Black Scholes $ ( \;)^+$ Notation Question Came across an exercise involving the Black Scholes formula which uses:
$E[(S_{T}-K)^{+}|F_{t}]$
What is the $\;(\;)^{+}\;$ in the expression? Is this some kind of operator?
Thanks.
 A: The expression $(x)^+$ is defined as $(\bullet)^+ : x \mapsto \max(0,x)$ for $x\in\mathbb{R}$.
A: $x^+$ 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 $(S_T - K)^+$.
A: 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.
A: It simply tells you that you take the positive payoff, that is the difference between current price and the strike price, otherwise if you get a negative, the result is zero. When you are dealing options, it shows you that when $S_T<K$, at time $T$ the option is worthless. 
