). At 20%, the Kelly fraction 2p - 1, that is +2.01%. At 50% it is -3.40%. After 100 bets the typical 20% bettor has about 7.5 times the start; the 50% bettor about 0.03 times.

    How can a bet with a positive edge lose money?

    A shopkeeper who stakes half his stock on every festival season, and wins more seasons than he loses, can still go under: one bad season halves him, and a good one only adds half back. Losses on a shrunken base are expensive to recover. The typical outcome of repeated betting is set by the average log return, not the average return, and at 50% the losses dominate the logs.

    Bet too much on a winning edge and you shrink-8%-4%0%+4%0%10%20%30%40%50%Fraction of capital bet each time20%: +2.0% a bet (Kelly)50%: -3.4% a betzero near 39%shrinking zone
    With a 60% chance of winning at even odds, growth per bet peaks at the 20% Kelly fraction, about +2.0% a bet, crosses zero near 39% and is -3.4% a bet at 50%, so oversizing a winning bet turns steady growth into steady decline.

    Where does the 20% come from?

    Set the slope of the growth curve to zero. For even odds the Kelly fractionThe fraction of capital to bet each time that maximises the long-run compound growth rate of the capital. is the probability of winning minus the probability of losing: 0.6 minus 0.4, or 20%. Betting less than Kelly gives up some growth for a much smoother ride; betting more than about twice Kelly gives up growth altogether. Here the curve crosses zero near 39%, just under twice Kelly.

    The relationship
    g(f)=pln⁡(1+f)+(1−p)ln⁡(1−f)f∗=2p−1=0.2g(f) = p\ln(1+f) + (1-p)\ln(1-f) \qquad f^* = 2p - 1 = 0.2
    fthe fraction of capital bet each time
    pthe probability of winning a bet, 0.6
    g(f)the expected log growth of capital per bet
    What it says in wordsLong-run growth is the probability-weighted average of the log of what each outcome does to your capital.

    For a risk manager the lesson is position sizing. A desk with a genuine edge can still destroy capital by running too much leverage, and the average P&L will look fine right up until the drawdowns compound. Real edges are also uncertain, which is why practitioners bet a fraction of Kelly.

    Where candidates lose it

    The trap is answering that 50% grows faster because each bet has a positive expected value and bigger bets earn more of it. That is true of the arithmetic average and false for the capital you actually end up with, which compounds.

    The second slip is knowing Kelly but not being able to show why 50% loses. Have the one-line check ready: six wins and four losses at 50% multiply capital by about 0.71.

    What the interviewer asks next

    • What is the Kelly fraction if you win 55% of the time at even odds?
    • What if the payoff is 2 to 1 with a 40% chance of winning?
    • Why do traders often bet half Kelly rather than full Kelly?