# Why do ATM call options have a delta of slightly bigger than 0.5 and not 0.5 exactly?

From the formula of the delta of a call option, i.e. $N(d1)$, where $d_1 = \frac{\mathrm{ln}\frac{S(t)}{K} + (r + 0.5\sigma^2)(T-t)}{\sigma\sqrt{T-t}}$, the delta of an ATM spot call option is slightly bigger than 0.5. However, this is unintuitive to me... can anyone explain why?

Also, is there any way to interpret what does $d_1$, and $d_2$ represent where $d_1$ is shown above and $d_2 = d_1 - \sigma(T-t)$

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Did you have look at: en.wikipedia.org/wiki/Black%E2%80%93Scholes#Interpretation? – Bob Jansen Nov 23 '12 at 16:16
Did you compute the delta numerically or did you use a close form solution ? Can you provide the details of your computation. Did you try to compute your delta with $r=0$ ? Is it simply because of the discounting on the strike ? – BlueTrin Nov 23 '12 at 16:44

In the no-arbitrage pricing the log return of the stock price does not have expected return $0$ but $r$, the risk free rate. This is strongly related to the pricing of forward contracts. There you could follow the steps to see that in the arbitrage free world the spot price grows with the risk-free rate in expectation.

Thus if you price an option then the probability (in the martingale measure) that the log return is positive is greater than $1/2$ if there are positive interest rates. If you calculate with a dividend yield then this yield is substracted from the risk-free rate.

All the things that I have said hold for the log-return. If you take the exponential: $$S_0 \exp( X_t ) = S_0 (1 + X_t + \frac12 X_t^2 + \cdots)$$ where $X_t$ is the log return process, and take the expectation then you get the terms $E[X_t] = r t$ for $E[X_t^2]/2 = t \sigma^2/2$ considering terms up to $2nd$ order.

In the Bachelier model, where the stock price is modelled as arithmetic Brownian motion, there you don't have this $\sigma^2$ term.

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The reason you find it counter intuitive is that because you think of it as the probability of ending in the money at the maturity which is not exactly right. Even with a interest rate of $0\%$ where the stock has no tendency to rise or fall you see that delta is slightly higher than $0.5$ for calls ( $\Delta = \sigma * \sqrt T$ ) and here is the reason:

Think of delta as number of shares to hedge. Even when $r=0\%$ because the stock moves with a geometric brownian motion higher stock prices have larger movements so in order to hedge against a call you need more shares than hedging against a put.

Alternatively, for dual delta which is the probablity of ending in the money you see that for at the money options where $r=0\%$ dual delta is $0.5$

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Using r=0 is a great simplification that shows the real "culprit" behind the greater than 0.5 delta. – Chan-Ho Suh Nov 24 '12 at 15:31