Malz Chapter 6. Log Math

[NNath]

Hi, Anyone who can solve this ?

In Chapter 6 page 217 Malz calculates value of spread as

Substituting the current market value of the debt (blue), we have

[D e^-rt − Put ]e^yt = D ---- A

so after taking logarithms, we have

y = 1/t log [(1 − e^-rt) D + put] --- B

I tried everything but I cannot get from A to B. Even tried B to A.

 

[ami44]

You can't get from A to B because there is no way. I don't know who made an error, but if you insert a plus sign into the first formula instead of a multiplication, you can get to B:
( D * e^-rt - Put ) + e^yt = D
=> e^yt = D - ( D * e^-rt - Put)
=> e^yt = D * (1 - e^-rt) + Put
=> y = 1/t * ln( D * (1 - e^-rt) + Put )

So, if i made no error, expression A with a plus inserted is equal to expression B. That means on the other hand, the original expression A with multiplication can not be equivalent to B.
But I did not look into Malz, so I don't know who is in the wrong here.

 

[QuantMan2318]

Dear @NNath, I don't have the exact formula that you have written, however, I may be able to explain the calculation of the spread of a corporate bond like how Malz intended.

Take a Zero Coupon Corporate Bond as D, the present value of the ZCB Corporate is D(Present) = D*e^(-r-z)*t where z is the spread that has to be further subtracted from the risk free rate r to get the higher discount rate for a corporate bond. We should now find z.

This present value is equal to the PV of the payoff we get when we take the weighted sums of the amount of the cash flows received when there is a default and when there is no default. Therefore, the default probability is 1-e^(-lambda)*t and the survival probability is e^(-lambda)*t, so the value of the Bond is

D*(e^(-lambda)*t)+ R*(1-e^(-lambda)*t), where R is the Recovery Amount in the event of default, discount this pay off at risk free rate r, to get the same thing as D(Present)
e^(-r*t)*[D*(e^(-lambda)*t)+R*(1-e^(-lambda)*t)] = D*e^(-r-z)*t ;

e^(-r*t)*[D*(e^(-lambda)*t)+R*(1-e^(-lambda)*t)] = D*e^(-r*t)*e^(-z*t);

taking logs on both sides

-r*t +Ln(D*(e^(-lambda)*t)+R*(1-e^(-lambda)*t))) = Ln(D) +(-r*t) + (-z*t) ;

-z*t = Ln(D*(e^(-lambda)*t)+R*(1-e^(-lambda)*t))) - Ln(D);

(where Malz uses D as $1 therefore the last term on the RHS becomes close to zero and only the first term on the RHS remains )

therefore,

z = (-1/t)*Ln(D*(e^(-lambda)*t)+R*(1-e^(-lambda)*t))) - Ln(D)

This is what Malz calculates using the Logs and Hazard rates. Perhaps this is equivalent to the formula that you have written

 

[NNath]

Thanks @ami44 and @QuantMan2318 , There may be some adjustment or approximation made and plus instead of multiplication not sure. @David Harper CFA FRM could you think of anything. Its a mistake made by Malz in expressing the spread.