Solution: Apply Itô — $d(e^{W_t})$ and $\mathbb{E}[e^{W_t}]$

Exercise: Apply Itô: $d(e^{W_t})$ and $\mathbb{E}[e^{W_t}]$

Part 1

With $V(W) = e^W$, $dX_t = dW_t$ (so $a = 0$, $b = 1$), and $V' = V'' = e^W$:

Part 2

Taking expectations and using that $\int_0^t e^{W_s}\,dW_s$ is a mean-zero martingale (under the usual integrability assumption, which holds because $\mathbb{E}[\int_0^T e^{2W_s}\,ds] < \infty$):

Solution: $m(t) = e^{t/2}$.

Part 3

$W_t \sim \mathcal{N}(0, t)$ has moment-generating function $\mathbb{E}[e^{\lambda W_t}] = e^{\lambda^2 t / 2}$. With $\lambda = 1$:

This matches part 2 exactly. Two routes, one answer — the Itô-calculus route is longer here but generalises to $V(W_t)$ for any smooth $V$, whereas the direct route requires knowing the distribution in closed form.

Part 4

To cancel the drift, set $\tilde Y_t = e^{\alpha W_t + \beta t}$ and apply Itô to $V(t, W) = e^{\alpha W + \beta t}$:

• $V_t = \beta V$
• $V_W = \alpha V$
• $V_{WW} = \alpha^2 V$

Itô gives $d\tilde Y_t = (\beta + \tfrac{1}{2}\alpha^2)\tilde Y_t\,dt + \alpha\tilde Y_t\,dW_t$. The drift vanishes iff $\beta = -\tfrac{1}{2}\alpha^2$. So the martingale correction is:

Takeaways

• The Itô correction for $e^{W_t}$ adds a $\tfrac{1}{2}$ drift — the same $\tfrac{1}{2}$ that appears in the log-normal moment formula.
• Two routes to the same expectation. Itô + ODE is the constructive route; the Gaussian MGF is the direct route. Quant finance uses both routinely.
• To make an exponential a martingale, subtract half its variance-accumulation. This single identity ($\mathbb{E}[e^{\alpha W_t - \alpha^2 t/2}] = 1$) drives the Doléans-Dade exponential, change-of-measure arguments, and the $-\tfrac{1}{2}\sigma^2$ term in log-GBM.