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Mathew C asked in Business & FinanceInvesting · 1 decade ago

More on Black and Scholes?

The hedge ratio is change in option price/change in stock price. When d1 of Black and Scholes formula gives this why should it be that it is diercted N(d1) as the hedge ratio?

Update:

Taranto has pulled a fast one again. Let me explain, ln(px/ps) is the growth rate of the stock price which is the growth rate in price for the option itself. t(krf+dellta/squreroot2) gives the impact of the dividends for the period and the cost of carry.The denominator d.squreroot t gives the mean standard deviation calculated like the Brownian formula for random walk. This is none other than the expected change in Stock price or growth. So d1 is nothing other than change in option price to change in stock price factored to their growth rate. In fact it has to be 1 or above otherwise the hedges will bomb. In my earlier question I have shown why Nd1 also bombs, since it is none other than taking the Normal value of another deviation. This is the biggest problem with this formula. What if it is the first partial derivative of the price function, the put-call parity equation. Please read my second question on this it shows how even Nd bombs. Nd is meaningless unless the rational is visibl

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  • Ranto
    Lv 7
    1 decade ago
    Favorite Answer

    The value d1 does not give the change in option value for a small change in asset price -- N(d1) does.

    In general, the delta of an option is the first partial derivative of the pricing function with respect to the underlying asset price. The delta is also called the hedge ratio. If, for example, the delta of an option is 0.75, then for a small change in the value of an asset, the option will change about 3/4 of that value.

    The Delta of a European Call Option is N(d1).

    Source(s): See page 346 of Hull's OPTIONS, FUTURES and OTHER DERIVATIVES (6th edition). Or look at the formula for Delta at: http://en.wikipedia.org/wiki/Black-Scholes
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