Risk Management puzzles, solved step by step

8%, lower than the 9.75% for independent loans. The chance of at least one default is 5% plus 5% minus the 2% where both default, which would otherwise be counted twice. Independent loans default together only 0.25% of the time, so they give 9.75%. Correlation makes any single default slightly less likely but a double default eight times more likely.

Why subtract the joint probability?

Think of a class where 5 students play cricket and 5 play football, and 2 play both. If you ask how many play at least one game, adding 5 and 5 counts the two all-rounders twice, so the answer is 8. The chance of at least one default is the sum of the single chances minus the chance of both, because the joint case sits inside each single case. Only when the joint case is tiny does adding the two chances get close.

Why does correlation cut the chance of any default but raise the chance of both?

Because the defaults are bunched into the same outcomes. When defaults tend to happen together, the bad outcomes overlap, so fewer outcomes contain a default at all, but the ones that do are worse. Independent loans default together only 5% of 5%, 0.25% of the time. Here it is 2%, eight times more often, and the implied default correlationThe correlation between two yes-or-no default outcomes, computed from the joint and single default probabilities. is about 0.37.

That is the lesson a credit portfolio manager takes from the puzzle. The expected number of defaults is 0.1 in both cases, because expectations add regardless of correlation. What correlation changes is the shape of losses: fewer mild years, more years where everything goes wrong at once. Capital is held for those years, which is why correlated books need more of it even when the expected loss is identical.

About 21%, and volatility would need to fall to about 6.2%. A loss means a return below zero, which is 8 points, or 0.8 standard deviations, under the mean. About 21.2% of a normal distribution lies below that, roughly one year in five. For a 10% chance, zero must sit 1.28 standard deviations below the mean, so volatility must be 8 divided by 1.28, about 6.2%.

How do a return target and a volatility target fix the chance of loss?

Think of a commute that takes 40 minutes on average but varies from day to day. Whether you are ever late for a 50 minute deadline depends on how much it varies, not only on the average. The chance of a losing year depends on how many standard deviations the expected return sits above zero: here 8 divided by 10, which is 0.8. Look up 0.8 in the normal table and about 21.2% of years fall below zero. The investor who hears 8% and thinks losses are rare is wrong one year in five.

What would you actually set as targets, and what is wrong with this model?

Set the targets as a pair, and state the trade-off. If the investor cannot tolerate losing more than one year in ten, then either volatility must come down to about 6.2%, which usually lowers the expected return too, or the loss tolerance must be stated over a longer horizon. Over five years the mean grows five times but the volatility only by the square root of five, so the chance of a losing five-year stretch is much lower. Asking about the horizon is the question a good risk manager raises first.

Then name the model's limits. Real returns have fatter left tails than a normal curve, so the chance of a large loss is understated; returns are not independent from year to year; and the 8% expected return is an assumption, not a promise. A drawdown limit, such as no more than a 15% fall from peak, is often more useful to an institution than a probability of a losing year.

One default expected, and about a 1.8% chance of four or more. The expected count is 50 x 2% = 1. The chance of none is 0.98 to the power 50, 36.4%; of exactly one, 37.2%; two, 18.6%; three, 6.1%. Those sum to 98.2%, so four or more is about 1.78%, roughly one year in 56.

Why is the expected count not enough to size the risk?

A school with 50 pupils, each with a 2% chance of being off sick on a given day, expects one absence. Most days it gets none, one or two; some days it gets four, and the class still has to run. An average of one default tells you what a normal year costs; it says nothing about how bad the bad year is, and capital exists for the bad year. Here, with Rs 10 crore in each name and a 60% loss on default, the expected loss is Rs 6 crore, but a four-default year costs Rs 24 crore.

How do you get 1.8% without a calculator?

Use the Poisson approximationFor many independent rare events, the count is close to a Poisson distribution with the same mean, so P(k) is about e to the minus mean times mean to the k over k factorial.. With a mean of 1, the chances of 0, 1, 2 and 3 are about 0.368, 0.368, 0.184 and 0.061. They add to about 0.981, so four or more is about 1.9%, within a whisker of the exact binomial 1.78%. Saying you are using the approximation, and why it works here, earns as much credit as the exact figure.

Then say the assumption that matters most. Independence is the weak link: names in the same sector or region default together in a downturn. With correlation, the expected count stays at one, but years with no defaults and years with many both become more common, and four or more can be several times more likely than 1.8%. That is why credit portfolio models spend most of their effort on correlation, not on the individual default probabilities.

About 69%. This is the gambler's ruin. With r = 0.48 / 0.52, starting 10 steps from each barrier, the chance of hitting the top first is 1 / (1 + r to the power 10). r to the 10 is about 0.449, so the answer is 1 / 1.449, or 69.0%. A 2-point daily edge becomes a 19-point edge because the race takes many days.

Why does a small daily edge grow so much?

A casino wins only a couple of points more than half its bets at roulette, yet over thousands of spins it almost never ends a month behind. An edge that barely shows in one step accumulates over many, because the random part grows with the square root of the number of steps while the edge grows in proportion to it. Reaching plus or minus 10 takes around 95 days on average here, long enough for the edge of 0.04 a day to push the walk decisively one way.

How do you check the answer, and what does it say about stop-losses?

Two sanity checks. At 0.50, r is 1 and the formula gives 50%, as symmetry demands. Widen both barriers to 20 and the answer at 0.52 rises to 83.2%. The wider the barriers relative to the daily step, the more the edge dominates the noise. That is the risk manager's reading: a tight stop-loss relative to daily volatility turns a trader with a genuine edge into a near coin flip, while a wide one lets the edge show but exposes more capital to a trader without one.

State the limits. The model assumes a constant edge, equal step sizes and independent days. Real P&L has fat tails, the edge decays, and a losing streak may itself be evidence that the edge was never there. The formula tells you how an edge would play out, not whether you have one; with drawdownThe fall from a running peak in cumulative profit, measured in rupees or as a percentage. limits, that second question is usually the harder one.

About 81.8%. Retail defaults make up 60% x 3% = 1.8% of the book and corporate defaults 40% x 1% = 0.4%, so 2.2% of all loans default. Given that a loan is among the 2.2%, the chance it is retail is 1.8 divided by 2.2, which is 9 in 11, or about 82%.

Why is the answer not simply 60%?

Suppose 60% of the cars on a road are small hatchbacks and 40% are trucks, but trucks are far more likely to have a broken tail light. If a policeman stops a car for a broken tail light, it is more likely to be a truck than the 40% share suggests. The stop tells you something. Learning that a loan defaulted is evidence, and it shifts the odds towards the segment that defaults more often.

What is the quickest way to do this in your head?

Use 1,000 loans. There are 600 retail loans, of which 18 default, and 400 corporate loans, of which 4 default. Out of 22 defaults, 18 are retail, so the answer is 18 over 22, and natural numbers make the base rate impossible to forget. This is Bayes' rule without the notation, and it is the fastest way to be right out loud.

For a risk team this is not an exam trick. The mix of a defaulted pool drives which collections team gets the work, which recovery rate applies and which model needs recalibrating. A 60% share of the book becomes an 82% share of the defaults.

10,100 trading days on average. Let E0 be the expected wait from no current run and E1 the wait after one exception. Then E0 = 1 + 0.99 E0 + 0.01 E1 and E1 = 1 + 0.99 E0. Solving gives E0 = (1 + p) / p squared = 1.01 / 0.0001 = 10,100, about 40 years of trading days if exceptions really are independent.

Why is the answer not simply one over p squared?

Waiting for two heads in a row with a fair coin takes six tosses on average, not four, because each time you get one head and then a tail, you are back to the start and the first head was wasted. A run has memory: after one exception you are one step from the finish, and a miss throws you back, so the waiting time needs one equation per state, not one probability. For VaR exceptions at 1% the extra is 100 days, the time spent waiting for each first exception.

Why would a risk manager care about a 40-year wait?

Because it turns an observation into a test. If the model is right and exceptions are independent, two in a row should take about forty years of trading to appear; seeing them twice in one year says the exceptions cluster, and clustering means the model misses changes in volatility. Backtesting frameworks test independence for exactly this reason, alongside the plain count of exceptions.

The limitation is the independence assumption, which is the thing being tested. Real exceptions tend to arrive in bursts when volatility jumps, so the observed wait is usually far shorter, and that gap is evidence against the model rather than bad luck.

About 27.2%. One manager beats the index five years running with probability one half to the fifth, 1 in 32. The chance that none of the ten does it is 31/32 to the power 10, about 0.728. So at least one streak appears 27.2% of the time with no skill anywhere. With 100 managers it is 95.8%.

Why is a streak among many managers weak evidence?

Ask a hall of 300 people to toss a coin five times, and about nine will throw five heads. Nobody would call them skilled tossers. The chance that one named person has a streak is small, but the chance that someone in a crowd has one is large, and a fund manager with a five-year record is usually picked from a crowd. Which question you are answering decides everything: the probability for manager G, or the probability for whoever turned out to have the streak.

How do you get there fast without a calculator?

Use the expected count first: ten managers times 1 in 32 is about 0.31 streaks. When the expected count is small, the chance of at least one is a little below it, so 31% is an upper bound and the exact answer is about 27%. The gap comes from the chance of two or more streaks, which the simple sum counts twice. Saying the bound and then the exact figure shows the interviewer you can check your own work.

What does an asset manager's risk team do with this?

It sets the bar for evidence. A five-year record of beating the index is common among hundreds of funds even if none has skill, so the team looks at how the returns were earned: the size of the edge against its volatility, whether it comes from one bet or many, and whether the process explains the outcome. Survivorship matters too: the managers whose streaks broke often closed their funds, so the crowd you see is already filtered toward winners.

About Rs 24.19. A game that ends on flip n pays 2 to the n with probability one half to the n, so each flip count adds exactly Rs 1. Uncapped, that sum is infinite. The client can pay only Rs 1 crore, which 2 to the 23 stays below, so 23 flip counts add Rs 23 and every longer game pays Rs 1 crore, adding about Rs 1.19.

Why is the uncapped game worth an infinite amount?

Picture a prize that doubles every time you survive another round, while the chance of surviving halves. Each round's prize times its chance is always the same Rs 1. In this game every possible length contributes exactly Rs 1 to the expected value, one Rs 1 for each flip count, forever, so the sum never stops. This is the St Petersburg paradox: the arithmetic says pay anything, yet nobody would pay even Rs 100.

How does the cap turn infinity into Rs 24?

Find where the cap bites: 2 to the 23 is Rs 83,88,608 and 2 to the 24 is over Rs 1 crore, so the first 23 flip counts pay in full. Those contribute Rs 23; every longer game pays the capped Rs 1 crore, and the chance of lasting past 23 flips is one half to the 23, so the tail adds Rs 1 crore divided by 2 to the 23, about Rs 1.19. The fair price is about Rs 24.19.

Why is this a counterparty risk question?

Because the value came entirely from payouts the client could never make. A promised payoff is worth only what the payer can actually pay, and most of this game's theoretical value sat in states where the payer would default. Raising the cap a thousandfold to Rs 1,000 crore only lifts the value to about Rs 34.16, because each doubling of capacity adds just one rupee. The same logic prices protection bought from a seller who could not survive the event it insures.