Equity Requirement using Merton Model / 7.b.1

[dthigale]

Hi David,

Could you please let me know what formula did you use to solve for equity needed.

I was trying to understand cell # 14.

Especially, I donot find a formula where we use the confidence level / critical z value

** Hopefully it will be helpful if we have the actual formulae on the worsheets.

I tried using the KVM Credit Monitor formula for PD since we know the PD ( here we donot use confidence level) . However I get different answer.

I may need your help checking the steps. That will be useful when solving other problems / reverse enginnering the formala where they have LN / N functions)

PD = N(( - ln(V/F) + (mu - 1 /2 Sigma^2) T) / sigma sqrtT)

3.4 = N ( - LN(V/10) + ( 0.05 - 1/2 (0.2)^2) / .2) : Here I am getting rid of N (1/ 0.2) = N (5) = apprx 1

3.4 = N (- LN (V/10) + .03)

LN 3.4 = - LN (V/10) + 0.03 << Am I right here in calculations to get rid of N on the right hand side?

1.22 = - LN (V/10) + 0.03

1.19 = - LN V- LN 10

1.19 = - LN V - 2.30

LN V = 1.11

V = exp (1.11) << Is this right?

V = 3.03. Your answer 13.98. I am sure I am making a big mistake somewhere...

Thanks David as always.

D.

 

[David Harper CFA FRM]

Hi Dinesh,

I think the mistake is where you take natural log, LN, to undo the N(.). The N(.) is standard normal cumulative, so its inverse is NORMSINV(). It's a good workout, that's for sure! Keeps us mindful of:
LN & EXP are inverses, and
NORMSINV() and NORMSDIST() are inverses

As in:
PD = N(-(ln(V/F)+ (mu - 0.5*sigma^2)*T) / (sigma *SQRT(T)))
PD = N(-(ln(V/10)+ (5% - 0.5*20%^2)*1) / (20% *SQRT(1)))
PD = N(-(ln(V/10)+ 3%*1) / (20%)), so that
3.4% = N(-(ln(V/10) + 3%) / (20%))

take inverse normal cumulative distribution (not natural log) of both sides:
N-1(3.4%) = -(ln(V/10) + 3%) / (20%)
N-1(3.4%) * 20% = -(ln(V/10)+ 3%)
-1* (N-1(3.4%) * 20% + 3%) = ln(V/10) + 3%
-1* (N-1(3.4%) * 20% + 3%) - 3% = ln(V/10)

now take exponential:
EXP (-1* (N-1(3.4%) * 20% + 3%) - 3% ) = V/10
EXP (-1* (NORMSINV(3.4%) * 20% + 3%) - 3% ) * 10 = V
EXP (-1* (NORMSINV(3.4%) * 20% + 3%) - 3% ) * 10 = $13.98

Hope this helps, David

 

[dthigale]

Hi David,

Thanks a lot. Yes its some heavy workout for me. I will try to digest it slowly.

In the NeededEquity worksheet of 7.b.1 cell # 14 you use confidence level / z value of 1.83.

LN (Firm Value) ( Cell#14) = =H13*H12*SQRT(H11)+LN(H9)-H10*H11+0.5*H12^2*H11)

Could you please give the original formula?

I suppose you are getting Z value from PD. Your calculations seem much easier compared to KVM Moodys formula.

Thanks again.

D.

 

[David Harper CFA FRM]

Hi Dinesh,

Since the normal is symmetric, I used N(1-PD) = N(1-3.4%) = 1.83 deviate; i.e., to "the right of mean"
note that's essentially the same as N(3.4%) = -1.83 so that N (-PD) = 1.83
(this is the only "semantic" difference between Stulz and de Servigny)

re: KMV Moody's, I am not current on their method, but I wouldn't expect their PD to be a simple function of the normal deviate. First, i don't think they do a parametric translation; Second, it's not heavy tailed. This is *Merton* not KMV (which is merely Merton-based)...the part company at translation of the deviate into a PD

Regarding cell 14, I used the same formula to solve for firm value:
N(1-PD) = (LN(V/F)+ (mu - 0.5*sigma^2)*T) / (sigma *SQRT(T))
N(1-PD) * (sigma *SQRT(T)) = (LN(V/F)+ (mu - 0.5*sigma^2)*T)
N(1-PD) * (sigma *SQRT(T)) - mu*T + 0.5*sigma^2*T = LN(V/F)
N(1-PD) * (sigma *SQRT(T)) - mu*T + 0.5*sigma^2*T = LN(V) - LN(F)
N(1-PD) * (sigma *SQRT(T)) - mu*T + 0.5*sigma^2*T - LN(F) = LN(V)
so EXP(...the expression above ...) = V = 13.98

David