Linear Stochastic Differential Equation

Question

Please could someone help me with the following proof:

Prove that $Y_t = e^{-2t} (Y_0 + 4 \int_0^t e^{2s}d B_s )$

is the solution to the homogeneous linear stochastic differential equation $ dY_t = -2 Y_t dt + 4dB_t $.

I am not at all sure how to go about this.

Answer 1

The standard stochastic differential equation (for a geometric case) is is $\mathrm{d}S_t = \mu \, S_t\,\mathrm{d}t + \sigma \, S_t\,\mathrm{d}W_t$ . What you have here is of the form $\mathrm{d}S_t = \mu \, S_t\,\mathrm{d}t + \sigma \, \mathrm{d}W_t$, with $\mu =-2$ and $\sigma=4$. The usual route is Ito's formula. Here it is a bit more complicated because $\sigma$ is divided by S, so it decreases at lower values of S.

Answer 2

This is a direct application of Ito lemma.

By definition of Ito integral, $Z_t = Y_0 + 4 \int_0^t \mathrm{e}^{2s} \mathrm{d}B_s$ satisfies $\mathrm{d}Z_t = 4 \mathrm{e}^{2 t} \mathrm{d} B_t$ with initial condition $Z_0 = Y_0$. Then we want to find the SDE for $Y_t = \mathrm{e}^{-2 t} Z_t$ and the Ito lemma is just the tool: $$ \mathrm{d}Y_t = \left( \mathrm{e}^{-2t} \right)^\prime Z_t \, \mathrm{d}t + \mathrm{e}^{-2 t} \mathrm{d} Z_t = \left(-2 \underbrace{\mathrm{e}^{-2t} Z_t}_{Y_t} \right) \mathrm{d}t + 4 \mathrm{d} B_t $$