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Why does an option lose time value faster as it approaches expiry
Here's another attempt at explanation: it's basically because parabolas are flat at the bottom. Let me explain.  As you might know, the variance of the log stock price in Black Scholes is vol^2 * T, in other words, variance of the log stock price is linear in time to expiry. Now, that means that the standard deviation of your log stock price is square root in time. This is consequential.  For normally distributed random variables, in 68% of cases we end up within one standard deviation. So, basically, we expect our log stock price to be within something something times square root of T. So, if your stock has a vol of 16%, it'll be plus/minus 32% in 4 years, plus/minus 16% for one year, plus/minus 8% for 3m, plus/minus 4% for 3-ish weeks, and plus/minus 1% for a business day. As you see, the decay is slow at first, but much more rapid as we get closer.  How does the square root function look? It's a sideways parabola. As we come closer to zero, the slope of the square root function goes to infinity. (That is related to the fact that Brownian motion is almost surely no-where differentiable - it just shoots off with infinite slope, returning immediately, of course :-) Another way of looking at it is the old traders rule of thumb that an at-the-money option is worth approximately S * 0.4 * vol * sqrt(T). (Just do a Taylor expansion of Black Scholes). Again, you have the square root of time to expiry in there, and as outlined above, as we get closer to zero, the square root drops slowly at first, and then precipitously.