Finance-Theory · Unit 8 · Video 5 · Interactive Practice

When Defaults Move Together: How Correlation Undid the AAA Tranche

IKey Formulas

FormulaNameWhat it needs
P(both default)=p2+ρ p(1−p)=0.01+0.09ρP(\text{both default}) = p^2 + \rho\,p(1-p) = 0.01 + 0.09\rhoJoint default with correlation ρ\rhoEach bond defaults with p=0.10p = 0.10; ρ\rho is the correlation between the two default indicators, 0≤ρ≤10 \le \rho \le 1. The video compares only the endpoints ρ=0\rho = 0 (independent) and ρ=1\rho = 1 (perfectly correlated); the formula in between is our model. Then P(exactly one)=0.18(1−ρ)P(\text{exactly one}) = 0.18(1-\rho) and P(both pay)=0.81+0.09ρP(\text{both pay}) = 0.81 + 0.09\rho
Senior=1,000×(1−P(both default))\text{Senior} = 1{,}000 \times \big(1 - P(\text{both default})\big)Senior tranche price (r=0r = 0)Paid in full unless the pool is empty: $990 when ρ=0\rho = 0, $900 when ρ=1\rho = 1
Junior=1,000×P(both pay)\text{Junior} = 1{,}000 \times P(\text{both pay})Junior tranche price (r=0r = 0)Paid only when neither bond defaults: $810 when ρ=0\rho = 0, $900 when ρ=1\rho = 1
Senior+Junior=2×900=1,800\text{Senior} + \text{Junior} = 2 \times 900 = 1{,}800The pool is conservedCorrelation changes how the two bonds' value is split, never the total: 990+810=900+900990 + 810 = 900 + 900

Key Insight: Independence was the one input the tranching example took for granted. Under perfect correlation the AAA senior tranche defaults ten times as often (10% instead of 1%) and the "toxic waste" junior becomes exactly as risky as one original bond, so $90 of value moves from senior to junior. A bank that went long the senior tranches and short the junior ones at the independent-defaults prices loses on both legs when correlation rises: $90 on each in the example.

IITurning the Correlation Dial

Raising the correlation moves probability into the both-default corner and shifts value from the senior tranche to the junior.

IIILong Senior, Short Junior

A position's P&L from rising correlation is proportional to junior units minus senior units, so long senior, short junior loses on both legs.

IVQuiz Questions

Problem 1 · The Senior Tranche Under Perfect Correlation

Given: Two bonds each pay $1,000 or nothing, and each defaults with probability 20%. Their defaults are perfectly correlated and the risk-free rate is zero. As in the video, the senior tranche is promised the first $1,000 of the pool's payoff and the junior tranche the second $1,000, and each tranche is priced at its expected payoff. Find the price of the senior tranche.

✅ Correct! The bonds pay together with probability 80% and default together with probability 20%, so the senior is worth 0.80×1,000=8000.80 \times 1{,}000 = 800 dollars, and the junior is worth exactly the same.
❌ That assumes independence. With independent defaults the pool is empty with probability 0.22=0.040.2^2 = 0.04, giving 0.96×1,000=9600.96 \times 1{,}000 = 960. Perfectly correlated bonds default together, so the pool is empty 20% of the time.
❌ That is the independent junior price. 0.82×1,000=6400.8^2 \times 1{,}000 = 640 is what the junior tranche is worth when defaults are independent. The question asks for the senior tranche under perfect correlation.
❌ That is the video's number. $900 goes with a 10% default probability. Here each bond defaults 20% of the time, so both pay only 80% of the time.
Show solution

With perfect correlation the bonds never split: either both pay (probability 80%, pool $2,000) or both default (probability 20%, pool $0). The senior tranche receives $1,000 in the first case and nothing in the second. With r=0r = 0 the price is the expected payoff:

Senior=0.80×1,000+0.20×0=800\text{Senior} = 0.80 \times 1{,}000 + 0.20 \times 0 = 800

so $800. The junior tranche has the same payoffs with the same probabilities and is also worth $800, the value of one original bond. With independent defaults the prices would have been $960 and $640.

Problem 2 · Where the Value Goes

Given: In the video's example (two $1,000 bonds, each defaulting 10% of the time, r=0r = 0), defaults change from independent to perfectly correlated. What happens to the combined value of the senior and junior tranches?

✅ Correct! 990+810=900+900=1,800990 + 810 = 900 + 900 = 1{,}800. The two tranches split the same two bonds, each still worth $900; correlation changes only the split.
❌ Not quite. A loss of $180 is what the risk manager's trade, long one senior and short one junior, loses in the example. The junior tranche itself rises in value, from $810 to $900.
❌ The junior moves too. It is paid whenever both bonds pay, which now happens 90% of the time instead of 81%, so its price rises from $810 to $900.
❌ Not quite. Each bond still defaults 10% of the time; correlation changes how the defaults line up, not how often each bond defaults. So the pool is still worth 2×0.9×1,000=1,8002 \times 0.9 \times 1{,}000 = 1{,}800 dollars.
Show solution

Independent: senior =0.99×1,000=990= 0.99 \times 1{,}000 = 990, junior =0.81×1,000=810= 0.81 \times 1{,}000 = 810.

Perfectly correlated: senior == junior =0.90×1,000=900= 0.90 \times 1{,}000 = 900.

990+810=1,800=900+900990 + 810 = 1{,}800 = 900 + 900

The total equals the two original bonds, each worth 0.9×1,000=9000.9 \times 1{,}000 = 900 dollars whatever the correlation. The senior loses $90 and the junior gains $90: value is transferred, not created or destroyed.

Problem 3 · Partly Correlated Defaults

Given: Two $1,000 bonds each still default 10% of the time, but their defaults are partly correlated: both default together with probability 4%. The risk-free rate is zero. Find the prices of the senior and junior tranches.

Price of the senior tranche?

Price of the junior tranche?

✅ Correct! Senior =0.96×1,000=960= 0.96 \times 1{,}000 = 960 and junior =0.84×1,000=840= 0.84 \times 1{,}000 = 840 (dollars), and 960+840=1,800960 + 840 = 1{,}800 as always.
❌ That uses independence. $990 assumes both default only 1% of the time. Here the pool is empty with probability 4%.
❌ That is the junior's price. The senior is also paid in full when exactly one bond defaults; it misses only when both default.
❌ That is the perfectly correlated price. $900 goes with both bonds defaulting together 10% of the time; 4% lies between the independent 1% and the perfectly correlated 10%.
❌ That is the senior's price. The junior is paid only when neither bond defaults.
❌ One single-default outcome is missing. A defaults alone with probability 0.10−0.04=0.060.10 - 0.04 = 0.06, and B defaults alone with probability 0.060.06 too, so exactly one default has probability 0.120.12, not 0.060.06.
❌ That uses independence. When defaults tend to coincide, the bonds also pay together more often: 84% of the time here, not 81%.
Show solution

Each bond defaults 10% of the time, and 4% of the time it does so together with the other, so each defaults alone with probability 0.10−0.04=0.060.10 - 0.04 = 0.06:

P(exactly one)=0.06+0.06=0.12,P(both pay)=1−0.04−0.12=0.84P(\text{exactly one}) = 0.06 + 0.06 = 0.12, \qquad P(\text{both pay}) = 1 - 0.04 - 0.12 = 0.84

The senior is paid unless both default; the junior only if both pay:

Senior=0.96×1,000=960,Junior=0.84×1,000=840\text{Senior} = 0.96 \times 1{,}000 = 960, \qquad \text{Junior} = 0.84 \times 1{,}000 = 840

So $960 and $840, which add to $1,800. In the model of the key formulas this is 0.01+0.09ρ=0.040.01 + 0.09\rho = 0.04, a correlation of ρ=13\rho = \tfrac{1}{3}.

Problem 4 · A Bigger Version of the Trade

Given: A desk buys 2 senior tranches and sells 1 junior tranche short, all at the independent-defaults prices of the video's example. Defaults then turn out to be perfectly correlated, and the positions are marked to market. Find the desk's total profit or loss.

✅ Correct! 2×(900−990)+(810−900)=−180−90=−2702 \times (900 - 990) + (810 - 900) = -180 - 90 = -270: both legs lose.
❌ Check the size. $180 is the loss on the video's one-for-one trade. Here the senior leg is two tranches, and it alone loses 2×(900−990)=−1802 \times (900 - 990) = -180.
❌ The short is not a hedge here. Treating it as one gives −180+90-180 + 90. But the junior price rises from 810 to 900, so buying it back costs more than the sale brought in: 810−900=−90810 - 900 = -90.
❌ Not quite. The tranches' total is unchanged only for someone holding equal amounts of both. This desk is on the losing side of the senior-to-junior transfer on both legs.
Show solution

Senior leg: bought 2 at $990, now worth $900 each:

2×(900−990)=−1802 \times (900 - 990) = -180

Junior leg: sold 1 short at $810; closing it means buying back at $900:

810−900=−90810 - 900 = -90

Total: −180+(−90)=−270-180 + (-90) = -270, a loss of $270. In general the position's value changes by 90ρ×(junior units−senior units)=90×1×(−1−2)=−27090\rho \times (\text{junior units} - \text{senior units}) = 90 \times 1 \times (-1 - 2) = -270.

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