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. 005A rating grade shows a 2% cumulative probability of default after one year and 5% after two years. What is the probability of default in year two for a borrower that survived year one?Credit risk arithmeticCoreBank credit riskQuant risk

    Try it first

    What is the year-two default probability for a survivor?

    Show the worked solution

    About 3.06%. Start with 10,000 borrowers. 200 default in year one, leaving 9,800. By the end of year two 500 have defaulted, so 300 did so in year two. For a borrower who reached the start of year two, the chance is 300 out of 9,800, which is 3.06%, a little above the 3% you get by subtracting.

    Why divide by the survivors?

    Think of a school where 2 of every 100 students leave in class nine and 5 in total have left by the end of class ten. If you are a class ten student today, your chance of leaving this year is measured against the 98 who are still in the room, not the 100 who started. A conditional probability of default is always measured against the borrowers who survived to the start of the period. The unconditional slice, 3% of the original pool, is the right number only if you are standing at the start of year one.

    Follow the survivors: the year-two rate is counted on who is still thereStart10,000 aliveEnd of year 19,800 alive-200 defaultEnd of year 29,500 alive-300 defaultcumulative 2%cumulative 5%Wrong base: 300 / 10,000= 3.00%Survivors only: 300 / 9,800= 3.06%
    Of 10,000 borrowers, 200 default in year one and 300 in year two, so a borrower who survives year one faces 300 defaults out of 9,800 survivors, 3.06%, not 300 out of 10,000.
    The relationship
    PD2∣1=C2−C11−C1=0.05−0.020.98≈3.06%PD_{2|1} = \frac{C_2 - C_1}{1 - C_1} = \frac{0.05 - 0.02}{0.98} \approx 3.06\%
    C_1, C_2cumulative default probabilities at one and two years
    1 - C_1the share still alive at the start of year two
    What it says in wordsTake the extra defaults in year two and divide by the share of borrowers still alive to default.

    Where does this matter on a credit desk?

    Whenever you price or provision a loan over several years. Expected loss in year two uses the marginal default of the original pool, while a hazard rateThe probability of default in a short period for a borrower that has survived to the start of it. used to model a surviving borrower uses the conditional figure. Mixing them up is a small error at 2% and 5%, 3.06% against 3.00%, but at high-yield default rates the gap widens: 20% and 35% cumulative gives 18.75% conditional against 15% by subtraction.

    Say the limitation too. Cumulative default tables are averages across many cohorts and economic cycles, so a borrower in a downturn year may face a higher rate than the table shows. The arithmetic is exact; the inputs are estimates.

    Where candidates lose it

    The trap is answering 3% by subtracting. It feels complete because the numbers are clean, but it answers a different question: what share of the original pool defaults in year two, not what a surviving borrower faces.

    Give 3.06%, then say why it differs from 3%. The interviewer is listening for the word survivors.

    What the interviewer asks next

    • If the year-two conditional default rate is the same as year one's 2%, what is the two-year cumulative rate?
    • Convert the 2% one-year figure into a constant hazard rate.
    • Why do cumulative default curves for high-yield grades often flatten in later years?
  2. 030A distressed one-year bond costs 80. It pays 100 at maturity with probability 85%, and if the issuer defaults, holders recover 40. What is the expected return, and at what default probability does the trade break even?Credit risk arithmeticCoreBank credit riskAsset manager risk

    Try it first

    Before you calculate: at what default probability does paying 80 stop making sense?

    Show the worked solution

    The expected return is 13.75%, and the trade breaks even at a default probability of 33.3%. The expected payoff is 0.85 x 100 plus 0.15 x 40, which is 91, against a price of 80. Breakeven solves 80 = 100 minus 60p: p is 20 over 60, one third. The real question is whether default odds could be that high.

    Why is the breakeven more useful than the expected return?

    The 15% is somebody's estimate, and estimates of default for a stressed issuer are soft. Expected return inherits every error in the 15%; the breakeven tells you how wrong the estimate can be before you lose money. It is like buying a second-hand car that needs repairs: rather than guess the repair bill, you ask how large the bill could be before the deal stops being worth it. Here default odds could more than double, from 15% to one in three, before the trade loses on average.

    The trade pays unless default odds exceed one in threePay 80todayno default, 85%default, 15%Paid 100+25.0% on 80Recover 40-50.0% on 800.85 x 100 + 0.15 x 40Expected 9113.75% over 80if 15% is right0%10%20%30%40%50%your 15%breakeven 33.3%default probability
    Paying 80 for a bond that returns 100 with 85% probability and 40 on default gives an expected payoff of 91, a 13.75% expected return, and the trade only loses on average if the default probability exceeds 33.3%.
    The relationship
    80=(1−p) 100+p 40  ⇒  p∗=100−80100−40=1380 = (1-p)\,100 + p\,40 \;\Rightarrow\; p^{*} = \frac{100 - 80}{100 - 40} = \frac{1}{3}
    pone-year default probability
    100 - 80the gain if the bond pays at par
    100 - 40the gap between par and recovery
    What it says in wordsBreakeven default probability is the discount to par divided by the loss given default measured from par.

    What would a credit risk manager add before approving the trade?

    Three things. First, the recovery is also a guess, and it moves the breakeven: if holders recover 20 rather than 40, breakeven default odds fall from one third to 25%, because each default now costs 60 from the purchase price instead of 40. Second, money has a time value. If you require an assumed 7% return from cash for the year, breakeven falls from 33.3% to 24.0%, because the bond must beat cash, not zero. Third, the payoff is lopsided, +25% or -50%, so position size matters more than the average: a book of such bonds survives, a single large position may not.

    Where candidates lose it

    Candidates compute 13.75% and stop, as if the 15% were a fact. The interviewer's follow-up is always what if the probability is wrong, and the candidate who has the one-third breakeven ready answers it before it is asked.

    The second trap is using the 20-point discount to par as the breakeven default rate. The loss on default is 40 from where you bought, not 20, and mixing the two gives nonsense.

    What the interviewer asks next

    • What default probability does the market price imply if investors demand a 7% return?
    • Recovery is uncertain, between 25 and 55. How does that change your view?
    • Why do distressed investors often care more about recovery analysis than default probability?
  3. 068A Rs 100 crore loan has a 2% probability of default and a 50% loss given default. What is the expected loss, and what is the standard deviation of the loss that capital has to cover?Credit risk arithmeticCoreBank credit risk

    Try it first

    How does the standard deviation of loss compare with the expected loss?

    Show the worked solution

    Expected loss is Rs 1 crore; the standard deviation of loss is Rs 7 crore. If the loan defaults, the bank loses 50% of Rs 100 crore, Rs 50 crore; otherwise it loses nothing. The average is 2% x 50 = Rs 1 crore, which the spread should cover. The standard deviation is 50 x the square root of 0.02 x 0.98, which is Rs 7 crore, the unexpected loss capital exists for.

    Why is the average loss the wrong thing to hold capital against?

    A shopkeeper insuring against a one-in-fifty chance of a Rs 5 lakh fire can budget Rs 10,000 a year for the premium, the average. But in the year the fire happens, the budget is useless; what saves the shop is savings. Expected loss is a cost of doing business and belongs in the price; unexpected lossThe spread of possible losses around the expected loss, usually measured by a standard deviation or a high percentile, which capital is held to absorb. is the surprise, and it is what capital is for.

    One loan, two outcomes: the average hides the size of the bad one01020304050Loss on the loan, Rs crore98%: no loss2%: lose 50Expected loss = 2% x 50 = 11 standard deviation = 7EL is priced into the spreadUL of 7 is what capital covers
    The loan either loses nothing, 98% of the time, or loses Rs 50 crore, 2% of the time. The expected loss is Rs 1 crore and the standard deviation is Rs 7 crore, so the surprise the bank must be able to absorb is seven times the average it prices in.

    How do you get Rs 7 crore quickly?

    Treat default as a coin that lands bad with probability p. Its standard deviation is the square root of p times (1 - p): the square root of 0.0196, which is 0.14. Multiply by the loss if it lands bad, Rs 50 crore, and you have Rs 7 crore. The ratio of seven to one is typical for a low-default loan, and it grows as the default probability falls: rarer defaults mean a mean closer to zero and a relatively wider spread.

    The relationship
    EL=PD×LGD×EAD=1UL=LGD×EAD×PD(1−PD)=50×0.14=7EL = PD \times LGD \times EAD = 1 \qquad UL = LGD \times EAD \times \sqrt{PD(1-PD)} = 50 \times 0.14 = 7
    PDprobability of default, 2%
    LGDloss given default, 50%
    EADexposure at default, Rs 100 crore
    What it says in wordsThe average loss is the default chance times the loss; the spread is the loss times the square root of p times one minus p.

    Say the limitation. For one loan, the standard deviation is a poor summary: the real outcome is 0 or 50, never 7. In a large, diversified portfolio losses spread out and the standard deviation becomes meaningful, and capital models set it at a high percentile of the portfolio loss rather than one standard deviation of a single loan.

    Where candidates lose it

    The trap is stopping at the expected loss of Rs 1 crore and treating it as the risk. It is the one number the bank should never be surprised by; the risk is the spread.

    The second slip is using the variance formula for a continuous variable, or forgetting the square root and quoting Rs 0.98 crore. Say the Bernoulli formula aloud, p times one minus p, then take the root.

    What the interviewer asks next

    • What happens to UL if the default probability falls to 0.5%?
    • The bank holds 100 such loans, independent of each other. What is the portfolio UL?
    • Why do defaults in a real portfolio not behave independently?