Credit Loss Distributions and Tail Outcomes

Why the Average Does Not Describe the Whole Credit Loss

Expected credit loss is a useful summary of the average loss over many repetitions of a specified model. It does not say how losses are distributed around that average, how often losses exceed a threshold, or how severe losses can be in the tail. For risk management, those distinctions matter: a low average can coexist with a small probability of a very large loss.

This tutorial treats loss as a random variable over a fixed horizon, such as one year. It focuses on interpreting the distribution and its tail outcomes. The calculations below use hypothetical scenario distributions; they are not observations from a real insurer, lender, or portfolio.

Definition: A credit loss distribution specifies the possible losses $L$ over a stated horizon and the probability of each outcome. In the discrete case, write the outcomes as $l_1,\ldots,l_n$, with probabilities $p_1,\ldots,p_n$, where $p_i\geq 0$ and $\sum_{i=1}^{n}p_i=1$. The cumulative distribution function is $F_L(x)=P(L\leq x)$.

Before calculating a metric, check that the horizon, loss amounts, and scenario probabilities are clear. The scenarios should be mutually exclusive and collectively exhaustive. In the examples, losses are measured in dollars at the same valuation date, so no discounting is applied. Losses are nonnegative. No assumption of a normal distribution is made.

Mean and Dispersion of a Discrete Loss Distribution

The expected loss is the probability-weighted average of the outcomes. It is a long-run average under the stated model, not a prediction that the realized loss will equal that amount. Variance measures squared dispersion around the mean; standard deviation expresses dispersion in the same units as loss.

E[L]=\sum_{i=1}^{n}p_i l_i,\qquad \operatorname{Var}(L)=\sum_{i=1}^{n}p_i\bigl(l_i-E[L]\bigr)^2 =E[L^2]-\bigl(E[L]\bigr)^2,\qquad \operatorname{SD}(L)=\sqrt{\operatorname{Var}(L)}.

The second variance expression follows by expanding the squared deviation and using $\sum_i p_i=1$:

\begin{aligned} \operatorname{Var}(L) &=E\bigl[(L-E[L])^2\bigr]\\ &=E[L^2]-2E[L]E[L]+(E[L])^2\\ &=E[L^2]-\bigl(E[L]\bigr)^2. \end{aligned}

Variance is in squared dollars, so it is usually less intuitive to interpret directly than standard deviation. Standard deviation is still only a measure of overall dispersion: by itself, it does not identify the probability or size of a particular tail loss. Two distributions can have the same mean and different standard deviations, or similar standard deviations and different tail shapes.

Worked Example: Calculating the Mean and Dispersion

A hypothetical credit position has a one-year loss of $0 with probability 0.90, $20,000 with probability 0.07, and $100,000 with probability 0.03. Calculate expected loss, variance, and standard deviation. The loss amounts are at the same valuation date, and no interest adjustment is required.

Method and probability check. The outcomes are mutually exclusive, and their probabilities sum to $0.90+0.07+0.03=1.00$. Use the probability-weighted mean, then the second-moment variance formula.

Expected loss.

E[L]=(0)(0.90)+(20{,}000)(0.07)+(100{,}000)(0.03) =0+1{,}400+3{,}000=\$4{,}400.

Second moment. Square each loss before weighting it:

E[L^2]=(0^2)(0.90)+(20{,}000^2)(0.07)+(100{,}000^2)(0.03) =0+28{,}000{,}000+300{,}000{,}000 =328{,}000{,}000\ \text{dollars}^2.

Variance and standard deviation.

\begin{aligned} \operatorname{Var}(L) &=328{,}000{,}000-(4{,}400)^2\\ &=328{,}000{,}000-19{,}360{,}000\\ &=308{,}640{,}000\ \text{dollars}^2,\\ \operatorname{SD}(L)&=\sqrt{308{,}640{,}000}=\$17{,}568.15. \end{aligned}

The standard deviation is much larger than the mean because the $100,000 outcome, although assigned only 0.03 probability, is far from the average. The calculation checks: $E[L^2]-(E[L])^2$ is positive, and the standard deviation is in dollars, not dollars squared. The common mistake this method avoids is reporting the variance itself as though it were a dollar amount.

Percentiles and Tail Averages

A percentile describes a point in the loss distribution. In this tutorial, Value at Risk at confidence level $p$, written $\operatorname{VaR}_p(L)$, uses the lower-quantile convention. For a discrete distribution, this definition matters because the cumulative probability can jump over a percentile.

Definition: For $0<p<1$, $\operatorname{VaR}_p(L)=\inf\{x:F_L(x)\geq p\}$. It is the smallest loss amount for which the cumulative probability of a loss at or below that amount is at least $p$. Thus, $\operatorname{VaR}_{0.95}(L)$ is a 95th-percentile loss under this convention; it is not the average loss in the worst 5% of outcomes.

A tail average summarizes losses in the worst $1-p$ fraction of outcomes. For a distribution with discrete outcomes, the mass at the percentile may need to be split: only the portion needed to make up the worst $1-p$ probability is included. One standard definition is the integrated quantile, also called Tail Value at Risk in this setting:

\operatorname{TVaR}_p(L)=\frac{1}{1-p}\int_p^1 \operatorname{VaR}_q(L)\,dq.

This definition gives an average over exactly the worst $1-p$ probability of outcomes, including a fractional share of an outcome at the cutoff when necessary. When the loss distribution is continuous, it agrees with the conditional mean of losses above the $p$-quantile. For a discrete distribution, simply taking $E[L\mid L>\operatorname{VaR}_p(L)]$ can exclude losses equal to the cutoff and therefore fail to represent the worst $1-p$ fraction.

Worked Example: Reading a Percentile and Tail Average

Use the one-year loss distribution from the previous example. Find the 95th-percentile VaR and the corresponding TVaR. State the amount and probability mass used in the tail calculation.

Method. First accumulate probability from the smallest loss upward. Then average exactly the worst 5% of probability mass.

The cumulative probabilities are $F_L(0)=0.90$, $F_L(20{,}000)=0.97$, and $F_L(100{,}000)=1.00$. Since 0.90 is below 0.95 and 0.97 is at least 0.95:

\operatorname{VaR}_{0.95}(L)=\$20{,}000.

The worst 5% consists of all the 3% probability at $100,000 and 2% of the probability at $20,000. The tail loss amount, probability-weighted, is $0.03(\$100{,}000)+0.02(\$20{,}000)=\$3{,}400$. Divide by the tail probability, 0.05:

\operatorname{TVaR}_{0.95}(L) =\frac{0.03(100{,}000)+0.02(20{,}000)}{0.05} =\frac{3{,}000+400}{0.05} =\$68{,}000.

Check. The included probability is $0.03+0.02=0.05$, as required, and the tail average lies between the included outcomes, $20,000 and $100,000. A common wrong calculation is $E[L\mid L>20{,}000]=\$100{,}000$. That conditions on only the 3% outcome strictly above VaR and omits the 2% share at the cutoff; it is not the average over the worst 5%.

Equal Averages Can Hide Different Tails

Expected loss alone can make two credit risks appear equivalent even when their severe outcomes are not. The following hypothetical distributions have the same one-year expected loss but different probabilities and sizes of loss. No independence or aggregation assumption is needed; each is a complete distribution for one modeled loss amount.

Worked Example: Comparing Two Distributions With the Same Mean

Distribution A has a loss of $0 with probability 0.90 and $10,000 with probability 0.10. Distribution B has a loss of $0 with probability 0.99 and $100,000 with probability 0.01. For each distribution, calculate expected loss, standard deviation, 95th-percentile VaR, and 95% TVaR. Dollar amounts are at the same valuation date.

Expected losses.

E[L_A]=0(0.90)+10{,}000(0.10)=\$1{,}000,\qquad E[L_B]=0(0.99)+100{,}000(0.01)=\$1{,}000.

Dispersion. For A, $E[L_A^2]=10{,}000^2(0.10)=10{,}000{,}000$ dollars squared. Therefore, $\operatorname{Var}(L_A)=10{,}000{,}000-1{,}000^2=9{,}000{,}000$, and $\operatorname{SD}(L_A)=\$3{,}000$.

For B, $E[L_B^2]=100{,}000^2(0.01)=100{,}000{,}000$ dollars squared. Therefore, $\operatorname{Var}(L_B)=100{,}000{,}000-1{,}000^2=99{,}000{,}000$, and $\operatorname{SD}(L_B)=\$9{,}949.87$.

Percentiles and tail averages. A has cumulative probability 0.90 at $0, so $\operatorname{VaR}_{0.95}(L_A)=\$10{,}000$; the worst 5% is entirely within the 10% loss outcome, so $\operatorname{TVaR}_{0.95}(L_A)=\$10{,}000$. B has cumulative probability 0.99 at $0, so $\operatorname{VaR}_{0.95}(L_B)=\$0$. Its worst 5% includes the 1% at $100,000 and 4% at $0:

\operatorname{TVaR}_{0.95}(L_B) =\frac{0.01(100{,}000)+0.04(0)}{0.05} =\$20{,}000.

Interpretation and check. Both expected losses are $1,000, but B has a higher standard deviation and a higher 95% tail average. Its 95th-percentile VaR is lower because at least 95% of its outcomes are zero; this does not mean B has the less severe tail. The comparison shows why a percentile and a tail average answer different questions, and why neither should be inferred from expected loss alone. The common mistake is to rank the risks only by their equal means or by VaR alone.

Tail Concentration and Interpretation

A useful interpretation question is how much of the expected loss comes from a rare severe scenario. A scenario can contribute a large share of the mean without occurring often: its contribution is the loss amount multiplied by its probability. That contribution is distinct from the scenario’s contribution to variance, which depends on its squared deviation from the mean.

Worked Example: Identifying a Severe Scenario’s Contribution

A hypothetical one-year credit loss distribution has outcomes $0 with probability 0.92, $15,000 with probability 0.05, $60,000 with probability 0.02, and $300,000 with probability 0.01. Calculate expected loss, the 95th-percentile VaR, the 95% TVaR, and the share of expected loss contributed by the $300,000 scenario. Losses are measured at one valuation date with no discounting.

Expected loss.

E[L]=0(0.92)+15{,}000(0.05)+60{,}000(0.02)+300{,}000(0.01) =750+1{,}200+3{,}000=\$4{,}950.

VaR and TVaR. The cumulative probabilities are 0.92 at $0, 0.97 at $15,000, 0.99 at $60,000, and 1.00 at $300,000. Thus $\operatorname{VaR}_{0.95}(L)=\$15{,}000$. The worst 5% contains all 1% at $300,000, all 2% at $60,000, and 2% of the 5% probability at $15,000:

\operatorname{TVaR}_{0.95}(L) =\frac{0.01(300{,}000)+0.02(60{,}000)+0.02(15{,}000)}{0.05} =\frac{3{,}000+1{,}200+300}{0.05} =\$90{,}000.

Share from the largest scenario. Its contribution to expected loss is $0.01(300{,}000)=\$3{,}000$, so its share is $3{,}000/4{,}950=0.6061$, or 60.61% to two decimal places. The calculation checks because the four probability contributions to expected loss sum to $4,950. Here a 1% scenario contributes over 60% of the mean, illustrating why a modest average does not by itself describe the severity of possible outcomes.

Common Interpretation Traps

Treating expected loss as the likely realized loss. The mean is a probability-weighted average, not necessarily an outcome in the distribution.
Confusing VaR with a tail average. VaR is a percentile threshold. TVaR averages the worst specified fraction of probability.
Ignoring probability mass at a discrete cutoff. For TVaR, include only the fraction of mass at VaR needed to complete the tail probability. Strictly conditioning on losses above VaR may select too little probability.
Using standard deviation as a complete tail description. It measures dispersion around the mean but does not specify a particular exceedance probability or tail average.
Mixing units. Variance is in squared dollars; standard deviation, VaR, and TVaR are in dollars. Keep the horizon fixed when comparing distributions.

Key takeaway: Read a credit loss distribution through several distinct summaries: expected loss for the probability-weighted average, standard deviation for overall dispersion, VaR for a percentile threshold, and TVaR for the average severity of the worst specified probability fraction. These metrics describe different features; none can be substituted for all the others.

Check Your Understanding

For each problem, use the stated discrete distribution and lower-quantile VaR convention. Monetary amounts are hypothetical one-year losses at the same valuation date.

A position has losses of $0, $5,000, and $25,000 with probabilities 0.80, 0.15, and 0.05. Calculate expected loss and $\operatorname{VaR}_{0.90}$. (Answer: expected loss = $2,000; $\operatorname{VaR}_{0.90}=\$5,000$.)
A distribution has losses of $0 with probability 0.95, $40,000 with probability 0.04, and $200,000 with probability 0.01. Find $\operatorname{VaR}_{0.95}$ and $\operatorname{TVaR}_{0.95}$, accounting for the mass at the cutoff. (Answer: $\operatorname{VaR}_{0.95}=\$0$; $\operatorname{TVaR}_{0.95}=\$72,000$.)
A loss is $0 with probability 0.90, $10,000 with probability 0.08, and $50,000 with probability 0.02. Calculate expected loss and the conditional mean loss given a positive loss. (Answer: expected loss = $1,800; $E[L\mid L>0]=\$18,000$.)
A hypothetical loss is $0 or $2,000, each with probability 0.50. Calculate its expected loss, variance, standard deviation, and $\operatorname{VaR}_{0.75}$. (Answer: expected loss = $1,000; variance = 1,000,000 dollars squared; standard deviation = $1,000; $\operatorname{VaR}_{0.75}=\$2,000$.)
A loss distribution has outcomes $0, $10,000, and $100,000 with probabilities 0.90, 0.08, and 0.02. Find $\operatorname{VaR}_{0.95}$ and $\operatorname{TVaR}_{0.95}$. (Answer: $\operatorname{VaR}_{0.95}=\$10,000$; $\operatorname{TVaR}_{0.95}=\$46,000$.)

What You'll Learn