Unit 10, lesson 74

Structural credit and the Merton model

How the Merton structural credit model treats equity as a call and debt as face minus a put on firm asset value, prices both at a terminal default boundary, and separates risk-neutral from physical default probability.

  • Video 74
  • Default models and credit default swaps

A firm’s assets will be worth a random amount on the date its debt of 100 falls due. If they are worth 80, the creditors take all of it and the shareholders get nothing, and if they are worth 140, the creditors get 100 and the shareholders keep 40. Can equity and debt be priced as options on the same asset value, and which default probability does that price contain? This lesson states the Merton model, derives the equity and debt values from two normal integrals, and separates the default probability used in prices from the one a forecast would need.

Key terms

Merton model

Definition A structural credit model describes default as an event determined by the value of a firm’s assets relative to its liabilities. The Merton model is the structural credit model with the following hypotheses.

  1. The firm’s assets have a market value AtA_t that can be traded or replicated in a frictionless market with a constant continuously compounded rate rr, so claims on the assets are priced as discounted Q\mathbb{Q}-expectations.
  2. Under Q\mathbb{Q} the asset value follows the geometric Brownian motion dAt=rAt dt+σAt dWtQdA_t = r A_t\,dt + \sigma A_t\,dW^{\mathbb{Q}}_t with A0>0A_0 > 0 and a constant σ>0\sigma > 0, and under P\mathbb{P} it follows the same equation with a constant drift μ\mu in place of rr and a P\mathbb{P}-Brownian motion WtW_t.
  3. The only liability is one zero-coupon debt with face F>0F > 0 due at T>0T > 0, and there are no asset payouts, coupons or dividends before TT.
  4. Limited liability and absolute priority hold, and there are no bankruptcy costs or taxes.
  5. Default can happen only at TT, and it happens exactly when AT<FA_T < F.

Under these hypotheses equity and debt pay

ET=(AT−F)+,DT=min⁡(AT,F)\begin{gathered} E_T = (A_T - F)^{+}, \\[0.4em] D_T = \min(A_T, F) \end{gathered}

at TT, and ET+DT=ATE_T + D_T = A_T in every state. The cards below derive the values

E0=A0Φ(d1)−Fe−rTΦ(d2),D0=A0−E0.\begin{aligned} E_0 &= A_0 \Phi(d_1) \\[0.4em] &\quad - F e^{-rT} \Phi(d_2), \\[0.4em] D_0 &= A_0 - E_0 . \end{aligned}

On this page EtE_t is the equity value and DtD_t the debt value, while E\mathbb{E} keeps its meaning as an expectation.

In plain words The model treats the whole firm as one asset and the capital structure as a split of its value at maturity. Creditors are paid first, up to the face, the shareholders keep the rest, and default means that the assets cannot cover the face on the due date.

Example The constructed firm of this lesson has assets A0=120A_0 = 120 and a face F=100F = 100 due in T=2T = 2 years, in millions of currency units, with r=0.03r = 0.03 and σ=0.25\sigma = 0.25 per square-root year. The model gives equity worth about 31.279831.2798 million, debt worth about 88.720288.7202 million and a risk-neutral default probability of about 0.3055120.305512, as the cards below compute. The two values add up to 120120 million, as the payoff identity requires.

Firm asset value

Definition The firm asset value AtA_t is the market value at time tt of the cash flows of the firm’s assets. Under the hypotheses of the Merton model it equals the total value Et+DtE_t + D_t of the claims on the firm. It is a modeled market value, not the asset total of a balance sheet, and it is not directly observed. Under Q\mathbb{Q} the geometric Brownian motion has the exact solution

ln⁡AT=ln⁡A0+(r−12σ2)T+σT Z,\begin{aligned} \ln A_T &= \ln A_0 + \bigl(r - \tfrac12 \sigma^2\bigr) T \\[0.4em] &\quad + \sigma \sqrt{T}\, Z , \end{aligned}

where Z=WTQ/TZ = W^{\mathbb{Q}}_T/\sqrt{T} is standard normal under Q\mathbb{Q}. The volatility σ\sigma is per square-root year and TT is in years, so σT\sigma\sqrt{T} has no unit. Before maturity the equity value is a smooth function Et=E(t,At)E_t = E(t, A_t) with ∂E/∂A=Φ(d1)\partial E/\partial A = \Phi(d_1), where d1d_1 uses the remaining time T−tT - t, and Itô’s lemma gives the instantaneous equity volatility

σE,t=σ At Φ(d1)Et.\sigma_{E,t} = \sigma\, \frac{A_t\, \Phi(d_1)}{E_t} .

It changes with AtA_t and tt, so equity is not itself a geometric Brownian motion. Inferring A0A_0 and σ\sigma from an observed equity value and equity volatility means solving this relation together with the equity formula, a nonlinear inverse problem that is stated here and not developed. Even a good fit leaves the physical drift μ\mu undetermined.

In plain words The asset value is what the firm’s future cash flows are worth in the market today, the total that creditors and shareholders share. Equity is the residual claim on it, so in percentage terms equity moves more than the assets do.

Example For the constructed firm the Q\mathbb{Q}-median of A2A_2 is 120 e(0.03−0.03125)⋅2≈119.70120\, e^{(0.03 - 0.03125) \cdot 2} \approx 119.70 million and its Q\mathbb{Q}-mean is 120 e0.06≈127.42120\, e^{0.06} \approx 127.42 million. With Φ(d1)≈0.805702\Phi(d_1) \approx 0.805702 and E0≈31.2798E_0 \approx 31.2798 million, the equity volatility at time 00 is 0.25⋅120⋅0.805702/31.2798≈0.77270.25 \cdot 120 \cdot 0.805702/31.2798 \approx 0.7727 per square-root year, about three times the asset volatility.

Default boundary

Definition The default boundary of a structural model is the level of the asset value at which default is declared. In the Merton model the boundary is the face FF, checked only at maturity, so the default event is

{AT<F},\{A_T < F\} ,

and on this event default happens at time TT. The tie AT=FA_T = F has Q\mathbb{Q}-probability zero under the lognormal law, and in that state the creditors are paid in full anyway, so it affects no value. A path that falls below FF before TT and ends above it is not a default. A first-passage model instead declares default at the first time the asset value reaches a boundary before TT. That changes the default event and leads to a barrier problem in place of the terminal option problem of this lesson, and first-passage credit models are outside this lesson.

In plain words In the Merton model only the asset value on the due date counts. The creditors are repaid in full exactly when the assets cover the face at maturity, whatever happened along the way.

Example Suppose the constructed firm’s assets fall from 120 to 90 million after one year and end at 105 million at T=2T = 2. There is no default, the creditors receive 100 million and the shareholders 5 million. A first-passage rule with boundary 100 million would have declared default during the first year, at the first time the assets reached 100 million, and the payoffs would then have been fixed at that earlier date.

Equity as a call option

Definition Under the hypotheses of the Merton model, equity pays ET=(AT−F)+E_T = (A_T - F)^{+} at TT, the payoff of a European call on the asset value with strike FF. Its value is

E0=A0Φ(d1)−Fe−rTΦ(d2),\begin{aligned} E_0 &= A_0 \Phi(d_1) \\[0.4em] &\quad - F e^{-rT} \Phi(d_2), \end{aligned}

where

d2=1σT[ln⁡A0F+(r−12σ2)T],d1=d2+σT.\begin{aligned} d_2 &= \frac{1}{\sigma\sqrt{T}}\Bigl[\ln\frac{A_0}{F} \\[0.4em] &\qquad + \bigl(r - \tfrac12\sigma^2\bigr) T\Bigr], \\[0.4em] d_1 &= d_2 + \sigma\sqrt{T} . \end{aligned}

To derive it, write the payoff as AT1{AT>F}−F1{AT>F}A_T \mathbf{1}_{\{A_T > F\}} - F \mathbf{1}_{\{A_T > F\}}. Solving ln⁡AT>ln⁡F\ln A_T > \ln F for ZZ gives {AT>F}={Z>−d2}\{A_T > F\} = \{Z > -d_2\}, so the second term is worth Fe−rTΦ(d2)F e^{-rT}\Phi(d_2). For the first term put v=σTv = \sigma\sqrt{T}, so that e−rTAT=A0evZ−v2/2e^{-rT} A_T = A_0 e^{vZ - v^2/2}. Completing the square gives evz−v2/2φ(z)=φ(z−v)e^{vz - v^2/2}\varphi(z) = \varphi(z - v), and the substitution y=z−vy = z - v gives

e−rT EQ[AT1{AT>F}]=A0∫−d2∞φ(z−v) dz=A0 Φ(d2+v)=A0Φ(d1).\begin{aligned} &e^{-rT}\,\mathbb{E}^{\mathbb{Q}}\bigl[A_T \mathbf{1}_{\{A_T > F\}}\bigr] \\[0.4em] &\quad = A_0 \int_{-d_2}^{\infty} \varphi(z - v)\,dz \\[0.4em] &\quad = A_0\, \Phi(d_2 + v) \\[0.4em] &\quad = A_0 \Phi(d_1) . \end{aligned}

With A0A_0, FF, rr and TT held fixed, the equity value is strictly increasing in σ\sigma, because ∂E0/∂σ=A0φ(d1)T>0\partial E_0/\partial \sigma = A_0 \varphi(d_1)\sqrt{T} > 0.

In plain words Shareholders hold the right, but not the obligation, to pay the face at maturity and keep the assets. Limited liability is the floor at zero of the call payoff. The two normal probabilities have different cutoffs because the first one weights the surviving states by their asset value.

Example For the constructed firm, ln⁡1.2≈0.182322\ln 1.2 \approx 0.182322 and (0.03−0.252/2)⋅2=−0.0025(0.03 - 0.25^2/2) \cdot 2 = -0.0025, so the bracket is about 0.1798220.179822. With σT=0.252≈0.353553\sigma\sqrt{T} = 0.25\sqrt{2} \approx 0.353553, this gives d2≈0.508612d_2 \approx 0.508612 and d1≈0.862166d_1 \approx 0.862166. Then

E0≈120⋅0.805702−94.176453⋅0.694488≈31.2798\begin{aligned} E_0 &\approx 120 \cdot 0.805702 \\[0.4em] &\quad - 94.176453 \cdot 0.694488 \\[0.4em] &\approx 31.2798 \end{aligned}

million. Raising the asset volatility to 0.400.40 with everything else fixed gives d2≈0.145525d_2 \approx 0.145525, d1≈0.711211d_1 \approx 0.711211 and E0≈38.8463E_0 \approx 38.8463 million, because the wider distribution of ATA_T makes the convex payoff more valuable.

Debt valuation

Definition Under the hypotheses of the Merton model, debt pays

DT=min⁡(AT,F)=F−(F−AT)+,\begin{aligned} D_T &= \min(A_T, F) \\[0.4em] &= F - (F - A_T)^{+} , \end{aligned}

the face minus the payoff of a European put on the asset value with strike FF. Its value is

D0=Fe−rT−Put0=A0Φ(−d1)+Fe−rTΦ(d2),\begin{aligned} D_0 &= F e^{-rT} - \mathrm{Put}_0 \\[0.4em] &= A_0 \Phi(-d_1) \\[0.4em] &\quad + F e^{-rT} \Phi(d_2) , \end{aligned}

where Put0=Fe−rTΦ(−d2)−A0Φ(−d1)\mathrm{Put}_0 = F e^{-rT}\Phi(-d_2) - A_0\Phi(-d_1), and the same number is A0−E0A_0 - E_0. The term A0Φ(−d1)=e−rT EQ[AT1{AT<F}]A_0\Phi(-d_1) = e^{-rT}\,\mathbb{E}^{\mathbb{Q}}\bigl[A_T \mathbf{1}_{\{A_T < F\}}\bigr] is the value of what the creditors take in the default states, and Fe−rTΦ(d2)F e^{-rT}\Phi(d_2) is the value of the full face in the survival states. The payoff min⁡(AT,F)\min(A_T, F) is below ATA_T when AT>FA_T > F and below FF when AT<FA_T < F, and both events have positive probability, so 0<D0<min⁡(A0,Fe−rT)0 < D_0 < \min(A_0, F e^{-rT}). The continuously compounded yield yy solves D0=Fe−yTD_0 = F e^{-yT}, and the credit spread is

y−r=−1Tln⁡D0Fe−rT>0.y - r = -\frac{1}{T} \ln \frac{D_0}{F e^{-rT}} > 0 .

The debt value depends on how much is missing in each default state, so a default probability alone does not determine it.

In plain words Creditors hold a default-free promise of the face and have, in effect, sold the shareholders a put on the firm’s assets. That put is the shortfall that limited liability leaves with the creditors, and its value depends on how far below the face the assets can fall, not only on how likely they are to fall below it.

Example For the constructed firm, D0≈120−31.2798=88.7202D_0 \approx 120 - 31.2798 = 88.7202 million. The default-free value of the face is 100 e−0.06≈94.1765100\, e^{-0.06} \approx 94.1765 million, so the put is worth about 5.45635.4563 million. The yield is −12ln⁡0.887202≈0.059841-\tfrac12 \ln 0.887202 \approx 0.059841, a spread of about 0.0298410.029841, or 298.41298.41 bp, over rr. In the default states the Q\mathbb{Q}-mean asset value is

A0erTΦ(−d1)Φ(−d2)≈81.04\frac{A_0 e^{rT}\Phi(-d_1)}{\Phi(-d_2)} \approx 81.04

million, so under Q\mathbb{Q} the creditors lose on average about 19 percent of the face when default occurs. With σ=0.40\sigma = 0.40 the debt value falls to about 81.153781.1537 million, while the asset value is still split between the same two claims.

Risk-neutral default probability

Definition Under the hypotheses of the Merton model, the risk-neutral default probability is

Q(AT<F)=Φ(−d2),\mathbb{Q}(A_T < F) = \Phi(-d_2),

and Φ(d2)\Phi(d_2) is the Q\mathbb{Q}-probability of survival, so a claim paying 11 at TT if the firm survives costs e−rTΦ(d2)e^{-rT}\Phi(d_2). Under P\mathbb{P}, with drift μ\mu, the same event has the probability

P(AT<F)=Φ(−d2P),d2P=d2+(μ−r)Tσ.\begin{aligned} \mathbb{P}(A_T < F) &= \Phi\bigl(-d_2^{\mathbb{P}}\bigr), \\[0.4em] d_2^{\mathbb{P}} &= d_2 + \frac{(\mu - r)\sqrt{T}}{\sigma} . \end{aligned}

The two laws are linked by Girsanov’s theorem. Take the constant market price of risk θ=(μ−r)/σ\theta = (\mu - r)/\sigma and the density dQ/dP=exp⁡(−θWT−12θ2T)d\mathbb{Q}/d\mathbb{P} = \exp(-\theta W_T - \tfrac12\theta^2 T) on FT\mathcal{F}_T, a true martingale density because Novikov’s condition holds for a constant θ\theta. Then WtQ=Wt+θtW^{\mathbb{Q}}_t = W_t + \theta t is a Q\mathbb{Q}-Brownian motion, and substituting dWt=dWtQ−θ dtdW_t = dW^{\mathbb{Q}}_t - \theta\,dt turns the drift μ\mu into rr. If μ>r\mu > r, then d2P>d2d_2^{\mathbb{P}} > d_2 and the physical default probability is the smaller of the two. The boundary and the payoffs are the same under both measures, and only the probability law changes.

In plain words The same event, assets below the face at maturity, has two probabilities. The risk-neutral one uses the drift rr and serves as a weight inside prices. A default forecast needs the physical drift μ\mu, which a fit of the model to prices does not supply.

Example For the constructed firm, Q(A2<100)=Φ(−0.508612)≈0.305512\mathbb{Q}(A_2 < 100) = \Phi(-0.508612) \approx 0.305512, and a claim paying 11 at year 2 if the firm survives costs e−0.06⋅0.694488≈0.654044e^{-0.06} \cdot 0.694488 \approx 0.654044. With μ=0.08\mu = 0.08, the shifted cutoff is d2P≈0.508612+0.282843=0.791455d_2^{\mathbb{P}} \approx 0.508612 + 0.282843 = 0.791455 and P(A2<100)≈0.214339\mathbb{P}(A_2 < 100) \approx 0.214339. The spread of 0.0298410.029841 per year is far below Φ(−d2)/T≈0.152756\Phi(-d_2)/T \approx 0.152756, because the creditors lose only part of the face in default and the spread is a yield difference, not a default rate. At σ=0.40\sigma = 0.40 the risk-neutral default probability rises to about 0.4421480.442148.

Check yourself

  1. A firm in the Merton model owes the face 100 at TT. Find the payoffs of equity, of debt and of the shortfall put when AT=70A_T = 70 and when AT=104A_T = 104. A path of the assets falls to 85 at an intermediate date and ends at 104. Is that a default?
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For AT=70A_T = 70, equity pays (70−100)+=0(70 - 100)^{+} = 0, debt pays min⁡(70,100)=70\min(70, 100) = 70 and the put pays (100−70)+=30(100 - 70)^{+} = 30, so debt is the face 100100 minus the put payoff 3030, and equity plus debt is 7070. For AT=104A_T = 104, equity pays 44, debt pays 100100 and the put pays 00, and the claims add up to 104104. The path that dips to 85 is not a default, because the Merton default event is AT<FA_T < F at maturity only, and its payoffs are those of AT=104A_T = 104. A first-passage model with boundary 100 would have declared default at the dip.

  1. In the Merton model take A0=100A_0 = 100, F=80F = 80, T=1T = 1, r=0.02r = 0.02 and σ=0.30\sigma = 0.30. Find d2d_2, d1d_1, the equity and debt values, the risk-neutral default probability and the credit spread, and check the debt value with the put.
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The bracket is ln⁡1.25+(0.02−0.045)≈0.223144−0.025=0.198144\ln 1.25 + (0.02 - 0.045) \approx 0.223144 - 0.025 = 0.198144, so d2≈0.198144/0.3≈0.660479d_2 \approx 0.198144/0.3 \approx 0.660479 and d1≈0.960479d_1 \approx 0.960479. With Φ(d1)≈0.831593\Phi(d_1) \approx 0.831593, Φ(d2)≈0.745527\Phi(d_2) \approx 0.745527 and 80 e−0.02≈78.41589480\, e^{-0.02} \approx 78.415894,

E0≈83.1593−58.4611≈24.6981,\begin{aligned} E_0 &\approx 83.1593 - 58.4611 \\[0.4em] &\approx 24.6981 , \end{aligned}

and D0≈100−24.6981=75.3019D_0 \approx 100 - 24.6981 = 75.3019. The put is worth 78.415894⋅0.254473−100⋅0.168407≈3.114078.415894 \cdot 0.254473 - 100 \cdot 0.168407 \approx 3.1140, and 78.4159−3.1140=75.301978.4159 - 3.1140 = 75.3019, the same debt value. The risk-neutral default probability is Φ(−d2)≈0.254473\Phi(-d_2) \approx 0.254473. The yield is −ln⁡(75.3019/80)≈0.060522-\ln(75.3019/80) \approx 0.060522, so the spread over rr is about 0.0405220.040522, or about 405 bp.

  1. An analyst reports the number 0.2544730.254473 from the previous question as the probability that the firm defaults within the year. Under which hypothesis would that be right? Compute the model’s physical default probability when the asset drift is μ=0.08\mu = 0.08.
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The number Φ(−d2)\Phi(-d_2) is the default probability under the pricing measure Q\mathbb{Q}, where the assets grow at the drift rr. It equals the physical probability only when μ=r\mu = r, that is, when the market price of risk θ=(μ−r)/σ\theta = (\mu - r)/\sigma is zero. With μ=0.08\mu = 0.08,

d2P=0.660479+0.06⋅10.3=0.860479,\begin{aligned} d_2^{\mathbb{P}} &= 0.660479 + \frac{0.06 \cdot 1}{0.3} \\[0.4em] &= 0.860479 , \end{aligned}

and P(A1<80)=Φ(−0.860479)≈0.194763\mathbb{P}(A_1 < 80) = \Phi(-0.860479) \approx 0.194763, smaller than the risk-neutral 0.2544730.254473 because μ>r\mu > r. Both numbers are model outputs under the stated hypotheses, and neither one divided by TT equals the spread of about 0.0405220.040522, which also reflects how much the creditors lose when default occurs.

How this lesson connects