# How to hedge a put under the Black-Scholes model?

To hedge a call, one would invest the option price proceeds into $\Delta_t*S_t + B_t = c_t$. (ok)

However, a put has negative delta, so I would short $\Delta_t*S_t$ and invest $p_t+\Delta_t*S_t>p_t$ into a risk-free bond?

It just seems a bit odd to me that I would invest more than the actual put price into a bond.

## 1 Answer

Assuming zero interest, the put option has the price \begin{align*} KN(-d_2)-S_0N(-d_1), \end{align*} and delta $-N(-d_1)$. When $N(-d_1)$ units of stocks are shorted and invested in bonds, the total value in bonds is $KN(-d_2)$, which is indeed greater than the option price. However, as you have shorted $N(-d_1)$ units of stocks, your portfolio value is \begin{align*} KN(-d_2) -N(-d_1) S_t \end{align*} at time $t$. That is, the portfolio value is not necessarily greater than the option price, and at the deal inception, is the same as the option price.

• Why do you think the bond value is not always greater than the option price? The put delta is always negative... – emcor Jun 15 '15 at 17:32
• @emcor: The total bond value is indeed greater than the option price, however, the hedged portfolio value is not necessarily greater. Note that, for call option, we have a long stock position, while, for a put option, we have a short position. The short position is still in our hedged portfolio, as it is our liability. – Gordon Jun 15 '15 at 17:37
• The portfolio value is always equal to the option price by definition, or why do you say "not necessarily greater"? – emcor Jun 15 '15 at 17:38
• @emcor: As time changes, it will not equal. At the deal inception (option trading time), it is. – Gordon Jun 15 '15 at 17:39
• With continuous trading it will be equal of course. – emcor Jun 15 '15 at 17:44