Distribution of $X_t := \int^\sqrt t _0 \sqrt{2u}\; dB_u$

Question

Let $(B_t)_{t≥0}$ be a Brownian motion. What is the (distribution of the) process $(X_t)_{t≥0}$ given by $X_t := \int^\sqrt t _0 \sqrt{2u}\; dB_u$?

Answer 1

HINT You are adding a bunch of zero-mean normals to each other, so the result must be a zero-mean normal. can you find the variance?

Answer 2

Find the SDE for $X_t$ by differentiating the given integral,

$$dX_t =(\sqrt t)'\sqrt{2t}\; dB_t = dB_t$$

which yields $X_t = B_t$.