European binary and bull call spread hedge

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Hi all

Why the european binary call can be mathematically priced as follows if they can be perfectly hedge by a bull call spread:

binary call = (c(K+h) - c(K))/h when h is going infinitely small. K is the strike of the binary, h is the strike price difference between the two calls of the bull call spread,c is the price of the calls of the call spread. I know it is related to the volatility skew in between the two strike prices of the spread, but I don't understand the mathematical equation.

Last edited:

Martinghoul

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2,690 276
It has nothing to do with skew. Just draw the diagram of a call spread payoff and look at it closely for a bit. It should become reasonably clear then.

Active member
183 0
It has nothing to do with skew. Just draw the diagram of a call spread payoff and look at it closely for a bit. It should become reasonably clear then.

Thanks!

How did he get the equation of (C(K+h) - C(k))/h with h -> 0?? Besides, it is defined in the book that if skew is taken into account, the equation becomes (C(K+h) - C(k))/h+ partial derivative of C with respect to vol times partial derivative of vol with repect to K. I just dont understand intuitively how this equation (C(K+h) - C(k))/h with h -> 0 (without skew effect) comes out?

Besides, I have attached the two pages that related to the skew distribution of a binary example, I really don't understand how come the left integral is equal the right one plus the risk neutrial drift?? And also how did he get the graphs between Bet and delta?? many thanks mate

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Last edited:

Shakone

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thanks mate. that's very good advice. But I still want to get more intuitive understanding about that.

There's another thing, in the attachment it says that in figure 17.7 the skewed distribution, the right integral represents the delta for a call and the left integral represents the delta for a put, which, I think, corresponds to the probablity being in the money. However, the attachment says the delta is not the probablity of being excercised, which is inconsistent with the statement delta is the integral of the distribution, right? thanks

Shakone

Senior member
2,458 665
thanks mate. that's very good advice. But I still want to get more intuitive understanding about that.

There's another thing, in the attachment it says that in figure 17.7 the skewed distribution, the right integral represents the delta for a call and the left integral represents the delta for a put, which, I think, corresponds to the probablity being in the money. However, the attachment says the delta is not the probablity of being excercised, which is inconsistent with the statement delta is the integral of the distribution, right? thanks

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