Let $B(t)$ be a Brownian Motion. Find all constants $A$ and $b$ such that $X(t)=\int_0^t(a+b\frac{u}{t})dB(u)$ is also a Brownian Motion.
First we know that if $f \in L^2[a,b]$ then $\int_a^bfdB(u)$ is a Gaussian random variable with mean $0$ and variance $||f||_{L^2[a,b]}^2$ and also if $X(t)$ is going to be a Brownian motion then $X(t)$ must be Gaussian with mean $0$ and variance $t$. So since $(a+b\frac{u}{t}) \in L^2[0,t]$ then this shows that $X(t)$ is Gaussian with mean $0$ and variance $t(a^2+ab+\frac{b^2}{3})$ which means that $(a^2+ab+\frac{b^2}{3})$ must equal $1$ for $X(t)$ to be a Brownian motion but this isnt enough im guessing?
Next we need to show that $X(0)=0$ almost surely. For this we may use integration by parts and the identity $\int_0^tB(u)du=\int_0^t(t-u)dB(u)$ to see that $$ X(t)=(a+b)B(t)-\frac{b}{t}\int_0^t(t-u)dB(u) $$ and as $t\to0^+$ we easily see the first term goes to $0$ and the second term also goes to $0$ after applying L'hospital's rule and the stochastic form of Leibniz's rule.
I also need to show that $X(t)$ has independent increments and $X(t)-X(s)$ is Gaussian with mean $0$ and variance $t-s$. I am unsure how to show these two things. Any help is appreciated.