Pricing of an Aggregate Stop-Loss contract Pricing of an Aggregate Stop-Loss contract when aggregate losses are lognormally distributed.

Assumptions

  1. Stock price distribution ST at time T from now is lognormally distributed; log(ST) has mean m and standard deviation s. This means the density function for ST is given by
    f(x) = (2p)-1/2(sx)-1exp æ
    ç
    è     		- 1
    2
    		 æ
    ç
    è     		ln(x)-m
    s
    		 ö
    ÷
    ø     		2

     
    		 ö
    ÷
    ø
  2. The aggregate stop loss attaches at k
  3. The aggregate stop loss pays max(ST-k,0) exactly T years from now.
  4. The discounting interest rate is r¢.

The nominal pure premium M for this contract can be computed as follows:

M
=
E( max
(ST-k,0))
=
ó
õ 		¥

k 
 (x-k)f(x)dx
=
(2p)-1/2s-1 ó
õ 		¥

k 
 (x-k)exp(-(ln(x)-m)2/2s2)dx/x
=
(2p)-1/2s-1 ó
õ 		¥

ln(k) 
 (ey-k)exp(-(y-m)2/2s2)dy
=
(2p)-1/2s-1 ó
õ 		¥

ln(k) 
 exp(-1/(2s2)[y2-2y(s2+m)+m2])dy
-k(2p)-1/2 ó
õ 		¥

(ln(k)-m)/s 
 exp(-z2/2)dz
=
(2p)-1/2s-1 ó
õ 		¥

ln(k) 
 exp(-1/(2s2)[y2-(s2+m)]2+(s2/2+m))dy
-k(1-F((ln(k)-m)/s))
=
exp(m+s2/2)(1-F((ln(k)-(m+s2))/s))-k(1-F((ln(k)-m)/s))
=
exp(m+s2/2)F((ln(1/k)+(m+s2))/s)-kF((ln(1/k)+m)/s)
To conclude, we have shown that the expected value of a lognormal(m, s) random variable, limited at k is given by
M(k) = exp(m+s2/2)F((ln(1/k)+(m+s2))/s)-kF((ln(1/k)+m)/s).
(1)

Derivation of Black Scholes formula-assuming the hard part

For the Black Scholes formula the analogs are:

  1. Stock price distribution ST at time T years from now is lognormally distributed. Precisely, the theory assumes that incremental returns over a short period of time dt are normally distributed with mean mdt and standard deviation sÖ{dt}. These assumptions imply that log(ST) has mean mT - s2 T/2 and standard deviation sÖT. The density function for ST is given by
    f(x) = (2p)-1/2(sx)-1 exp æ
    ç
    è     		- 1
    2
    		 (ln(x)-(mT-s2T/2))2
    s2T
    		 ö
    ÷
    ø
  2. The option exercise price is k
  3. The current stock price is S0
  4. The option pays max(ST-k,0) at expiration, T years from now.
  5. The discounting interest rate is r¢.

First, ignore the discounting factor e-r¢T. Next note that the nominal pure premium M is given by

M : = E( max
(ST-k,0)) = S0 E( max
(ST/S0-k/S0,0)).
Now substituting
m¬ m-s2T/2,
k¬ k/S0,
and
s¬ sÖT/2
into (1) above gives
emTS0F(ln(S0/k)+(m+s2/2)T/sÖT) -kF(ln(S0/k)+(m-s2/2)T/sÖT).

We can now distinguish two situations. In the actuarial (realistic) situation, the stock is assumed to earn an instantaneous return m > r, where r is the risk free rate of return. Moreover, since the option is risky, the actuary would discount at an interest rate r¢ greater (buying) or less (selling) than the risk free rate. This gives the following ``actuarial'' price for the call option:

e(m-r¢) TS0F(ln(S0/k)+(m+s2/2)T/sÖT)
-e-r¢TkF(ln(S0/k)+(m-s2/2)T/sÖT).

Comparing this equations with the Black Scholes result shows that the latter is assuming

  1. Discount at the risk free rate of return: r¢ = r
  2. Stock earns the risk free rate of return: m = r

Making these two substitutions yields

S0F(ln(S0/k)+(r+s2/2)T/sÖT)
-e-rTkF(ln(S0/k)+(r-s2/2)T/sÖT).
which is exactly the Black Scholes formula.

Why does the expected return change?

Black Scholes assumes that the stock price satisfies the stochastic differential equation
dSt = mSt dt + sSt dWt
where Wt is a Brownian motion. Roughly, this means dWt is normally distributed with mean zero and standard deviation Ö[dt].

If the price were deterministic and satisfied

dSt = mSt dt
then the price process would be given by
St = exp(mt).
Adding the stochastic component results in the somewhat surprising solution
St = exp((m-s2/2)t + sWt),
which is not quite what you would expect. To see this is the correct solution, differentiate, to get:
dSt = S
t
 dt + S
W
 dWt + 1
2
 2 S
W2
 (dWt)2 + ... .
The omitted higher order terms are all smaller than dt and can be omitted. However, the magic of Brownian motion is that (dW)2 = dt: BM moves randomly, but at the same speed all the time-see discussion below. The claimed result is now a straight-forward substitution given:
S
t
 = (m-s2/2) St,
S
W
 = sSt
and
2 S
dW2
 = s2 St.

Fundamental Fact about Brownian Motion

It is a fundamental fact about Brownian motion that (dWt)2 = dt and since it is so important in the derivation of Black-Scholes, here is an indication of why it is true. The notation is actually shorthand for ò0t(dWt)2 = t, where the integral is computed as
lim
 n
å
 k = 1 
 (Wtk-Wtk-1)2
with the limit taken over finer and finer partitions {0 = t0 < ¼ < tn = t} of [0,t], see Panjer Section A.4 or Karatzas and Shreve (all references are in the paper). Without loss of generality we may assume that {tk} is an evenly spaced partition of [0,t], since any partition has an evenly spaced refinement. Suppose tk = kt/ n. Then Wtk-Wtk-1 is normally distributed with mean zero and variance t/n, from the definition of Brownian motion. Thus
Wtk-Wtk-1 =   ___
Öt/ n
 
 Z
where Z is a standard normal random variable and so
(Wtk-Wtk-1)2 = t/ nc
where c is chi-squared with one degree of freedom. The characteristic function of c is (1-2is)-1/2. Therefore the characteristic function of the sum above is given by
(1-2ist/n)-n/2
which tends to exp(ist) as n®¥, since (1+x/n)n®exp(x) as n®¥. This proves the result because exp(ist) is the characteristic function of a random variable which takes the value t almost certainly.

File translated from TEX by TTH, version 2.34.
On 30 Apr 2000, 11:55.