Worked Problem — Speculative Grade or Default

Problem. Using the grade-by-default table in §1.2 below — 30.00% of obligors are speculative grade, 2.00% default, and 1.60% are both — what is the probability that a randomly chosen obligor is speculative grade or defaults? And what is the probability that two independent obligors, each defaulting with probability 2.00%, include at least one default?

Solution.

1. Union. $P(SG \cup D) = 0.3000 + 0.0200 - 0.0160 = 0.3040$. Without the subtraction the answer would be 0.3200, counting the 1.60% of speculative-grade defaulters twice.
2. At least one of two. The complement is "neither defaults": $0.98 \times 0.98 = 0.9604$, so $P(\text{at least one}) = 1 - 0.9604 = 0.0396$ — slightly less than the $0.04$ that adding the two probabilities would give, by exactly the $0.0004$ chance that both default.

Answer: 30.40%, and 3.96%.

Worked Problem — Reading Conditional Probabilities off a Joint Table

Problem. Using the table above, compute the probability that an obligor defaults given that it is speculative grade, and the probability that a defaulted obligor was speculative grade. Are grade and default independent?

Solution — step by step.

1. Default given speculative. Divide the cell by the row total: $0.0160 \div 0.3000 = 0.053333$, or 5.3333%.
2. Default given investment grade. $0.0040 \div 0.7000 = 0.005714$, or 0.5714%. Speculative obligors default about 9.33 times as often.
3. Speculative given default. Now divide by the column total: $0.0160 \div 0.0200 = 0.800000$, or 80.00%.
4. Independence check. If grade and default were independent, the cell would equal the product of its margins: $0.7000 \times 0.0200 = 0.0140$. The actual cell is 0.0040. They are not independent.

Answer: 5.3333% and 80.00%; the two are not independent.

Steps 1 and 3 are the pair candidates reverse under time pressure. $P(\text{default} \mid \text{speculative}) = 5.33\%$ and $P(\text{speculative} \mid \text{default}) = 80.00\%$ describe the same cell and are not remotely the same number, because they are divided by different totals.

Worked Problem — Independent Within Each State, Dependent Overall

Problem. The economy is good with probability 0.80 and bad with probability 0.20. Each of two obligors defaults with probability 1.00% in a good year and 6.00% in a bad one, and given the state their defaults are independent. Find each obligor's default probability, the probability both default, and whether the two defaults are unconditionally independent.

Solution.

1. Each obligor. By total probability, $0.80 \times 0.01 + 0.20 \times 0.06 = 0.008 + 0.012 = 0.0200$.
2. Both default. Within each state the joint probability is the product: $0.01^2 = 0.0001$ in a good year and $0.06^2 = 0.0036$ in a bad one. Weighting: $0.80 \times 0.0001 + 0.20 \times 0.0036 = 0.00008 + 0.00072 = 0.0008$.
3. Test for independence. If the defaults were unconditionally independent, both would default with probability $0.02^2 = 0.0004$. The actual figure is $0.0008$, twice that.
4. Conditional probability. Given that one obligor defaults, the other defaults with probability $0.0008/0.02 = 4.00\%$, double its unconditional 2.00%.

Answer: each defaults with probability 2.00%, both with 0.08%, and the defaults are dependent — independent within each state of the economy, but not across them. One default is evidence that the year is bad, and a bad year raises the other's chance. The default correlation implied is $(0.0008 - 0.0004)/(0.02 \times 0.98) = 0.0204$, small as a correlation and large as a doubling of the joint tail.

Worked Problem — From a Mass Function to an Event and a CDF

Problem. For the 20-name portfolio, the PMF gives $P(0) = 0.6676$, $P(1) = 0.2725$, $P(2) = 0.0528$ and $P(3) = 0.0065$. Compute the CDF at 0, 1, 2 and 3, the probability of one or two defaults, and the probability of three or more.

Solution.

1. CDF. Running sums: $F(0) = 0.6676$, $F(1) = 0.6676 + 0.2725 = 0.9401$, $F(2) = 0.9401 + 0.0528 = 0.9929$, $F(3) = 0.9929 + 0.0065 = 0.9994$.
2. One or two defaults. Add the masses of the outcomes in the event: $0.2725 + 0.0528 = 0.3253$ — or, from the CDF, $F(2) - F(0) = 0.9929 - 0.6676 = 0.3253$.
3. Three or more. The complement of "two or fewer": $1 - F(2) = 1 - 0.9929 = 0.0071$.

Answer: CDF 0.6676, 0.9401, 0.9929, 0.9994; P(1 or 2) = 0.3253; P(3 or more) = 0.0071.

Worked Problem — What a Flag Is Worth

Problem. The screen has 90% sensitivity and 90% specificity, applied to a book with a 2.00% default rate. An obligor is flagged. What is the probability that it actually defaults?

Solution — step by step. Work in frequencies over 10,000 obligors; it is faster than the formula and much harder to get wrong.

1. Split the population. At a 2.00% rate, 200 will default and 9,800 will not.
2. True positives. The screen catches 90% of the 200: $200 \times 0.90 = 180$.
3. False positives. It wrongly flags 10% of the 9,800: $9,800 \times 0.10 = 980$.
4. Total flagged. $180 + 980 = 1,160$ names get flagged.
5. The answer. Of those, 180 genuinely default: $180 \div 1,160 = 0.155172$.

Answer: 15.52%. A flag from a 90%-accurate model is wrong 84.48% of the time — more than five times out of six.

Nothing is broken. The model does exactly what it advertises. The 980 false positives simply come from a much larger pool than the 180 true positives — there are 49 healthy obligors for every defaulter, so even a 10% error rate on the healthy names swamps a 90% hit rate on the defaulters.

# Ravensmoor's obligor default screen, as a frequency table over 10,000 names.
base, sens, spec, pop = 0.0200, 0.90, 0.90, 10_000

defaulters = pop * base
healthy = pop - defaulters
true_pos = defaulters * sens
false_pos = healthy * (1 - spec)
flagged = true_pos + false_pos

print(f"obligors                {pop:,}")
print(f"  will default          {defaulters:,.0f}")
print(f"  will not              {healthy:,.0f}")
print(f"flagged by the screen   {flagged:,.0f}")
print(f"  correctly (true +)    {true_pos:,.0f}")
print(f"  wrongly   (false +)   {false_pos:,.0f}")
print(f"P(default | flagged)    {true_pos / flagged:.6f}   = {true_pos / flagged:.2%}")
print(f"P(flagged)              {flagged / pop:.4f}")
# Output:
# obligors                10,000
#   will default          200
#   will not              9,800
# flagged by the screen   1,160
#   correctly (true +)    180
#   wrongly   (false +)   980
# P(default | flagged)    0.155172   = 15.52%
# P(flagged)              0.1160

Worked Problem — Probability From a Density

Problem. (a) $X$ is uniform on $[0, 0.5]$, with density 2. What is $P(0.1 < X < 0.3)$? (b) An obligor's time to default $T$, in years, is exponential with density $f(t) = \lambda e^{-\lambda t}$ and $\lambda = 0.02$. What is the probability it defaults within five years?

Solution.

1. Uniform. Probability is area under the density: $\int_{0.1}^{0.3} 2\,dx = 2 \times 0.2 = 0.40$.
2. Exponential. $P(T \le 5) = \int_0^5 0.02\,e^{-0.02t}\,dt = \left[-e^{-0.02t}\right]_0^5 = 1 - e^{-0.10} = 1 - 0.904837 = 0.095163$.

Answer: 0.40, and 9.52%. Both are CDF values: the uniform's CDF is $2x$ on the interval and the exponential's is $1 - e^{-\lambda t}$, which is how the exam usually presents them — the integration has been done once and for all.

Worked Problem — Four Moments of a Quarterly Loss

Problem. Ravensmoor's operational-risk team models the quarterly loss from one process as a discrete variable: $0M with probability 0.8200, $6M with 0.1300, $12M with 0.0400 and $18M with 0.0100. Compute the mean, variance, standard deviation, skewness and kurtosis.

Solution — step by step.

1. Check it is a PMF. $0.8200 + 0.1300 + 0.0400 + 0.0100 = 1.0000$. Good.
2. Mean. $0(0.82) + 6(0.13) + 12(0.04) + 18(0.01) = 0 + 0.78 + 0.48 + 0.18 = 1.4400$, so $1.44M.
3. Variance. Weight each squared deviation from 1.44: $(0-1.44)^2(0.82) + (6-1.44)^2(0.13) + (12-1.44)^2(0.04) + (18-1.44)^2(0.01) = 1.700352 + 2.703168 + 4.460544 + 2.742336 = 11.6064$.
4. Standard deviation. $\sqrt{11.6064} = 3.406817$, so $3.41M — more than twice the mean.
5. Skewness. The cubed deviations are $-2.985984$, $94.818816$, $1,177.583616$ and $4,541.308416$; weighted by the probabilities they sum to $-2.448507 + 12.326446 + 47.103345 + 45.413084 = 102.394368$. Dividing by $\sigma^3 = 3.406817^3 = 39.540877$ gives 2.589583.
6. Kurtosis. The fourth powers weighted by probability sum to $3.525850 + 56.208594 + 497.411319 + 752.040674 = 1,309.186437$. Dividing by $\sigma^4 = 11.6064^2 = 134.708521$ gives 9.718661, an excess of 6.718661.

Answer: mean $1.44M, standard deviation $3.41M, skewness 2.5896, kurtosis 9.7187.

Read the shape rather than the numbers. The distribution is strongly right-skewed and extremely fat-tailed, which is the signature of every operational-loss series: almost all quarters cost nothing, and the average is driven by rare large events. A mean of $1.44M describes no quarter that actually occurs — the process either loses nothing, which happens 82% of the time, or loses several multiples of the mean.

# The four population moments of a discrete loss variable, from first principles.
outcomes = [0.0, 6.0, 12.0, 18.0]          # $M lost in a quarter
probs    = [0.8200, 0.1300, 0.0400, 0.0100]

assert abs(sum(probs) - 1) < 1e-12, "a PMF must sum to one"

mean = sum(x * p for x, p in zip(outcomes, probs))
var  = sum((x - mean) ** 2 * p for x, p in zip(outcomes, probs))
sd   = var ** 0.5
skew = sum((x - mean) ** 3 * p for x, p in zip(outcomes, probs)) / sd ** 3
kurt = sum((x - mean) ** 4 * p for x, p in zip(outcomes, probs)) / sd ** 4

print(f"mean      {mean:.4f}")
print(f"variance  {var:.4f}")
print(f"std dev   {sd:.6f}")
print(f"skewness  {skew:.6f}")
print(f"kurtosis  {kurt:.6f}   (excess {kurt - 3:.6f})")
# Output:
# mean      1.4400
# variance  11.6064
# std dev   3.406817
# skewness  2.589583
# kurtosis  9.718661   (excess 6.718661)

Worked Problem — The Fund's Monthly Loss Quantiles

Problem. The Core Equity Fund returns 11.00% a year with a volatility of 18.00%. Assuming returns are normal and independent across months, find the monthly mean and volatility, then the 5% and 1% monthly return quantiles.

Solution — step by step.

1. Monthly mean. Means scale linearly with time: $11.00 \div 12 = 0.916667\%$.
2. Monthly volatility. Variances scale linearly, so volatilities scale with the square root of time: $18.00 \div \sqrt{12} = 5.196152\%$.
3. The 5% quantile. The standard normal's 5% point is $-1.644854$: $0.916667 + (-1.644854)(5.196152) = -7.630243\%$.
4. The 1% quantile. The standard normal's 1% point is $-2.326348$: $0.916667 + (-2.326348)(5.196152) = -11.171391\%$.

Answer: a monthly mean of 0.9167% and volatility of 5.1962%; the 5% quantile is −7.6302% and the 1% quantile is −11.1714%.

Step 2 is where marks are lost. Dividing the volatility by 12 rather than by $\sqrt{12}$ gives 1.50%, understating monthly risk by a factor of 3.46, and the resulting loss quantiles would be wrong by the same factor. Means scale with $t$; volatilities scale with $\sqrt{t}$.

Worked Problem — One Bad Month Against the Median

Problem. Ten monthly returns of a credit sleeve are $1.2$, $0.8$, $-0.5$, $1.5$, $0.9$, $1.1$, $-0.3$, $0.6$, $1.0$ and $-18.0$ percent, the last from a single default. Compare the mean and median, and the standard deviation and interquartile range, with and without that month. Use positions $(n+1)p$ for the quartiles.

Solution.

1. Mean. The ten returns sum to $-11.7$, so the mean is $-1.17\%$. Without the default month: $6.3/9 = 0.70\%$.
2. Median. Sorted, the ten are $-18.0, -0.5, -0.3, 0.6, 0.8, 0.9, 1.0, 1.1, 1.2, 1.5$; the median averages the fifth and sixth, $(0.8 + 0.9)/2 = 0.85\%$. Without the default month it is the fifth of nine, $0.90\%$.
3. Quartiles. The 25th percentile sits at position $11 \times 0.25 = 2.75$: $-0.5 + 0.75(0.2) = -0.35$. The 75th at $8.25$: $1.1 + 0.25(0.1) = 1.125$. The IQR is $1.475$ points.
4. Standard deviation. With the default month, $5.95\%$; without it, $0.67\%$.

Answer: one month moves the mean by 1.87 points and the standard deviation ninefold, while the median moves by 0.05. That is robustness. Whether it is a virtue depends on the question: the median describes a typical month well and says nothing about the default — which, for a credit sleeve, is the thing being measured.

Worked Problem — A Management Fee Is a Shift, Not a Scaling

Problem. The Core Equity Fund charges a flat management fee of 1.20% a year, accrued monthly, deducted from the gross return. Using the monthly figures from §1.7, state the net monthly mean, volatility and 5% quantile.

Solution — step by step.

1. Identify the transformation. Net $=$ gross $-$ fee, so $Y = a + bX$ with $b = 1$ and $a = -1.20 \div 12 = -0.10$.
2. Mean. $0.916667 - 0.100000 = 0.816667\%$.
3. Volatility. Since $b = 1$, volatility is unchanged at $5.196152\%$.
4. 5% quantile. The quantile shifts by $a$ along with everything else: $-7.630243 - 0.100000 = -7.730243\%$.

Answer: mean 0.8167%, volatility 5.1962% unchanged, 5% quantile −7.7302%.

A flat fee makes the fund worse without making it any less volatile, so every risk-adjusted ratio falls — the numerator drops and the denominator does not. A performance fee behaves entirely differently, because it is a function of the return rather than a constant, and therefore is not a linear transformation at all.

Worked Problem — Rescaling a Skewed Loss, and Turning It Upside Down

Problem. The quarterly operational loss $X$ of §1.6 has mean $1.44M, standard deviation $3.4068M, skewness 2.5896, kurtosis 9.7187, median $0M and 95th percentile $6M. (a) A subsidiary's loss is modelled as $Y = 0.40 + 1.25X$: a quarter more exposure plus a fixed $0.40M of handling cost. Give the same six figures for $Y$. (b) A reinsurer that pays the whole loss sees its own result as $Z = -X$. What happens to the skewness and kurtosis?

Solution.

1. Mean and median shift and scale. Mean $0.40 + 1.25 \times 1.44 = 2.20$, so $2.20M; median $0.40 + 1.25 \times 0 = 0.40$, so $0.40M.
2. Spread scales by $|b|$ only. Standard deviation $1.25 \times 3.4068 = 4.2585$, so $4.2585M; the 95th percentile, a quantile, moves with the line: $0.40 + 1.25 \times 6 = 7.90$, so $7.90M.
3. Shape is untouched. Skewness and kurtosis are dimensionless and $b > 0$, so they stay at 2.5896 and 9.7187.
4. Reversing the sign. For $Z = -X$, $b = -1$: the standard deviation is still $3.4068M, the kurtosis still 9.7187, but the skewness becomes −2.5896, because the long tail now points the other way.

Answer: mean $2.20M, sd $4.2585M, median $0.40M, 95th percentile $7.90M, skewness and kurtosis unchanged; for the reinsurer, skewness flips to −2.5896 and nothing else about the shape changes. The fixed handling cost is the part candidates mis-handle: it shifts every location measure and touches no measure of spread or shape.

# The fund's monthly moments, a normal quantile, and what a fee does to them.
from scipy.stats import norm

r_ann, vol_ann = 11.00, 18.00
r_m = r_ann / 12
vol_m = vol_ann / 12 ** 0.5

print(f"monthly mean        {r_m:.6f}%")
print(f"monthly volatility  {vol_m:.6f}%")
for p in (0.05, 0.01):
    print(f"{p:.0%} quantile        {r_m + norm.ppf(p) * vol_m:.6f}%")

# a flat fee of 1.20% a year, charged monthly: Y = a + bX with a = -0.10, b = 1
fee_m = 1.20 / 12
print()
print(f"after a {fee_m:.2f}% monthly fee:")
print(f"  mean              {r_m - fee_m:.6f}%   (shifts by the fee)")
print(f"  volatility        {vol_m:.6f}%   (UNCHANGED - a shift is not a scaling)")
print(f"  5% quantile       {r_m - fee_m + norm.ppf(0.05) * vol_m:.6f}%")
# Output:
# monthly mean        0.916667%
# monthly volatility  5.196152%
# 5% quantile        -7.630243%
# 1% quantile        -11.171391%
#
# after a 0.10% monthly fee:
#   mean              0.816667%   (shifts by the fee)
#   volatility        5.196152%   (UNCHANGED - a shift is not a scaling)
#   5% quantile       -7.730243%

Worked Problem — A Poisson Count, a Normal Tail and an Exponential Median

Problem. (a) Operational-loss events arrive at an average of 3 a quarter, Poisson. What is the probability of no events, and of five or more? (b) The fund's monthly return is normal with mean $0.916667\%$ and volatility $5.196152\%$. Given $\Phi(-1.1387) = 0.1274$, what is the probability of losing more than 5% in a month? (c) A time to default is exponential with $\lambda = 0.02$ a year. What are its mean and median?

Solution.

1. Poisson, zero. $e^{-3} = 0.049787$.
2. Poisson, five or more. The complement of four or fewer. The masses for 0 to 4 are $0.049787$, $0.149361$, $0.224042$, $0.224042$ and $0.168031$ — each the previous one times $3/k$ — summing to $0.815263$. So $P(\ge 5) = 1 - 0.815263 = 0.184737$.
3. Normal. $z = (-5 - 0.916667)/5.196152 = -1.1387$, and $\Phi(-1.1387) = 0.1274$: a 12.74% chance in any month.
4. Exponential. Mean $1/0.02 = 50$ years; median $\ln 2/0.02 = 0.693147/0.02 = 34.66$ years. The median is well below the mean, because the exponential is right-skewed.

Answer: 4.98% and 18.47%; 12.74%; mean 50 years, median 34.66 years.

Worked Problem — A Price, a Recovery Rate and a Default Count

Problem. (a) A price relative $X$ (one year's price divided by today's) is lognormal, with $\ln X$ normal with mean 0.05 and standard deviation 0.20. Find the mean and median of $X$, and the probability the price falls, given $\Phi(-0.25) = 0.4013$. (b) A recovery rate is beta with $\alpha = 2$ and $\beta = 3$. Find its mean and standard deviation. (c) Fifteen independent obligors each default with probability 3.00%. Find the expected number of defaults, its variance, and the probability of two or more.

Solution.

1. Lognormal mean. $e^{0.05 + 0.20^2/2} = e^{0.07} = 1.0725$ — a 7.25% expected gain.
2. Lognormal median. $e^{0.05} = 1.0513$, below the mean: the lognormal is right-skewed, so its mean is pulled above its median by the long upper tail.
3. Price falls. $X < 1$ exactly when $\ln X < 0$, so $P = \Phi((0 - 0.05)/0.20) = \Phi(-0.25) = 0.4013$.
4. Beta. Mean $2/(2 + 3) = 0.40$; variance $\dfrac{2 \times 3}{5^2 \times 6} = \dfrac{6}{150} = 0.04$, so the standard deviation is $0.20$. A beta is bounded by 0 and 1, which is why it suits a recovery rate.
5. Binomial moments. Mean $15 \times 0.03 = 0.45$; variance $15 \times 0.03 \times 0.97 = 0.4365$.
6. Two or more. $P(0) = 0.97^{15} = 0.633251$ and $P(1) = 15 \times 0.03 \times 0.97^{14} = 0.45 \times 0.652836 = 0.293776$, so $P(\ge 2) = 1 - 0.633251 - 0.293776 = 0.072973$.

Answer: lognormal mean 1.0725, median 1.0513, P(fall) 40.13%; beta mean 0.40, sd 0.20; binomial mean 0.45, variance 0.4365, P(2 or more) 7.30%. The lognormal result is the one candidates miss: an asset with an expected gain of 7.25% finishes below where it started 40% of the time, and half the time it gains less than 5.13% — the mean is not the typical outcome.

import numpy as np
from math import exp, log
from scipy import stats

print(f"P(SG or D)          {0.30 + 0.02 - 0.016:.4f}")
print(f"P(at least 1 of 2)  {1 - 0.98 ** 2:.4f}")
print(f"P(D), total prob.   {0.004 / 0.70 * 0.70 + 0.016 / 0.30 * 0.30:.4f}")
b = stats.binom(20, 0.02)
print(f"binomial CDF        {b.cdf(0):.4f} {b.cdf(1):.4f} {b.cdf(2):.4f} {b.cdf(3):.4f}")
print(f"P(1 or 2 defaults)  {b.pmf(1) + b.pmf(2):.4f}   P(3+) {b.sf(2):.4f}")
print(f"exponential P(T<=5) {1 - exp(-0.02 * 5):.6f}   median {log(2) / 0.02:.2f}")
po = stats.poisson(3)
print(f"Poisson P(0) {po.pmf(0):.6f}   P(<=4) {po.cdf(4):.6f}   P(>=5) {po.sf(4):.6f}")
z = (-5 - 11.00 / 12) / (18.00 ** 2 / 12) ** 0.5
print(f"normal z {z:.4f}   P(R < -5%) {stats.norm.cdf(z):.4f}")
r = np.array([1.2, 0.8, -0.5, 1.5, 0.9, 1.1, -0.3, 0.6, 1.0, -18.0])
print(f"ten months: mean {r.mean():.2f}  median {np.median(r):.2f}  sd {r.std(ddof=1):.2f}")
print(f"nine months: mean {r[:-1].mean():.2f}  median {np.median(r[:-1]):.2f}  sd {r[:-1].std(ddof=1):.2f}")
print(f"lognormal mean {exp(0.07):.4f}  median {exp(0.05):.4f}  P(X<1) {stats.norm.cdf(-0.25):.4f}")
print(f"beta(2,3) mean {2 / 5:.2f}  sd {(6 / 150) ** 0.5:.2f}")
bi = stats.binom(15, 0.03)
print(f"binomial(15, 3%) P(0) {bi.pmf(0):.4f}  P(1) {bi.pmf(1):.4f}  P(2+) {bi.sf(1):.4f}")
# Output:
# P(SG or D)          0.3040
# P(at least 1 of 2)  0.0396
# P(D), total prob.   0.0200
# binomial CDF        0.6676 0.9401 0.9929 0.9994
# P(1 or 2 defaults)  0.3253   P(3+) 0.0071
# exponential P(T<=5) 0.095163   median 34.66
# Poisson P(0) 0.049787   P(<=4) 0.815263   P(>=5) 0.184737
# normal z -1.1387   P(R < -5%) 0.1274
# ten months: mean -1.17  median 0.85  sd 5.95
# nine months: mean 0.70  median 0.90  sd 0.67
# lognormal mean 1.0725  median 1.0513  P(X<1) 0.4013
# beta(2,3) mean 0.40  sd 0.20
# binomial(15, 3%) P(0) 0.6333  P(1) 0.2938  P(2+) 0.0730

Worked Problem — Calibrating a Regime Mixture to the Fund

Problem. Model the fund's monthly return as a mixture: a calm regime with 4.00% volatility 90% of the time, and a stressed regime the other 10%. Choose the stressed volatility so that the mixture reproduces the fund's own monthly volatility of 5.196152%. Then find the mixture's kurtosis.

Solution — step by step.

1. Set up the variance condition. $0.90(4.00)^2 + 0.10\sigma_s^2 = 5.196152^2 = 27.000000$.
2. Solve for the stressed variance. $0.90(16) = 14.40$, so $0.10\sigma_s^2 = 27.000000 - 14.400000 = 12.600000$ and $\sigma_s^2 = 126.000000$.
3. Stressed volatility. $\sqrt{126} = 11.224972\%$ — a little over 2.8 times the calm regime.
4. Kurtosis. The numerator is $3[0.90(16)^2 + 0.10(126)^2] = 3[230.40 + 1587.60] = 5454.00$; the denominator is $27^2 = 729$. So $K = 5454 \div 729 = 7.481481$.

Answer: a stressed volatility of 11.2250% gives a mixture with kurtosis 7.4815 against the normal's 3.

# Two normal regimes whose mixture reproduces the fund's monthly volatility.
from scipy.stats import norm

vol_m = 18.00 / 12 ** 0.5
w, s_calm = 0.10, 4.00
s_stress = ((vol_m ** 2 - (1 - w) * s_calm ** 2) / w) ** 0.5

var = (1 - w) * s_calm ** 2 + w * s_stress ** 2
kurt = 3 * ((1 - w) * s_calm ** 4 + w * s_stress ** 4) / var ** 2

print(f"stressed-regime vol  {s_stress:.6f}%")
print(f"mixture volatility   {var ** 0.5:.6f}%   (the fund's {vol_m:.6f}%)")
print(f"mixture kurtosis     {kurt:.6f}   (each component is 3)")

def mix_cdf(x):
    return (1 - w) * norm.cdf(x, 0, s_calm) + w * norm.cdf(x, 0, s_stress)

lo, hi = -60.0, 60.0
for _ in range(200):
    mid = (lo + hi) / 2
    lo, hi = (mid, hi) if mix_cdf(mid) < 0.01 else (lo, mid)
q_mix = (lo + hi) / 2
q_norm = norm.ppf(0.01, 0, var ** 0.5)

print(f"1% mixture quantile  {q_mix:.6f}%")
print(f"1% normal quantile   {q_norm:.6f}%")
print(f"the normal is short by {abs(q_mix) - abs(q_norm):.6f} pp")
# Output:
# stressed-regime vol  11.224972%
# mixture volatility   5.196152%   (the fund's 5.196152%)
# mixture kurtosis     7.481481   (each component is 3)
# 1% mixture quantile  -14.471310%
# 1% normal quantile   -12.088058%
# the normal is short by 2.383252 pp