I am studying for an exam next week and got stuck on this problem.
$W = {W(t)|0 \leq t \leq \infty}$ is a Wiener process. Set for $0\leq t \leq 1$
$W^o(t) = W(t) - t\cdot W(1)$
Find the probability $P(W^o(0.4) > 0.5)$
Once I find out the mean and variance, I am able to calculate the probability. But I can't follow how they calculate the variance. The following is the solution given by the examiner:
$W^o(t) = W(t) − t · W(1) = (1 − t)W(t) − t · (W(1)-W(t))$
$W(t) ∈ N(0, t), W(1) − W(t) ∈ N(0, 1 − t)$
$E[W^o(t)] = (1 − t)E [W(t)] − t · E [W(1) − W(t)] = 0.$
$Var [W^o(t)] = (1-t)^2t+t^2(1-t)=t-2t^2+t^3+t^2-t^3=t*(1-t)$
Thus we have $W^o(t) ∈ N(0, t(1 − t)).$ QED
I have tried using the definition of variance: $Var[W^o(t)] = E[(W^o(t)-\mu)^2] = / \mu = 0 /E[W^o(t)^2] = E[W^o(t)W^o(t)] = R_x(t,t) = min(t,t) = t$
My guess is that I am making a fallible claim that $W^o(t)$ is a brownian processes and therefore am I able to assume $E[W^o(t)W^o(t)] = R_x(t,t)$. But once I substitute $W^o(t)$ for the given above, I just end up with expectations and variances I am unable to calculate.
I'll modify the notation a bit to make everything clearer.
$W = \left(W_t\right)_{t\ge 0}$ is a Brownian motion with $W_0=0$ and we define $$A_t = W_t - tW_1,\qquad t\in[0,1].$$
Now, notice that we may write $$A_t = W_t - t(W_1 - W_t + W_t) = (1-t)W_t + t(W_1 - W_t).$$ At the light of this, $A_t$ corresponds to a convex combination of two independent centered Gaussian variables, $W_t$ and $W_1-W_t$. Hence, $A_t$ is a centered Gaussian variable and $$\mathbb V[A_t] = (1-t)^2\mathbb V[W_t] + t^2\mathbb V[W_1 - W_t] = (1-t)^2t + t^2(1-t) = (1-t)t.$$
In particular, $A_{2/5}\sim\mathrm N\left(0, 6/25\right)$. From this, we may compute $$\mathbb P(A_{2/5} > 1/2) = \mathbb P\left(\frac{A_{2/5} - 0}{\sqrt 6 / 5} > \frac 5{2\sqrt 6}\right) = 1 - \Phi\left(\frac 5{2\sqrt 6}\right) = 0.153.$$