Suppose That It Is Now Zero And Zero Is Some Later Time ✓ Solved

Asuppose That It Is Now 0 And 0 Is Some Later Time Ca

(a) Suppose that it is now \(\tau=0\) and \(t>0\) is some later time. Carefully show that the probability in a risk-neutral world satisfies the equation, \(P(S_t > K) = N(d_2)\). Here, \(S_t\) is a stock’s price at time \(t\), \(N(\cdot)\) is the cumulative distribution function of a standard normal random variable, and \(d_2 = \frac{\ln(s_0/K) + (r - \frac{\sigma^2}{2}) t}{\sigma \sqrt{t}}\).

(b) Consider a European call with expiration date \(T\), underlying stock price \(S_0\), volatility \(\sigma\), risk-free rate \(r\), and strike price \(K\). If \(C(0) =\) its price at time \(0\), then compute \(\lim_{\sigma \to 0} C(0)\).

Sample Paper For Above instruction

Part (a): Demonstrating \(P(S_t > K) = N(d_2)\) in a Risk-Neutral Framework

Under the risk-neutral measure, the dynamics of the stock price \(S_t\) follow the stochastic differential equation:

dS_t = r S_t dt + \sigma S_t dW_t,

where \(W_t\) is a standard Wiener process. The solution to this SDE with initial stock price \(S_0\) is:

S_t = S_0 \exp\left( \left(r - \frac{\sigma^2}{2}\right)t + \sigma W_t \right).

Since \(W_t\) is normally distributed with mean 0 and variance \(t\), the logarithm of \(S_t\) is normally distributed:

\ln S_t = \ln S_0 + \left(r - \frac{\sigma^2}{2}\right)t + \sigma W_t,

which implies that:

\ln S_t \sim \mathcal{N}\left(\ln S_0 + \left(r - \frac{\sigma^2}{2}\right)t, \sigma^2 t\right).

To find \(P(S_t > K)\), we rewrite as:

P(S_t > K) = P(\ln S_t > \ln K).

Expressing in terms of the normal distribution, we set:

Z = \frac{\ln S_t - \left(\ln S_0 + \left(r - \frac{\sigma^2}{2}\right)t \right)}{\sigma \sqrt{t}} \sim \mathcal{N}(0,1).

Therefore:

P(S_t > K) = P\left( Z > \frac{\ln K - \ln S_0 - \left(r - \frac{\sigma^2}{2}\right)t }{\sigma \sqrt{t}} \right).

This probability can be written in terms of the standard normal CDF as:

P(S_t > K) = 1 - N\left( \frac{\ln K - \ln S_0 - \left(r - \frac{\sigma^2}{2}\right)t }{\sigma \sqrt{t}} \right).

By the Black–Scholes model, the variable:

d_2 = \frac{\ln(S_0/K) + (r - \frac{\sigma^2}{2})t}{\sigma \sqrt{t}},

then the probability becomes:

P(S_t > K) = N(d_2).

This completes the demonstration that in a risk-neutral world, the probability that the stock exceeds the strike \(K\) at time \(t\) is \(N(d_2)\).

Part (b): Limit of European Call Price as \(\sigma \to 0\)

The classical Black–Scholes formula for a European call option is:

C(0) = S_0 N(d_1) - K e^{-r T} N(d_2),

where

d_1 = \frac{\ln (S_0/K) + \left(r + \frac{\sigma^2}{2}\right) T}{\sigma \sqrt{T}}, \quad

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

As \(\sigma \to 0\), the behavior of \(d_1\) and \(d_2\) depends on the relative position of \(S_0\) and \(K\).

Case 1: \(S_0 > K\)

When \(S_0 > K\), \(\ln(S_0/K) > 0\). Then, as \(\sigma \to 0\),

d_1 \to +\infty, \quad d_2 \to +\infty.

Because \(N(d) \to 1\) as \(d \to +\infty\), the call price tends to:

C(0) \to S_0 - K e^{-r T}.

Case 2: \(S_0

When \(S_0

d_1 \to -\infty, \quad d_2 \to -\infty.

Since \(N(d) \to 0\) as \(d \to -\infty\), the call price tends to:

C(0) \to 0.

Case 3: \(S_0 = K\)

This is a boundary case where \(\ln (S_0/K) = 0\), then:

d_1 = \frac{0 + (r + \frac{\sigma^2}{2}) T}{\sigma \sqrt{T}} \to +\infty,

\end{pre>

and the call price tends to:

C(0) \to S_0 - K e^{-r T} = 0.

Summary: In the limit as volatility tends to zero, the European call price converges to its intrinsic value:

\boxed{

\lim_{\sigma \to 0} C(0) = \max(S_0 - K e^{-r T}, 0).

}

This aligns with economic intuition: with no volatility, the option’s value equals the present value of its deterministic payoff.

References

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