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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?

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    $\begingroup$ The evolution of the delta may not be monotonic as a function of the remaining time to maturity $T$. In your example assuming no carry (zero risk-free rate, no divs), one can show that $\Delta$ first decreases (roughly up to $T=30$), then increases to reach the asymptotic value of $1$. $\endgroup$
    – Quantuple
    Sep 14, 2017 at 8:38

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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).

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    $\begingroup$ Something wrong with the math here I think... Please check. $\endgroup$
    – nbbo2
    Sep 14, 2017 at 8:07
  • $\begingroup$ @noob2 only thing I could find wrong was misplaced factor of $T$ in second formula. If you think something else is wrong let me know why. $\endgroup$
    – spaceisdarkgreen
    Sep 14, 2017 at 13:44
  • $\begingroup$ Thanks. Now I don't completely understand how $d_2$ showed up in the last formula. $\endgroup$
    – nbbo2
    Sep 14, 2017 at 14:10
  • $\begingroup$ @noob2 Sorry, edited again. I did not get it right the second time either. It's just from $d_2 =( \log(S/K) -\frac{1}{2}\sigma^2T)/\sqrt{\sigma^2T}.$ I wouldn't have even added the second equation but I though it was neat that it would switch from increasing to decreasing in $T$ right when the probability of finishing in the money hit fifty percent. $\endgroup$
    – spaceisdarkgreen
    Sep 14, 2017 at 14:52

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