To solve this problem, we need to find the value of the constant $c$ such that the stochastic process $X(t) = Z(t)^3 + ctZ(t)$ has zero drift, where $Z(t)$ represents a standard Brownian motion.

Step-by-Step Solution

1. Identify the form of $X(t)$: Given: $X(t) = Z(t)^3 + ctZ(t)$ where $Z(t)$ is a standard Brownian motion.

2. Compute $dX(t)$ using Ito's Lemma: Since $X(t)$ is a function of $Z(t)$, we will apply Ito's Lemma to find the differential $dX(t)$. Recall that Ito's Lemma for a function $f(Z(t), t)$ where $f(Z(t), t) = Z(t)^3 + ctZ(t)$ is given by: $df(Z(t), t) = \frac{\partial f}{\partial t} \, dt + \frac{\partial f}{\partial Z} \, dZ + \frac{1}{2} \frac{\partial^2 f}{\partial Z^2} \, dZ^2$ For a standard Brownian motion $Z(t)$, we know: $dZ^2 = dt$

3. Calculate the partial derivatives of $f(Z, t) = Z^3 + ctZ$:

   • First partial derivative with respect to $t$: $\frac{\partial f}{\partial t} = cZ$

   • First partial derivative with respect to $Z$: $\frac{\partial f}{\partial Z} = 3Z^2 + ct$

   • Second partial derivative with respect to $Z$: $\frac{\partial^2 f}{\partial Z^2} = 6Z$

4. Substitute into Ito's Lemma: Now, we substitute the derivatives into Ito's formula to find $dX(t)$: $dX(t) = \left(cZ\right) dt + \left(3Z^2 + ct\right) dZ + \frac{1}{2} \left(6Z\right) dt$

5. Simplify the expression: Combine terms involving $dt$ and $dZ$: $dX(t) = \left(cZ + 3Z\right) dt + \left(3Z^2 + ct\right) dZ$ Simplify further: