So I want to calculate the effect of default risk on the required interest of bonds (not on the price of bonds as that is normalized to one).

I thought of using the consumption capital asset model and calculating the following.

$$ U = u(C_1) + \beta (1-p)E_tu(C_2) + \beta p u(C_2) $$

Where the following applies:

$$ p = no \; default ,\; (1-p) = default $$

and the following budget constraints apply:

$$C_1 = Y_1 - T_1 - S_1$$ $$C_2 = Y_2 - T_2 + (1+r)S_1$$

I got to the following result using Lagrange, such that the consumer maximizes utility by deciding on how much to invest:

$$1 = \frac{\beta E_t[U'(C_2)p(1+r)]}{(u'(C_2)}$$

However, I want to get it ultimately in an equation that uses covariances but I do not know how.


1 Answer 1


To make sense of your question, your $U$ should be $$ U = u(C_1) + \beta (1-p)E_1 [ u(C_2) ] + \beta p E_1[u(Y_2 - T_2)]. $$ (This definition of $U$ assumes that the default event is independent of other sources of uncertainty in the economy.)

If the bond defaults, saving $S_1$ from the first period is wiped out. If there is uncertainty in period-2 production/endowment/transfer/etc, then period-2 consumption is still random in the event of default.

Assuming that agent only chooses consumption $(C_1)$ (therefore $C_2$ in the event of no default), then real interest rate $r$ is determined by $$ 1 = E_1[\beta \frac{u'(C_2)}{u'(C_1)} (1-p) (1 + r)]. $$ This is just an extension of the standard Euler equation for interest rate $$ 1 = E_1[\beta \frac{u'(C_2)}{u'(C_1)} (1 + r)] $$ when $p = 0$.

If equilibrium consumption remains the same (e.g. equals exogenous endowment), then higher probability of default $p$ leads to higher interest rate $r$, as you would expect. High probability of default incentivizes agent to consume today instead of saving for tomorrow. To keep him from borrowing and clear the bond market, equilibrium interest rate must be higher. Equivalently, higher default risk means agent demands higher liquidity premium on bonds.

If $p$ and $r$ are small, $(1-p)(1+r) \approx e^{r-p}$, which implies that $r$ moves one-to-one with respect to $p$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.