If one draws a graph showing how the value of a bull call spread depends on volatility, one would find that at first this dependence is positive (the greater the volatility, the greater the value of the spread), and then it becomes negative (the greater the volatility, the smaller the value of the spread). So this simple strategy demonstrates a non-monotonous dependence on volatility.If one does Monte Carlo simulation, it is easy to see that the expected payoff of the bull call spread does really behave like that.But is there a simpler and more intuitive explanation / demonstration for this behaviour?

it depends on which strike is dominant.at infinite vol both strikes are worth same. thats the limiting value of zero for the spread. at low vols and depending on the strikes and maturities the lower strike call dominates - hence you are long vol on a net basisat high vols and again depending on the strikes, the upper strike dominates so you get short vol. the lower strike becomes less vol sensitive whilts due to the vol convexity of the OTM strike, that becomes more sensitive.

I guess the most simple explanation you can give: Image you are long this spread, the stock is below the strike and volatility is zero - then your chance of getting a non-zero payoff (ignoring the drift) is zero. Increasing the volatility increases your probability of ending up with a non-zero payoff at maturity so you are long vega below the lower strike. Apply the same logic above the upper strike: if volatility is zero you are guaranteed to get the max. payoff at maturity while increasing volatility increases your risk of ending up with something less than that..

This explanation relies on the strategy being composed of 2 call options with 2 strikes.But what if I show you an arbitrary payoff function, such as “Payoff (S) = sin (S) + cos (S)”, and ask you to analyze whether it is long or short vol? You do not have strikes and vanilla calls, you just have a payoff. And I thought that there must be a universal method of analyzing volatility sensitivity.I was thinking of using the future stock price probability density function, calculating the expected payoff by way of numerical integration and seeing the “weight” of different price ranges in the derivative expected payoff.What other approaches could one suggest?

Here are some likely general answers under GBM. Suppose an arbitrary payoff function w(S) and option value V(T,S,sigma).1. T small.For sigma^2 T small, then dV(T,S,sigma)/d sigma always has the same sign as w''(S).The way you prove this kind of thing (taking GBM as an example) is to differentiatethe Black-Scholes PDE with respect to sigma^2 and analyze the new PDE you get with Feynman-Kac. So, it is just the local convexity of the payoff for small times -- i.e., the convexity near the stock price youare interested in.2. T largeIf sigma^2 T is large, and r < sigma^2/2 (the usual case), then S(T) concentrates near zero.So, it is probably the convexity of w(S) near S=0 that will determine dV/dsigma for T >> 1.This would be true regardless of S0.There may not be simple general answers if T is not small or large and w''(S) changes sign.