Risk Management puzzles, solved step by step

Puzzles: 100
Traced to a firm: 17
Topics: 13
Hard: 30

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011Two loans each have a 5% one-year probability of default and a 2% chance of defaulting together. What is the probability that at least one defaults, and how does it compare with the 9.75% you would get if they were independent?Probability and base ratesCoreQuant riskBank credit risk
Try it first
What is the probability that at least one of the two loans defaults?
Show the worked solution
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.
Out of 400 equally likely outcomes, correlated loans put 8 cells in both-default and 32 cells in any-default, 8%, while independent loans put only 1 cell in both-default and spread defaults over 39 cells, 9.75%.
The relationship
$P(A \cup B) = P(A) + P(B) - P(A \cap B) = 5\% + 5\% - 2\% = 8\%$
P(A), P(B) each loan's probability of default, 5%
P(A and B) the chance both default together, 2%
What it says in wordsAdd the chances of each loan defaulting, then take away the overlap you counted twice.
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.
Where candidates lose it
The first trap is answering 10%, adding the two 5% figures and forgetting the overlap. The second is using the independent formula, 9.75%, when the question has handed you a joint probability that is not 0.25%.
The quieter miss is stopping at 8% and not saying what it means. The interviewer wants to hear that correlation shifts risk from single defaults to joint defaults.
What the interviewer asks next
What is the probability that exactly one loan defaults?
What joint default probability would make the two loans perfectly correlated?
With 100 such loans, how does correlation change the distribution of the number of defaults?
022An institutional investor asks for an 8% expected annual return with 10% volatility. Assuming returns are normal, what is the chance of a losing year, and what volatility would keep that chance below 10%?Probability and base ratesCoreMSCIAnonymous interview candidate in · 2013
Try it first
Roughly how often does this portfolio lose money in a year?
Show the worked solution
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.
With an 8% expected return and 10% volatility, 21.2% of the return distribution falls below zero, while cutting volatility to 6.2% narrows the curve until exactly 10% of years show a loss.
The relationship
$P(R < 0) = N\!\left(-\frac{\mu}{\sigma}\right) = N(-0.8) \approx 21.2\% \qquad \sigma_{\max} = \frac{\mu}{1.2816} \approx 6.2\%$
mu the expected annual return, 8%
sigma the annual volatility
N the standard normal cumulative distribution
1.2816 the number of standard deviations that leaves 10% in the lower tail
What it says in wordsDivide the expected return by the volatility, and the normal table tells you how often returns fall below zero.
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.
Where candidates lose it
The trap is assuming that a positive expected return makes losing years rare. At 0.8 standard deviations above zero, they happen about one year in five.
The second miss is solving for volatility with the wrong number from the normal table. For a 10% tail you need 1.28 standard deviations, not 1.645, which is the 5% tail.
What the interviewer asks next
What is the chance of a negative return over five years with the same targets, assuming independent years?
The investor adds a limit of no more than a 15% loss in any year. What volatility does that imply at 99% confidence?
Why might a pension fund care more about a drawdown limit than a volatility target?
Asked at MSCI, Risk Management, Anonymous interview candidate in, 2013 (Wall Street Oasis): What risk-return targets would you set for an institutional investor?
036A bond portfolio holds 50 names, each with a 2% one-year default probability, and defaults are independent. What is the expected number of defaults in a year, and what is the chance of four or more?Probability and base ratesCoreBank credit risk
Try it first
Expected defaults are 1. Roughly how likely is a year with four or more?
Show the worked solution
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.
Across 50 independent names at 2% each, one default is the most likely outcome at 37.2%, yet four or more defaults still happen 1.78% of the time, a tail that turns a Rs 6 crore expected loss into a Rs 24 crore bad year.
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.
The relationship
$P(X \ge 4) = 1 - \sum_{k=0}^{3} \binom{50}{k}(0.02)^k(0.98)^{50-k} \approx 1 - 0.982 = 0.018$
X the number of defaults in the year
\binom{50}{k} the number of ways to pick which k names default
What it says in wordsFour or more is one minus the chance of zero, one, two or three.
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.
Where candidates lose it
Candidates give the expected count and stop, or say four defaults is basically impossible because the average is one. The interviewer asked for the tail precisely because averages do not size capital.
The second trap is overconfidence in the 1.8%. Offer the independence caveat before being asked; it shows you know which assumption the answer is most sensitive to.
What the interviewer asks next
If defaults are correlated, what happens to the chance of zero defaults?
How many names would you need for the chance of four or more to exceed 10%?
Each name has a different default probability. How does that change your method?
061A loan book is 60% retail loans with a 3% default rate and 40% corporate loans with a 1% default rate. A loan picked at random has defaulted. What is the probability it was a retail loan?Probability and base ratesCoreBank credit riskRisk GCC
Try it first
Which number is closest?
Show the worked solution
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.
Retail loans are 60% of the book and default 3% of the time, contributing 1.8% of all loans as defaults; corporate loans contribute 0.4%. Of the 2.2% that default, 1.8 is retail, so a defaulted loan is retail with probability 81.8%.
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.
The relationship
$P(R\mid D) = \frac{P(D\mid R)P(R)}{P(D\mid R)P(R)+P(D\mid C)P(C)} = \frac{0.03 \times 0.6}{0.018+0.004} = \frac{9}{11}$
P(R) the retail share of the book, 60%
P(D|R) the retail default rate, 3%
P(D|C) the corporate default rate, 1%
What it says in wordsThe chance a default came from retail is retail's share of all defaults.
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.
Where candidates lose it
The trap is answering 60%, the share of the book, and ignoring the fact that the loan defaulted. That is base rate thinking run backwards: the prior is treated as the answer.
The other slip is answering 75% by comparing the default rates alone, 3 against 1. That forgets the book is not split evenly. Weight both: 1.8 against 0.4.
What the interviewer asks next
If the book were 50:50, what would the answer be?
Retail LGD is 60% and corporate LGD is 30%. What share of the loss comes from retail?
Two loans default. What is the chance both are retail?
086Ten fund managers each have a 50% chance of beating the index in any year, with no skill at all. What is the probability that at least one of them beats it five years running?Probability and base ratesCoreAsset manager risk
Try it first
Roughly how likely is it that at least one of ten coin-flipping managers has a five-year streak?
Show the worked solution
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.
Among ten coin-flipping managers, one happens to beat the index all five years; the chance for any single manager is only 1 in 32, but at least one of ten does it 27.2% of the time and at least one of 100 does it 95.8% of the time.
The relationship
$P(\text{at least one}) = 1 - \left(1 - \tfrac{1}{32}\right)^{10} = 1 - 0.728 = 0.272$
1/32 one manager beating the index five years in a row by chance
10 the number of managers
What it says in wordsWork out the chance that nobody has a streak, then take it away from one.
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.
Where candidates lose it
The fast wrong answer is 3.1%, the chance for one named manager. The question asks about any of ten, and the interviewer is testing whether you notice the difference.
The other miss is adding ten times 3.1% and calling it 31%. That overcounts the cases with two streaks; give it as a bound and then the exact figure.
What the interviewer asks next
How many managers would you need before a ten-year streak by luck is more likely than not?
A fund's marketing shows five straight years of beating the index. What questions do you ask?
How does survivorship bias change these numbers?

P(A), P(B)	each loan's probability of default, 5%
P(A and B)	the chance both default together, 2%

mu	the expected annual return, 8%
sigma	the annual volatility
N	the standard normal cumulative distribution
1.2816	the number of standard deviations that leaves 10% in the lower tail

X	the number of defaults in the year
\binom{50}{k}	the number of ways to pick which k names default

P(R)	the retail share of the book, 60%
P(D\|R)	the retail default rate, 3%
P(D\|C)	the corporate default rate, 1%

Try it first

Why subtract the joint probability?

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

Where candidates lose it

What the interviewer asks next

Try it first

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

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

Where candidates lose it

What the interviewer asks next

Try it first

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

How do you get 1.8% without a calculator?

Where candidates lose it

What the interviewer asks next

Try it first

Why is the answer not simply 60%?

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

Where candidates lose it

What the interviewer asks next

Try it first

Why is a streak among many managers weak evidence?

How do you get there fast without a calculator?

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

Where candidates lose it

What the interviewer asks next