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I put together some charts to understand how option greeks change. Can someone please explain why the delta of an ATM call option is 1 when vol is close to zero?

I get that an increase in vol for OTM or ITM options will increase the chance that they expire ITM or OTM. But what is the explanation for how the delta of an ATM option changes with volatility?

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  • $\begingroup$ It is a time value of money effect. With no volatility a stock with $S_0=X$ will appreciate at $r$ and end up ITM at $T$. If you set $r=0$ the ITM delta will go to $\frac{1}{2}$. Alternatively if you let $T$ go to zero the effect also disappears. $\endgroup$ – noob2 Nov 30 '18 at 2:54
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I don't know how complex is your model, but the intuitive answer is that if you have 0 vol and a positive drift (interest rate > dividend rate) your option is actually "in the money forward."

Interestingly an ATM option with extremely high volatility will also have a delta close to 1. That's a classic interview question btw

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We need to distinguish between centered delta and non-centered values like $\delta^+$ obtained using a small positive bump in price.

When $\sigma \rightarrow 0 $, $\delta^+ = 1$, $\delta^- = 0$.

So your graph looks consistent with $\delta^+$, the delta for a small positive perturbation.

The centered delta for an ATM call is $\delta=1/2$.

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