Calculating probability of options with normal/lognormal distribution: does time make a difference?

Question

I'm trying to calculate the probability of a calendar spread resulting in a profit at expiration, when estimating it is modeled as a lognormal distribution, by getting:

P(a <= x <= b) = CDF(b) - CFA(a)

where a and b are the breakevens at expiration.

But there is something that I don't understand:

1. Which value shall I use as variance? The IV of the ATM option for near expiration? The IV of the stock/index underneath?
2. Does time really matter? I mean, since lognormal distribution (as defined in scipy/numpy libraries) only requires mean and variance values, time does not matter unless you consider that volatility depends on t. If I get mean and variance for 2 calendars, one with front mont expiring in a week and another one expiring in a year, time should matter somehow, making the distribution PDF wider, and therefore affecting the results of the CDF. What am I missing here?

Answer 1

If $S_t$ is stochastic process and follow geometric Brownian motion with following SDE: $$dS_t=\mu S_t dt + \sigma S_t dW_t$$ then $S_T$ follows lognormal distribution, such that: $$S_T|S_t \sim logN\left(lnS_t+ (\mu - \frac{\sigma^2}{2})(T-t), \quad \sigma^2(T-t)\right)$$ or $$lnS_T|S_t \sim N\left(lnS_t+ (\mu - \frac{\sigma^2}{2})(T-t), \quad \sigma^2(T-t)\right)$$

As you may see, more you will go into the future, both the drift and volatility increase directly in proportionate to $(T-t)$ for log of stock price. This is natural phenomenon. You may think like this, the variability shown by stock price in one year(ie $T-t=1$) is much more than variability shown in one minute or one day(ie $T-t=\frac{1}{365}$).