The Distance to Default (DD) according to the Merton / Black-Scholes model is given by: $DD = \frac{\ln(A/L) + (\mu - \sigma_A^2 / 2) T}{\sigma_A \sqrt{T}}$

For Company X:

• A = $800,000, L = $500,000, $\sigma_A = 0.25$, $\mu = 0.07$, T = 3
• $\ln(A/L) = \ln(800,000/500,000) = \ln(1.6) \approx 0.4700$
• $(\mu - \sigma_A^2 / 2)T = (0.07 - 0.25^2 / 2) \times 3 = (0.07 - 0.03125) \times 3 = 0.11625$
• $\sigma_A \sqrt{T} = 0.25 \times \sqrt{3} \approx 0.4330$
• $DD_X = \frac{0.4700 + 0.11625}{0.4330} \approx 1.354$

For Company Y:

• A = $1,000,000, L = $600,000, $\sigma_A = 0.20$, $\mu = 0.05$, T = 4
• $\ln(A/L) = \ln(1,000,000/600,000) = \ln(1.667) \approx 0.5108$
• $(\mu - \sigma_A^2 / 2)T = (0.05 - 0.20^2 / 2) \times 4 = (0.05 - 0.02) \times 4 = 0.12$
• $\sigma_A \sqrt{T} = 0.20 \times \sqrt{4} = 0.40$
• $DD_Y = \frac{0.5108 + 0.12}{0.40} \approx 1.577$

For Company Z:

• A = $900,000, L = $400,000, $\sigma_A = 0.30$, $\mu = 0.06$, T = 2
• $\ln(A/L) = \ln(900,000/400,000) = \ln(2.25) \approx 0.8109$
• $(\mu - \sigma_A^2 / 2)T = (0.06 - 0.30^2 / 2) \times 2 = (0.06 - 0.045) \times 2 = 0.03$
• $\sigma_A \sqrt{T} = 0.30 \times \sqrt{2} \approx 0.4243$
• $DD_Z = \frac{0.8109 + 0.03}{0.4243} \approx 1.982$

The lower the Distance to Default (DD), the higher the probability of default. Ranking from most likely (lowest DD) to least likely (highest DD): Company X (1.354), Company Y (1.577), Company Z (1.982).