dstefan
07-01-2008, 05:48 PM
Chapter 4, Problem 10
Use risk--neutral pricing to price a supershare, i.e., an option that pays \left( \max(S(T)-K,0) \right)^2 at the maturity of the option.
In other words, compute
V(0) ~=~ e^{-r T} ~E_{RN}[\left( \max(S(T)-K,0) \right)^2]
where the expected value is computed with respect to the risk-neutral distribution of the price S(T) of the underlying asset at maturity T, which is assumed to follow a lognormal process with drift r and volatility \sigma. Assume that the underlying asset pays no dividends, i.e., q=0.
Solution:
Recall that
S(T) ~=~ S(0) \mbox{exp}\left(\left( r-\frac{\sigma^2}{2} \right)T + \sigma \sqrt{T} Z\right)
and note that
S(T) ~\geq~ K \quad \Longleftrightarrow \quad Z ~\geq~ \frac{\ln \left( \frac{K}{S(0)} \right) - \left( r - \frac{\sigma^2}{2} \right)~T}{\sigma \sqrt{T}} ~=~ - d_2
Then,
\label{ch4_ex10_1}
C(0) = e^{-rT} \frac{1}{\sqrt{2 \pi}} \int_{-d_2}^{\infty} \left( S(0) \mbox{exp}\left( \left( r-\frac{\sigma^2}{2} \right)T + \sigma \sqrt{T} x \right) - K \right)^2 e^{-\frac{x^2}{2}} dx
= K^2 e^{-rT} \frac{1}{\sqrt{2 \pi}} \int_{-d_2}^{\infty} e^{-\frac{x^2}{2}} ~dx -~ 2 K S(0) \frac{e^{-rT}}{\sqrt{2 \pi}} \int_{-d_2}^{\infty} \mbox{exp}\left( \left( r-\frac{\sigma^2}{2} \right)T + \sigma \sqrt{T} x - \frac{x^2}{2}\right) ~dx +~ S^2(0) \frac{e^{-rT}}{\sqrt{2 \pi}} \int_{-d_2}^{\infty} \mbox{exp}\left( (2r-\sigma^2) T + 2 \sigma \sqrt{T} x - \frac{x^2}{2} \right) ~dx
When pricing a plain vanilla call using risk-neutrality, we showed that
C_{BS}(0) = S(0) \frac{e^{-rT}}{\sqrt{2 \pi}} \int_{-d_2}^{\infty} \mbox{exp}\left( \left( r-\frac{\sigma^2}{2} \right)T + \sigma \sqrt{T} x - \frac{x^2}{2} \right) dx -~ K e^{-rT} \frac{1}{\sqrt{2 \pi}} \int_{-d_2}^{\infty} e^{-\frac{x^2}{2}} dx = S(0) N(d_1) ~-~ K e^{-rT} N(d_2)
We notice that
e^{-rT} \frac{1}{\sqrt{2 \pi}} \int_{-d_2}^{\infty} e^{-\frac{x^2}{2}} dx ~=~ e^{-rT} N(d_2)
\frac{e^{-rT}}{\sqrt{2 \pi}} \int_{-d_2}^{\infty} \mbox{exp}\left( \left( r-\frac{\sigma^2}{2} \right)T + \sigma \sqrt{T} x - \frac{x^2}{2} \right) dx ~=~ N(d_1) and conclude that
\label{ch4_ex10_4}
C(0) = K^2 e^{-rT} N(d_2) ~-~ 2 K S(0) N(d_1) + S^2(0) \frac{e^{-rT}}{\sqrt{2 \pi}} \int_{-d_2}^{\infty} \mbox{exp}\left( (2r-\sigma^2) T + 2 \sigma \sqrt{T} x - \frac{x^2}{2}\right) dx
The integral from (\ref{ch4_ex10_4}) is computed by completing the square as follows:
\label{ch4_ex10_5}
S^2(0) \frac{e^{-rT}}{\sqrt{2 \pi}} \int_{-d_2}^{\infty} \mbox{exp}\left( (2r-\sigma^2) T + 2 \sigma \sqrt{T} x - \frac{x^2}{2} \right) ~dx = S^2(0) \frac{e^{-rT}}{\sqrt{2 \pi}} \int_{-d_2}^{\infty} \mbox{exp}\left( - \frac{(x - 2\sigma \sqrt{T})^2}{2} ~+~ (2r+\sigma^2)T \right) ~dx
= S^2(0) e^{(r+\sigma^2)T}~\frac{1}{\sqrt{2 \pi}} \int_{-d_2}^{\infty} \mbox{exp}\left( - \frac{(x - 2\sigma \sqrt{T})^2}{2} \right) ~dx = S^2(0) e^{(r+\sigma^2)T} ~\frac{1}{\sqrt{2 \pi}} \int_{-(d_2+2\sigma \sqrt{T})}^{\infty} \mbox{exp}\left(-\frac{y^2}{2}\right) ~dy = S^2(0) e^{(r+\sigma^2)T} N(d_2+2\sigma \sqrt{T})
we used the substitution y = x - 2\sigma \sqrt{T} and the fact that
\frac{1}{\sqrt{2 \pi}} \int_{-a}^{\infty} \mbox{exp}\left(-\frac{y^2}{2}\right) ~dy ~=~ \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^a \mbox{exp}\left(-\frac{y^2}{2}\right) ~dy ~=~ N(a)
From (\ref{ch4_ex10_4}) and (\ref{ch4_ex10_5}), we conclude that
C(0) ~=~ K^2 e^{-rT} N(d_2) ~-~ 2 K S(0) N(d_1) ~+~ S^2(0) e^{(r+\sigma^2)T} N(d_2+2\sigma \sqrt{T})
Use risk--neutral pricing to price a supershare, i.e., an option that pays \left( \max(S(T)-K,0) \right)^2 at the maturity of the option.
In other words, compute
V(0) ~=~ e^{-r T} ~E_{RN}[\left( \max(S(T)-K,0) \right)^2]
where the expected value is computed with respect to the risk-neutral distribution of the price S(T) of the underlying asset at maturity T, which is assumed to follow a lognormal process with drift r and volatility \sigma. Assume that the underlying asset pays no dividends, i.e., q=0.
Solution:
Recall that
S(T) ~=~ S(0) \mbox{exp}\left(\left( r-\frac{\sigma^2}{2} \right)T + \sigma \sqrt{T} Z\right)
and note that
S(T) ~\geq~ K \quad \Longleftrightarrow \quad Z ~\geq~ \frac{\ln \left( \frac{K}{S(0)} \right) - \left( r - \frac{\sigma^2}{2} \right)~T}{\sigma \sqrt{T}} ~=~ - d_2
Then,
\label{ch4_ex10_1}
C(0) = e^{-rT} \frac{1}{\sqrt{2 \pi}} \int_{-d_2}^{\infty} \left( S(0) \mbox{exp}\left( \left( r-\frac{\sigma^2}{2} \right)T + \sigma \sqrt{T} x \right) - K \right)^2 e^{-\frac{x^2}{2}} dx
= K^2 e^{-rT} \frac{1}{\sqrt{2 \pi}} \int_{-d_2}^{\infty} e^{-\frac{x^2}{2}} ~dx -~ 2 K S(0) \frac{e^{-rT}}{\sqrt{2 \pi}} \int_{-d_2}^{\infty} \mbox{exp}\left( \left( r-\frac{\sigma^2}{2} \right)T + \sigma \sqrt{T} x - \frac{x^2}{2}\right) ~dx +~ S^2(0) \frac{e^{-rT}}{\sqrt{2 \pi}} \int_{-d_2}^{\infty} \mbox{exp}\left( (2r-\sigma^2) T + 2 \sigma \sqrt{T} x - \frac{x^2}{2} \right) ~dx
When pricing a plain vanilla call using risk-neutrality, we showed that
C_{BS}(0) = S(0) \frac{e^{-rT}}{\sqrt{2 \pi}} \int_{-d_2}^{\infty} \mbox{exp}\left( \left( r-\frac{\sigma^2}{2} \right)T + \sigma \sqrt{T} x - \frac{x^2}{2} \right) dx -~ K e^{-rT} \frac{1}{\sqrt{2 \pi}} \int_{-d_2}^{\infty} e^{-\frac{x^2}{2}} dx = S(0) N(d_1) ~-~ K e^{-rT} N(d_2)
We notice that
e^{-rT} \frac{1}{\sqrt{2 \pi}} \int_{-d_2}^{\infty} e^{-\frac{x^2}{2}} dx ~=~ e^{-rT} N(d_2)
\frac{e^{-rT}}{\sqrt{2 \pi}} \int_{-d_2}^{\infty} \mbox{exp}\left( \left( r-\frac{\sigma^2}{2} \right)T + \sigma \sqrt{T} x - \frac{x^2}{2} \right) dx ~=~ N(d_1) and conclude that
\label{ch4_ex10_4}
C(0) = K^2 e^{-rT} N(d_2) ~-~ 2 K S(0) N(d_1) + S^2(0) \frac{e^{-rT}}{\sqrt{2 \pi}} \int_{-d_2}^{\infty} \mbox{exp}\left( (2r-\sigma^2) T + 2 \sigma \sqrt{T} x - \frac{x^2}{2}\right) dx
The integral from (\ref{ch4_ex10_4}) is computed by completing the square as follows:
\label{ch4_ex10_5}
S^2(0) \frac{e^{-rT}}{\sqrt{2 \pi}} \int_{-d_2}^{\infty} \mbox{exp}\left( (2r-\sigma^2) T + 2 \sigma \sqrt{T} x - \frac{x^2}{2} \right) ~dx = S^2(0) \frac{e^{-rT}}{\sqrt{2 \pi}} \int_{-d_2}^{\infty} \mbox{exp}\left( - \frac{(x - 2\sigma \sqrt{T})^2}{2} ~+~ (2r+\sigma^2)T \right) ~dx
= S^2(0) e^{(r+\sigma^2)T}~\frac{1}{\sqrt{2 \pi}} \int_{-d_2}^{\infty} \mbox{exp}\left( - \frac{(x - 2\sigma \sqrt{T})^2}{2} \right) ~dx = S^2(0) e^{(r+\sigma^2)T} ~\frac{1}{\sqrt{2 \pi}} \int_{-(d_2+2\sigma \sqrt{T})}^{\infty} \mbox{exp}\left(-\frac{y^2}{2}\right) ~dy = S^2(0) e^{(r+\sigma^2)T} N(d_2+2\sigma \sqrt{T})
we used the substitution y = x - 2\sigma \sqrt{T} and the fact that
\frac{1}{\sqrt{2 \pi}} \int_{-a}^{\infty} \mbox{exp}\left(-\frac{y^2}{2}\right) ~dy ~=~ \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^a \mbox{exp}\left(-\frac{y^2}{2}\right) ~dy ~=~ N(a)
From (\ref{ch4_ex10_4}) and (\ref{ch4_ex10_5}), we conclude that
C(0) ~=~ K^2 e^{-rT} N(d_2) ~-~ 2 K S(0) N(d_1) ~+~ S^2(0) e^{(r+\sigma^2)T} N(d_2+2\sigma \sqrt{T})