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Came across an exercise involving the Black Scholes formula which uses:


What is the $\;(\;)^{+}\;$ in the expression? Is this some kind of operator?


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up vote 2 down vote accepted

$x^+$ is the "positive part" of $x$: $x^+ = \max\{0, x\}$. That is, $x^+ = x$ when $x$ is positive, and $x^+ = 0$ when $x$ is negative.

The reason why it appears here is this: suppose you hold a call option on a stock with strike price $K$, and at expiration the stock price is $S_T$. If $S_T \ge K$ (i.e. $S_T - K \ge 0$) then you can exercise the option, buy a share at the strike price $K$ and immediately sell it at the market price $S_T$, for a profit of $S_T - K$. But if $S_T < K$ (i.e. $S_T - K < 0$) then it would be stupid to exercise the option because you would lose money. So you should instead let it expire, and your profit is 0. Thus we can combine these two cases and write the profit as $(S_T - K)^+$.

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The expression $(x)^+$ is defined as $(\bullet)^+ : x \mapsto \max(0,x)$ for $x\in\mathbb{R}$.

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When deriving the BS equation it might be useful to keep in mind that:

$$ (S_T-K)^+=(S_T-K)1_{S_T{\ge}K} $$ Where: $$ 1_{S_T{\ge}K}=\begin{cases}1,&\text {if}&S_T~{\ge}~K\\0,&\text{if}&S_T<K\end{cases} $$

This makes it simpler to find the expectation in your question.

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