Prove that the standard deviation of the value $V(T)$ at time of any portfolio $(x,y)$ at time $T$ in a one-step binomial is given by $\sigma_V=x\sigma_S$, where $\sigma_S$ is the standard deviation of the stock value at time $T$.
I know we should use the property of the expectation and the variance to do this but I have no idea where to start. Can someone help me please?
You can use: $${\rm Var}\left[V(T) \right] = {\rm Var}\left[xS(T)+yA(T) \right] $$ $$=x^2{\rm Var}\left[S(T) \right] + y^2{\rm Var}\left[A(T) \right] + 2xy{\rm Cov}\left[A(T),S(T)\right]$$
You appear to be assuming that $A(T)$ is deterministic, that is, ${\rm Var}\left[A(T) \right]=0$ and ${\rm Cov}\left[A(T),S(T)\right]=0$. This will indeed lead to $$ \sigma_{V(T)} = |x| \sigma_{S(T)}.$$