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What is the final explicit solution for S obtained in this clip?

The final explicit solution for the geometric Brownian motion process SS is S=c1eσW+(μ−12σ2)tS = c_1 e^{\sigma W + \left(\mu - \frac{1}{2}\sigma^2\right)t}, where c1c_1 is an integration constant determined by the initial condition.

Conditions

  • μ\mu and σ\sigma are constant.
  • The ansatz S=f(W,t)S=f(W,t) is valid.
  • The ODE for ϕ(t)\phi(t) is solved.

Reasoning, step by step

  1. Solve the ODE for ϕ(t)\phi(t) to get ϕ(t)=c1e(μ−12σ2)t\phi(t) = c_1 e^{\left(\mu - \frac{1}{2}\sigma^2\right)t}.
  2. Substitute ϕ(t)\phi(t) back into f(W,t)=ϕ(t)eσWf(W,t) = \phi(t)e^{\sigma W}.
  3. Combine the exponents to get S=c1eσW+(μ−12σ2)tS = c_1 e^{\sigma W + \left(\mu - \frac{1}{2}\sigma^2\right)t}.
  4. Identify c1=S0c_1 = S_0 using the initial condition at t=0t=0.

Example

The bottom line on the board writes 'S=f(W,t)=c1exp⁡(σW+(μ−1/2σ2)t)S = f(W,t) = c_1 \exp (\sigma W + (\mu - 1/2 \sigma^2)t)'.

Common misconceptions

  • Forgetting the −12σ2-\frac{1}{2}\sigma^2 term in the exponent, which arises from Itô's lemma.
  • Assuming c1c_1 is always 1, rather than being determined by the initial value S0S_0.

Watch the explanation

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Answers are generated from source material and independently checked. Consult the original video or creator if something is unclear.