# When Will N(d1) Become the Probability of Exercising an Option?

A proof by comparing change of measure with the Black-Scholes formula.

Written on September 26, 2018

Consider the value of a vanilla European call option:

\begin{align} Call(0) &= \mathbb{ E^Q } \left[ \frac{ { \left [ S(T)-K \right ] }^{ + } }{ B(T) } \right] \\ &=\mathbb{ E^Q } \left[ \frac{ { \left [ S(T)-K \right ] } { \textbf{ 1 } _ { S(T)>K } } }{ B(T) } \right] \\ &={ \mathbb{ E^Q } \left[ \frac{ { S(T) } { \textbf{ 1 } _ { S(T)>K } } }{ B(T) } \right] } - { \mathbb{ E^Q } \left[ \frac{ { K } { \textbf{ 1 } _ { S(T)>K } } }{ B(T) } \right] } \\ &={ \mathbb{ E^Q } \left[ \frac{ { S(T) } { \textbf{ 1 } _ { S(T)>K } } }{ B(T) } \right] } - { K \ \mathbb{ E^Q } \left[ \frac{ { \textbf{ 1 } _ { S(T)>K } } }{ B(T) } \right] } \\ &={ \mathbb{ E^Q } \left[ \frac{ { S(T) } { \textbf{ 1 } _ { S(T)>K } } }{ B(T) } \right] B(0) } - { K \ \mathbb{ E^Q } \left[ \frac{ { \textbf{ 1 } _ { S(T)>K } } }{ B(T) } \right] B(0) } \\ &={ \mathbb{ E^S } \left[ \frac{ { S(T) } { \textbf{ 1 } _ { S(T)>K } } }{ S(T) } \right] S(0) } - { K \ \mathbb{ E^{ Q^T } } \left[ \frac{ { \textbf{ 1 } _ { S(T)>K } } }{ P(T,T) } \right] P(0,T) } \\ &={ \mathbb{ E^S } \left[ { \textbf{ 1 } _ { S(T)>K } } \right] S(0) } - { K \ \mathbb{ E^{ Q^T } } \left[ { \textbf{ 1 } _ { S(T)>K } } \right] P(0,T) } \\ &={ \mathbb{ P^S } \left[ S(T)>K \right] S(0) } - { K \ \mathbb{ P^{ Q^T } } \left[ S(T)>K \right] P(0,T) } \\ &={ \mathbb{ P^S } \left[ S(T)>K \right] S(0) } - { \mathbb{ P^{ Q^T} } \left[ S(T)>K \right] K P(0,T) } \end{align}

Assuming constant interest rate:

$P(0,T)=e^{ -rT }$

we have:

$Call(0)={ \mathbb{ P^S } \left[ S(T)>K \right] S(0) } - { \mathbb{ P^{ Q^T } } \left[ S(T)>K \right] K e^{ -rT } }$

Compare this with the Black-Scholes formula:

$Call(0)={ N(d_1)S(0) }-{ N(d_2)K e^{ -rT } }$

Since both formulas hold true for any choice of the parameters, we have:

$N(d_1)=\mathbb{ P^S } \left[ S(T)>K \right]$ $N(d_2)=\mathbb{ P^ { Q^T } } \left[ S(T)>K \right]$

In other words, $N(d_1)$ is the probability of exercising the option under the stock measure, while $N(d_2)$ is the probability of exercising the option under the T-forward measure (and of course, the risk-neutral measure as well).