# Delta of a standardized at-the-money 30-day put option

The plot below depicts the delta of a standardized at-the-money 30-day put option on the S&P500 tracker SPY over a 14-year period. This is data from OptionMetrics and standardized prices are calculated using linear interpolation from the volatility surface

My question is: Why does delta increase (i.e. decrease in absolute value) during the 2008 financial crisis?

Delta of a put option over time, whose characteristics are constant. I.e. the underlying option characteristics is modeled so that it is perpetually at the money and 30 days from expiry

• I deleted my answer as I'm really not sure how that plot is created and if my answer correctly answers the question. I had assumed the strike equals the share price, but that likely isn't true for this chart and might be leading to the results you are seeing.
– John
Apr 11, 2014 at 15:11
• Actually your assumption that strike equals the share price is true: the underlying option characteristics is modeled so that it is perpetually at the money. - I added some clarification in the question Apr 11, 2014 at 15:13
• Okay, I undeleted my answer. I had to do an example to convince myself I was right again. However, there could be changes in the dividend yield that I effectively assumed away (higher dividend yields in the financial crisis should make the put delta increase away from -0.5).
– John
Apr 11, 2014 at 15:28

## 1 Answer

Delta increases as volatility increases.

In particular, the formula for delta of a put is $$\Delta=-exp\left(-qt\right)\Phi\left(-d_{1}\right)$$ with $$d_{1}\equiv\frac{ln\left(S/K\right)+\left(r-q+\frac{\sigma^{2}}{2}\right)t}{\sigma\sqrt{t}}$$ setting $S=K$ and $r=q$ you would get $$d_{1}\equiv\frac{\sigma}{2}\sqrt{t}$$ By the chain rule, an increase in $\sigma$ leads to an increase in $d_{1}$, which leads to a decrease in the $\Phi(-d_{1})$ term but an increase in $\Delta$.

• The delta for a put option would be different, with a negative in front of d_1 and a negative infront of exp. Apr 11, 2014 at 14:35
• Sorry, I didn't read it that carefully. Was answering for calls. I don't think it should change the answer though.
– John
Apr 11, 2014 at 14:45
• Thanks. And would you know intuitively; why an increased volatility decreases the sensitivity of the put option price to an underlying asset's price change? Apr 11, 2014 at 14:53
• At a mathematical level, stock prices in the Black-Scholes model are assumed to be log normally distributed. The distribution of a log normal variable is not symmetric. The mean of the log normal distribution increases as the volatility increases. This means there's a greater chance a call option will be in the money at expiration and less of a chance the put will be in the money.
– John
Apr 11, 2014 at 15:43