The probability of default using the Merton model is based on the actual (physical) probability measure, calculated as $N(-d_2)$, where: $d_2 = \frac{\ln(V/K) + (\mu - \sigma^2/2)T}{\sigma\sqrt{T}}$

Given:

• $V = 110$
• $K = 130$
• $T = 3$
• $\mu = 0.20$
• $\sigma = 0.30$

Calculate $d_2$: $\ln(110/130) = -0.16705$ $\mu - \sigma^2/2 = 0.20 - 0.30^2 / 2 = 0.20 - 0.045 = 0.155$ Numerator = $-0.16705 + (0.155 \times 3) = -0.16705 + 0.465 = 0.29795$ Denominator = $0.30 \times \sqrt{3} = 0.5196$$d_2 = 0.29795 / 0.5196 = 0.5734$

The probability of default is $N(-d_2) = N(-0.5734) = 1 - N(0.5734)$. Using standard normal probabilities, $N(0.5734)$ is approximately $0.7168$, so the probability of default$\approx 1 - 0.7168 = 0.2832$.