Given $Z(t)=B(t)Y$, with $B(t)$ standard brownian motion, also $t \in (0,1)$ and $Y \sim N(0,1)$. $B(t)$ and $Y$ are independent. Compute $E[Z(t)]$ , $Var[Z(t)]$ and $Cov[Z(t),Z(s)]$ . Prove also that $Z(t)$ is a brownian motion. It's easy to obtain $$ E[Z(t)]=0 \ \ \ \ \ \ \ \ Var[Z(t)]=t $$ Then I suppose that $t>s$ and compute
$$ Cov[Z(t),Z(s)]=E\big\{[B(t)Y][B(s)Y]\big\} =\\ =E\big\{Y^2B(s)[B(t)-B(s)]+Y^2B^2(s) \big\} = \\ E[Y^2B^2(s)]=s $$ For a generic $s$ we get
$$ Cov[Z(t),Z(s)]=min\{t,s\} $$
Here comes the doubt. To prove that Z(t) is also a brownian motion I have to prove the independence of the increments. For $t>s$ $$ E[Z(t)-Z(s)]=0 \\ Var[Z(t)-Z(s)]=s $$
To prove that generic increments are mutually independent is sufficient to prove that $Cov[Z(t)-Z(s),Z(s)-Z(0)]=0$? Thank you for the answer. :)
You computed $\mathbb{E}(Z(t))$ , $Var(Z(t))$ and $Cov(Z(t),Z(s))$ correctly.
But what you've been asked to prove is not true, i.e. $Z$ is not a Brownian motion. To see this, just note that $Z(t)$ is the product of two Normal random variables, so it cannot be normal.