# Why “ Even if the underlying asset price remains unchanged, the option delta for an in-the-money option increases as expiration nears”

I saw this line on some website but can not understand it. Can anyone explain it?

"Even if the underlying asset price remains unchanged, the option delta for an in-the-money option increases as expiration nears; the opposite is true for an out-of-the-money option."

In the context the simplest BSM equation, the delta has the formula:

$$\Delta = \frac{1}{\sqrt{2 \pi}} \int_{\frac{1}{\sigma\sqrt{T}} \left( \ln \left( \frac{K}{S_0} \right) - \left( r + \frac{\sigma^2}{2} \right) T \right)}^{+\infty}\exp \left( -\frac{x^2}{2} \right) \mathrm{d}x.$$

By doing some mathematical analysis, when ITM, $\Delta$ shall increase when $T$ decreases only if

$$T\leq\frac{ \left( r + \sigma^2/2 \right)}{\ln \left( S_0 / K \right)}.$$

When OTM, $\Delta$ shall decrease whenever $T$ decreases, unconditionally. So this result is different from what I saw online and it confuses me.

• I think the statement should read "Even if the underlying forward asset price remains unchanged, the option delta for an in-the-money option increases as expiration nears; the opposite is true for an out-of-the-money option." – dm63 Jan 21 '17 at 20:19
• try recasting it in terms of the forward asset price F – dm63 Jan 21 '17 at 20:19

## 3 Answers

@Kiwiakos gave you the intuition. Here is the corresponding analysis that you asked for. The European plain vanilla call delta is given by

\begin{equation} \frac{\partial C_0}{\partial S_0} = \mathcal{N} \left( d_+ \right), \end{equation}

where

\begin{equation} d_+ = \frac{1}{\sigma \sqrt{T}} \left( \ln \left( \frac{S_0}{K} \right) + \left( r - \frac{1}{2} \sigma^2 \right) T \right) \end{equation}

Differentiating again w.r.t. $T$ yields

\begin{eqnarray} \frac{\partial^2 C_0}{\partial S_0 \partial T} & = & \mathcal{N}' \left( d_+ \right) \frac{\partial d_+}{\partial T}\\ & = & \mathcal{N}' \left( d_+ \right) \frac{1}{2 \sigma T \sqrt{T}} \left( -\ln \left( \frac{S_0}{K} \right) + \left( r - \frac{1}{2} \sigma^2 \right) T \right). \end{eqnarray}

This term is negative (delta is increasing as the time-to-maturity becomes shorter) if

\begin{equation} -\ln \left( \frac{S_0}{K} \right) + \left( r - \frac{1}{2} \sigma^2 \right) T < 0 \qquad \Leftrightarrow \qquad S_0 > K \exp \left\{ \left( r - \frac{1}{2} \sigma^2 \right) T \right\} := S^*(T) \end{equation}

This is in accordance with the formula you provided. In the limit we have

\begin{equation} \lim_{T \downarrow 0} S^*(T) = K \end{equation}

as expected.

Regarding the statement that you quoted that says that delta increases for in-the-money options as the time-to-maturity decreases. This is roughly true as you see from the above though not fully accurate.

Note that the option delta is given by $\Delta= N(d_1)$, where \begin{align*} d_1 = \frac{\ln\frac{S_t}{K}+\big(r+\frac{\sigma^2}{2}\big)(T-t)}{\sigma\sqrt{T-t}}. \end{align*} Let $\tau = T-t$ be the residual option maturity. Then, for $K <S_t$, \begin{align*} \frac{\partial d_1}{\partial \tau} &= \frac{(r+\frac{\sigma^2}{2}\big) \tau - \ln\frac{S_t}{K}}{2\sigma \tau^{\frac{3}{2}}}. \end{align*} Therefore, as expiration nears, or as $\tau \rightarrow 0$, $\frac{\partial d_1}{\partial \tau} < 0$, that is, the option delta increases as the residual option maturity $\tau$ decreases (i.e., expiration nears).

• Hi Gordon thanks for the answer. I can only mark one correct answer but your answer is correct to me too. – Ethan Jan 22 '17 at 17:34
• Thanks @Ethan. My answer is similar to that of LocalVolatility and also employed insights from Kiwiakos (i.e, $S_t$ does not enough time to be smaller than $K$. – Gordon Jan 22 '17 at 17:48

If the call is ITM, ie $K<S$, as expiry approaches the likelihood that the option will be exercised increases, as there is now less time for it to go OTM. Delta is the position that the hedger is holding, therefore she will need to hold more of the stock.

• This explanation sounds intuitive and reasonable. But some of my analysis must have gone wrong. When I look at the mathematical formula of $\Delta$, I can not find a mathematical deduction of this result. But your explanation makes sense to me. – Ethan Jan 21 '17 at 18:35