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If instead of a regular Wiener process $W_t$, we had a process of the form $X_t=g(t)W_{at}$ where $g$ is continuous and deterministic and $a$ is a deterministic scalar, then what is the autocorrelation function $R_{X}(t,s)$?

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1 Answer 1

Let $a>0$. Note that

$$\mathbb{E}(X_t) = g(t) \cdot \mathbb{E}(W_{at})=0$$

Moreover,

$$\mathbb{E}(X_t \cdot X_s) = g(t) \cdot g(s) \cdot \mathbb{E}(W_{at} \cdot W_{as}) = g(t) \cdot g(s) \mathbb{E}((W_{at}-W_{as}) \cdot W_{as} + W_{as}^2) \\ = 0 + g(t) \cdot g(s) \mathbb{E}(W_{as}^2)= g(t) \cdot g(s) \cdot (as) $$

for all $s \leq t$ (using the independence of the increments and $W_r \sim N(0,r)$). So

$$R(s,t) = \frac{\mathbb{E}((X_t-\mathbb{E}X_t) \cdot (X_s - \mathbb{E}X_s))}{\sigma_t \cdot \sigma_s} = \frac{g(t) \cdot g(s) \cdot (as)}{\sigma_t \cdot \sigma_s}=c \cdot \frac{s}{\sqrt{t \cdot s}} = c \cdot \sqrt{\frac{s}{t}}$$

for all $s \leq t$ where $$\sigma_t^2 = \mathbb{E}(X_t^2)=g(t)^2 \cdot (a \cdot t) \\ c := \frac{g(t) \cdot g(s)}{\sqrt{g(t)^2 \cdot g(s)^2}} = \text{sgn} (g(t)) \cdot \text{sgn} (g(s))$$

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