Suppose $W(t)$ is a Wiener process (which satisfies the following properties):
- $W(0)=0$
- $W(T) - W(t) \sim \mathcal{N}(0, T-t)$ for any $t<T$
- For any $0<t_{1}<...t_{j}<....<t_{n}$, increments $W(t_{j})$, $W(t_{j+1}) - W(t_{j})$ are independent random variables.
I would like to find the distribution of $2W(1.5) - W(2)$ and I am getting conflicting answers.
My first method. Since $W(1.5) \sim \mathcal{N}(0, 1.5)$, then we know that $2W(1.5) \sim \mathcal{N}(0, 6)$ using the fact that for any random variable $X$, the expectation and variance of $2X$ are respectively $E(2X) = 2E(X)$ and $Var(2X) = 4Var(X)$ . Thus, we have:
$$2W(1.5) - W(2) = W(6) - W(2) = W(4) \sim \mathcal{N}(0,4)$$ using property 2 above.
My second method. Let $Y = W(1.5) - (W(2) - W(1.5))$. Then $W(1.5) \sim \mathcal{N}(0, 1.5)$, $W(2) - W(1.5) \sim \mathcal{N}(0, 0.5)$ and these are independent, hence the difference $Y \sim \mathcal{N}(0,2)$.
(My last step is using standard fact that if two random variables, say $X_{1}$ and $X_{2}$, are independent, then $E(X_{1}-X_{2}) = E(X_{1}) - E(X_{2})$ and $Var(X_{1} - X_{2}) = Var(X_{1}) + Var(X_{2})$ ).
Where am I going wrong?
The linear combination of Gaussian variables is normally distributed. It remains to calculate mean and variance. Let $s<t$. $E[aW_s-W_t]=aE[W_s]-E[W_t]=0$. For the variance $$V[aW_s-W_t]=a^2V[W_s]+V[W_t]-2a\textrm{Cov}(W_s,W_t)$$ By definition, $V[W_s]=s,V[W_t]=t$. Now we calculate the covariance. Since the means are zero: $$\begin{aligned}\textrm{Cov}(W_s,W_t)&=E[W_tW_s]=\\ &=E[((W_t-W_s)+W_s)W_s]=\\ &=E[(W_t-W_s)W_s]+E[W_s^2]=\\ &=E[W_t-W_s]E[W_s]+s=s\end{aligned}$$ So $$aW_s-W_t\sim \mathcal{N}(0,a^2s+t-2as)$$