Pricing and Hedging a Binary Put in the Black–Scholes Model

# Is my solution on Black Scholes correct?


# Is my solution on Black Scholes correct?












Question: Consider in the context of the Black Scholes model with stock price dynamics $d S_{t}=\mu S_{t} d t+\sigma S_{t} d W_{t}$, initial stock price $S_{0}>0$ and with money market account $B_{t}=\exp (r t)$ for $r>0$ a so-called binary put option with payoff $h\left(S_{T}\right)=1_{\left\{S_{T} \leq K\right\}}$ for some $K>0$. Use the risk neutral pricing formula to compute the price of the option at $t=0$ and the stock position of the corresponding hedging portfolio.

### This is what I've tried:

The price of a binary put option with payoff $h(S_T) = 1_{\{S_T \leq K\}}$ can be obtained by taking the risk-neutral expectation of the binary payoff under the Black-Scholes model.

Using the Black-Scholes formula, the stock price $S_T$ at expiration follows a lognormal distribution, given by: $$S_T = S_0 \cdot e^{(r - \frac{1}{2} \sigma^2)T + \sigma \sqrt{T} Z},$$ where $S_0$ is the initial stock price, $r$ is the risk-free interest rate, $\sigma$ is the stock price volatility, $T$ is the time to expiration, and $Z$ is a standard normal random variable.

To calculate the expectation, we evaluate the probability that $S_T \leq K$ under the lognormal distribution: $$\mathbb{E}[h(S_T)] = \mathbb{P}(S_T \leq K).$$

Taking the natural logarithm of both sides, we get: $$\ln(S_T) = \ln(S_0) + (r - \frac{1}{2} \sigma^2)T + \sigma \sqrt{T} Z.$$

Rearranging the inequality, we have: $$\ln\left(\frac{S_T}{S_0}\right) \leq \ln\left(\frac{K}{S_0}\right).$$

Substituting the expression for $\ln(S_T)$, we obtain: $$\ln\left(\frac{S_0 \cdot e^{(r - \frac{1}{2} \sigma^2)T + \sigma \sqrt{T} Z}}{S_0}\right) \leq \ln\left(\frac{K}{S_0}\right).$$

Simplifying further, we have: $$(r - \frac{1}{2} \sigma^2)T + \sigma \sqrt{T} Z \leq \ln\left(\frac{K}{S_0}\right).$$

Since $Z$ is a standard normal random variable, the probability $\mathbb{P}(S_T \leq K)$ is equivalent to the probability $\mathbb{P}\left((r - \frac{1}{2} \sigma^2)T + \sigma \sqrt{T} Z \leq \ln\left(\frac{K}{S_0}\right)\right)$.

We can now use the standard normal cumulative distribution function $\Phi(\cdot)$ to calculate the probability: $$\mathbb{E}[h(S_T)] = \mathbb{P}(S_T \leq K) = \Phi\left(\frac{\ln\left(\frac{S_0}{K}\right) + (r - \frac{1}{2} \sigma^2)T}{\sigma \sqrt{T}}\right).$$

Therefore, the price of the binary put option at time $t = 0$ is given by the risk-neutral expectation: $$\text{Price} = e^{-rT} \cdot \mathbb{E}[h(S_T)] = e^{-rT} \cdot \Phi\left(\frac{\ln\left(\frac{S_0}{K}\right) + (r - \frac{1}{2} \sigma^2)T}{\sigma \sqrt{T}}\right).$$

### updated:

To calculate the price of the binary put option at time $t = 0$, assuming a risk-free interest rate of zero, we can use the cumulative distribution function of $-d_2$:

$$\text{Price} = e^{-rT} \cdot \mathbb{P}(S_T \leq K) = e^{-rT} \cdot \Phi(-d_2),$$

where $\Phi(\cdot)$ is the cumulative distribution function of a standard normal random variable.

To compute $d_2$, we first calculate $d_1$ using the formula:

$$d_1 = \frac{\ln\left(\frac{S_0}{K}\right)}{\sigma \sqrt{T}} + \frac{1}{2}\sigma \sqrt{T}.$$

Then, we can obtain $d_2$ as:

$$d_2 = d_1 - \sigma \sqrt{T}.$$

Finally, we substitute $d_2$ into the expression for the price of the binary put option:

$$\text{Price} = e^{-rT} \cdot \Phi\left(-\left(d_1 - \sigma \sqrt{T}\right)\right).$$

## Answer by Frido (score 2, accepted)

https://quant.stackexchange.com/a/75912

Too long for a comment so posting as answer.

I am not sure your update makes sense, it seems to me that you're just `making' it work.

In your first attempt, everything made sense until $$ \mathbb P(rT - \tfrac12 \sigma^2T + \sigma \sqrt T Z \leq \log K/S_0) $$ and after that you reached a wrong conclusion without detailing the in between steps.

The point is that $$ rT - \tfrac12 \sigma^2T + \sigma \sqrt T Z \leq \log K/S_0 $$ is equivalent to $$ Z \leq \frac{\log K/S_0 - rT + \tfrac12 \sigma^2T}{\sigma \sqrt T} = -d_2 $$ And using this and the fact that $Z$ is a standard normal distributed it follows that the price of the binary put is $$ e^{-rT} N(-d_2) $$