How Time to Expiration Can Raise or Lower an ITM Call's Delta

Summary

The document analyzes how a European in-the-money call's Black–Scholes delta changes when time to expiration increases. Delta is the standard normal cumulative probability evaluated at d1, so its sensitivity to time depends on how d1 changes, not simply on the greater chance that the option could move out of the money. Differentiating d1 gives a condition involving spot relative to strike, the risk-free rate, and volatility.

When the rate is zero, the derivative can be expressed using d2; its sign is negative when the risk-neutral probability of finishing in the money exceeds one half. With nonzero rates, the threshold is adjusted. Thus, longer maturity does not imply a universal direction for delta. The analysis assumes the Black–Scholes framework and fixed spot, strike, rate, and volatility; it distinguishes delta from the probability of expiring in the money.

Key ideas

• An increase in time to expiration does not always move an in-the-money call's delta in the same direction.
• The time sensitivity of delta follows from differentiating d1 in the Black–Scholes formula.
• With a zero interest rate, the derivative is negative when the risk-neutral probability of finishing in the money is above one half.
• Nonzero rates modify the condition, and the result assumes other model inputs remain fixed.

Tags

# ITM call delta when T increases


# ITM call delta when T increases












for an expiring European in-the-money (ITM) call (delta = 0.9), if $T$ increases from 1 to 30, what should delta be now?

Let's say $K = 100$, $S_0 = 105$, $\sigma = 10%$.

Intuitively I think the delta would decrease since now there is more time for the option to move out of money, and we are less certain it would end ITM.

However BS pricing formula's $\mathcal{N} \left( d_1 \right)$ section seem to suggest that increased T would have a positive effect?

## Answer by spaceisdarkgreen (score 1)

https://quant.stackexchange.com/a/36027

Remember $$ d_1 = \frac{\log(S/K) + \left(r+\frac{1}{2}\sigma^2\right)T}{\sigma\sqrt{T}}$$ so the first term is decreasing in $T.$

Let's take the derivative: $$\frac{\partial \delta}{\partial T} = N'(d_1)\frac{\partial d_1}{\partial T} = \frac{e^{-d_1^2/2}}{\sqrt{2\pi}}\frac{1}{2\sigma\sqrt T}\left(-\frac{1}{T}\log (S/K) + r+\frac{1}{2}\sigma^2\right)$$

We see that it's possible for this quantity to be negative and it's less positive / more negative (for everything else fixed) when $S/K$ becomes large.

Also observe that for $r=0$ we can rewrite it as: $$ \frac{\partial \delta}{\partial T}=-\frac{e^{-d_1^2/2}}{\sqrt{2\pi}}\frac{1}{2T}d_2.$$

Recall that $N(d_2)$ is the probability of finishing in the money, so we see the derivative is negative whenever there is a more than $50\%$ chance of finishing in the money (when $r$ is non-zero, this is adjusted a bit).

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

This summary was written by Stratmill's research agent from the original; it is not a copy of the source.