1. \(X_t = B_t^2\)

\(dg/dt(t,X_t)dt =0dt\),

\(dg/dx(t,X_t)dX_t = 2B_tdB_t\),

\(1/2*d^2g/dx^2(t,X_t) \cdot (dX_t)^2 = 1/2* 2 \cdot dB_t\cdot dB_t = dt\)

so we have \(2B_tdB_t + dt\) as the answer.

  1. \(X_t = 2 + t + e^{B_t}\)

\(dg/dt(t,X_t)dt =1dt\),

\(dg/dx(t,X_t)dX_t = e^{B_t}dB_t\),

\(1/2*d^2g/dx^2(t,X_t) \cdot (dX_t)^2 = 1/2* e^{B_t} \cdot dB_t\cdot dB_t = 1/2* e^{B_t}dt\)

Thus, we have \(dt + e^{B_t}dB_t + 1/2* e^{B_t}dt\)

  1. \(X_t = B_1^2(t) + B_2^2(t)\) where \((B_1,B_2)\) is 2-dimensional.

\(dg/dt(t,X_t)dt =0dt\),

\(dg/dx_1(t,X_t)dX_1 = 2B_1dB_1(t)\),

\(dg/dx_2(t,X_t)dX_2 = 2B_2dB_2(t)\),

\(1/2*d^2g/(dx_1dx_1)(t,X_t) \cdot (dX_1)(dX_1) = 1/2* 2 \cdot dB_1\cdot dB_1 = 1dt\),

\(1/2*d^2g/(dx_1dx_2)(t,X_t) \cdot (dX_1)(dX_2) = 1/2*d^2g/(dx_2dx_1)(t,X_t) \cdot (dX_2)(dX_1) = 1/2* 0 \cdot dB_1\cdot dB_2 = 0\),

\(1/2*d^2g/(dx_2dx_2)(t,X_t) \cdot (dX_2)(dX_2) = 1/2* 2 \cdot dB_2\cdot dB_2 = 1dt\)

So we have \(2d_t + 2B_1dB_1(t) + 2B_2dB_2(t)\)

  1. \(X_t = (t_0 + t, B_t)\) (\(B_t\) is 1-dimensional)

For the first element \(t_0+t\), we have

\(dg/dt(t,X_t)dt =1dt\),

\(dg/dx(t,X_t)dX_t = 0dB_t\),

\(1/2*d^2g/dx^2(t,X_t) \cdot (dX_t)^2 = 1/2* 0 \cdot dB_t\cdot dB_t =0\)

so we have \(1dt\).

For the second element, we would have

\(dg/dt(t,X_t)dt =0dt\),

\(dg/dx(t,X_t)dX_t = 1dB_t\),

\(1/2*d^2g/dx^2(t,X_t) \cdot (dX_t)^2 = 1/2* 0 \cdot dB_t\cdot dB_t = 0dt\)

so we have \(1dB_t\).

In vector form, that’s \(\begin{bmatrix}1 \\ 0 \end{bmatrix}d_t + \begin{bmatrix}0 \\ 1 \end{bmatrix}dB_t\)

  1. \(X_t = (B_1(t) + B_2(t) + B_3(t), B_2^2(t) - B_1(t)B_3(t))\) where \((B_1, B_2, B_3)\) is 3-dimensional

For the first element \((B_1(t) + B_2(t) + B_3(t))\), we have

\(dg/dt(t,X_t)dt =0dt\),

\(dg/dx_1(t,X_t)dX_1 = 1dB_1(t)\),

\(dg/dx_2(t,X_t)dX_2 = 1dB_2(t)\),

\(dg/dx_3(t,X_t)dX_3 = 1dB_3(t)\),

\(1/2*d^2g/(dx_1dx_1)(t,X_t) \cdot (dX_1)(dX_1) = 1/2* 2 \cdot dB_1\cdot dB_1 = 1dt\),

and \(1/2 * \sum_{i,j}d^2g_k/dx_idx_j(t,X)dX_idX_j\) is 0 for all variations of \(x_i\) and \(x_j\).

Thus we have \(dB_1(t) + dB_2(t) + dB_3(t)\)

For the second element \((B_2^2(t) - B_1(t)B_3(t))\), we have \(dg/dt(t,X_t)dt =0dt\),

\(dg/dx_1(t,X_t)dX_1 = -B_3(t)dB_1(t)\),

\(dg/dx_2(t,X_t)dX_2 = 2B_2dB_2(t)\),

\(dg/dx_3(t,X_t)dX_3 = B_1dB_3(t)\),

\(d^2g_k/(dx_1dx_1) = 0\),

\(d^2g_k/(dx_1dx_2) = 0\),

\(d^2g_k/(dx_1dx_3) = -1dB_1dB_3\),

\(d^2g_k/(dx_2dx_1) = 0\),

\(d^2g_k/(dx_2dx_2) = 2dB_2dB_2\),

\(d^2g_k/(dx_2dx_3) = 0\),

\(d^2g_k/(dx_3dx_1) = -1dB_3B_1\),

\(d^2g_k/(dx_3dx_2) = 0\),

\(d^2g_k/(dx_3dx_3) = 0\).

Since \(dB_1dB_3\) and \(dB_2dB_2\) evaluate to \(d_t\), we have \(-B_3(t)dB_1(t) + 2B_2dB_2(t) + B_1dB_3(t)\).