Correct Answer: C

To find the probability that the market factor ends up at a quantile (or worse) associated with a portfolio loss of 0.01, we need to solve for the market factor return threshold ($\bar{m}$) and then find its cumulative probability using the standard normal cumulative distribution function $\Phi(\bar{m})$.

Given the formula: $\Phi^{-1}(\bar{x}) = \frac{k - \beta \bar{m}}{\sqrt{1 - \beta^2}}$

We are given the following values:

• Portfolio loss threshold $\bar{x} = 0.01$
• $\Phi^{-1}(0.01) \approx -2.3263$ (The text uses -2.33 for approximation, but using the more precise standard normal value -2.3263 avoids rounding errors).
• Default probability is 1.0%, so $k = \Phi^{-1}(0.01) \approx -2.3263$.
• Correlation is 0.64, so $\beta = \sqrt{0.64} = 0.80$.

Substituting these values into the equation: $-2.3263 = \frac{-2.3263 - 0.80 \bar{m}}{\sqrt{1 - 0.80^2}}$

We know that $\sqrt{1 - 0.80^2} = \sqrt{1 - 0.64} = \sqrt{0.36} = 0.60$.

$-2.3263 = \frac{-2.3263 - 0.80 \bar{m}}{0.60}$ $-2.3263 \times 0.60 = -2.3263 - 0.80 \bar{m}$ $-1.39578 = -2.3263 - 0.80 \bar{m}$ $`$0.80 \bar{m} = -2.3263 + 1.39578 $$ $$0.80` \bar{m} = -0.93052 $$ \bar{m} = \frac{-0.93052}{0.80} = -1.16315 $$

Now, we find the probability that the market factor is at this quantile or worse by applying the cumulative normal distribution function: $P(M \le \bar{m}) = \Phi(-1.16315)$

Looking up the standard normal distribution, $\Phi(-1.16315) \approx 0.12238$, or $12.24%$.

(Note: If we use the exact value -2.33 from the prompt throughout the calculation: $-2.33 \times 0.6 = -1.398$ $0.8\bar{m} = -2.33 + 1.398 = -0.932 $$ \bar{m} = -1.165 $$ \Phi(-1.165) \approx 0.1220 $, which is still closest to option C).