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How do I prove the following equation:

P(X=100)≤(P(X=110)-P(X=90))/2

I am not sure how to start and whether it involves using the Black-Sholes formula or not (something like this: https://www.youtube.com/watch?v=LM6iMfHbQDs&t=939s). Also, please note that this is an option price of a put, not a probability.

Thank you!

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  • $\begingroup$ I think what you want is actually $P(X=100)≤\frac{P(X=110) + P(X=90))}{2}$.That is, the convexity of the option price. $\endgroup$ – Gordon Aug 18 '20 at 18:31
  • $\begingroup$ I think yes, there is a mistake in the sign. $\endgroup$ – S_Star Aug 18 '20 at 19:53
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The pair of puts has to be more valuable. Consider the payoff at expiry - it looks like this:

Put Portfolio Payoff

The pair of puts pay off the same or more in every scenario (usually the same, but more when you're between 90 or 110) so they MUST be more valuable.

This is actually related to the convexity of the option payoff profile... it's duscussed further in this question and answers

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  • $\begingroup$ Hi, much more clearer. So this is basically a butterfly spread and I just need to calculate the payoff, right? I thought that this involves doing some integrals ... $\endgroup$ – S_Star Aug 18 '20 at 7:14
  • $\begingroup$ It's definitely true that you can compare BS fomulae for different options and strikes. But if all you want to show is that the K=90 and the K=110 puts must sum to give more than twice the price of the K=100 put, then the above is sufficient. At expiry, there is a positive probability of the portfolio being worth more and NO chance it is worth less. Since today's price is the discounted future price in the RN measure (ie. $C(0) = \delta(0,t){\mathbb E}[C(t)]$), this means the portfolio MUST, ALWAYS be more expensive. $\endgroup$ – StackG Aug 18 '20 at 7:48
  • $\begingroup$ Hi, is it possible to help me this too? quant.stackexchange.com/questions/57434/… $\endgroup$ – S_Star Aug 18 '20 at 13:23

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