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I am reading about "fake Brownian motion," on page 73 of this link it defines the process

$$X_t^{(a)} = \sqrt{t}(N_1 \cos(B_{a + \log(t)}) + N_2 \sin(B_{a+log(t)}))$$

where $N_1$ and $N_2$ are independent $\mathcal{N}(0,1)$ distributed random variables, $(B_t)_{t\ge0}$ is a standard Brownian motion, and we assume $N_1, N_2, B_t$ are independent. It says that "$X_t^{(a)} \sim \mathcal{N}(0,t)$ for every $t > e^{-a}$ which can easily be seen by just computing expectation and variance."

I can compute the expectation and variance, but I am not sure how this shows that $X^{(a)}_t$ is normally distributed. I know the linear combination of normal random variables are also normally distributed, but with the trig functions it's not clear to me that this should be normally distributed. Can someone help me see that it is normal?

scott_s
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    If $N_1$ and $N_2$ are independent standard normals and $B$ is any random variable independent of $(N_1,N_2)$, then $Z:=N_1\cos(B)+N_2\sin(B)\sim\mathcal N(0,1)$. An easy way to see this is to compute the characteristic function of $Z$, by first conditioning on $B$. – John Dawkins Feb 28 '21 at 23:53
  • Oh, I think I see now! Thanks, John! – scott_s Mar 01 '21 at 05:26

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