Question

Find u(X,t) and v(X,t) where dXₜ = udt+vdWₜ and where f is a bounded, differentiable function. (a) Xt =W²ₜ (b) Xt =1+t+eᵂᵗ (c) Xt=f(t)Wₜ

Answer 1

For the given stochastic differential equations (SDEs):

(a)
`\(dX_t = 2W_tdt + 2dW_t\), the solutions are \(u(X,t) = 2\) and \(v(X,t) = 2\).`

(b)
`\(dX_t = (1 + e^(W_t))dt + e^(W_t)dW_t\), the solutions are \(u(X,t) = 1 + e^(X_t)\) and \(v(X,t) = e^(X_t)\).`

(c)
`\(dX_t = f(t)dW_t\), the solutions are \(u(X,t) = f(t)\) and \(v(X,t) = 0\).`

Explanation:

In the case of
`\(X_t = W_t^2\), the given SDE \(dX_t = 2W_tdt + 2dW_t\) can be expressed as \(u(X,t) = 2\) and \(v(X,t) = 2\)`. The term
`\(2W_tdt\)` represents the deterministic part, and
`\(2dW_t\)` represents the stochastic part.

For
`\(X_t = 1 + t + e^(W_t)\), the SDE \(dX_t = (1 + e^(W_t))dt + e^(W_t)dW_t\) yields \(u(X,t) = 1 + e^(X_t)\) and \(v(X,t) = e^(X_t)\). The term \((1 + e^(W_t))dt\) is the deterministic part, and \(e^(W_t)dW_t\)` is the stochastic part.

In the case of
`\(X_t = f(t)W_t\), the SDE \(dX_t = f(t)dW_t\) gives \(u(X,t) = f(t)\) and \(v(X,t) = 0\). Here, \(f(t)dW_t\)` represents the stochastic part with no additional deterministic component.

These solutions provide the expressions for the coefficients
`\(u(X,t)\) and \(v(X,t)\)` in the given SDEs. The deterministic parts involve the functions
`\(f(t)\), \(1 + e^(W_t)\), and \(2\)`, while the stochastic parts involve the increments
`\(dW_t\) or \(dW_t^2\)`, depending on the specific form of the SDE.