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Let $Y_t = e^{-\alpha t}W_{\beta \ \exp({2\alpha t})}$, where $W_{s} \ \text {is Wiener prosses,} \quad 0\le t, \quad \alpha , \beta\in \mathbb R^1$. Find $\text{Cov}(Y_t , Y_s).$

Here is my solution. But I'm not sure and asking for verifications.

$\text{Cov}(Y_t , Y_s) = e^{-\alpha t}e^{-\alpha s} \min(\beta e^{ {2\alpha t}},\beta e^{ {2\alpha s}}) = \beta e^{-\alpha|t - s|}$

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    $\begingroup$ what is $W_{f(t)}$ supposed to mean ? $\endgroup$ – Ahmad Bazzi May 18 at 10:16
  • $\begingroup$ @Ahmad It's Wiener process $\endgroup$ – Helen May 18 at 10:20
  • $\begingroup$ Your answer is correct. $\endgroup$ – Kavi Rama Murthy May 18 at 11:48
  • $\begingroup$ @Kavi Rama Murthy Thank you! $\endgroup$ – Helen May 18 at 11:52
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Since $Y_t$ has mean $0$ and for a Wiener process holds $\Bbb E [W_r W_u] = r\wedge u$ we have that $$\text{Cov}(Y_t, Y_s) = \Bbb E [Y_t Y_s] = \Bbb E [e^{-\alpha t} W_{\beta \exp (2\alpha t)} e^{-\alpha s} W_{\beta \exp (2\alpha s)}] = e^{-\alpha t} e^{-\alpha s} \Bbb E [ W_{\beta \exp (2\alpha t)} W_{\beta \exp (2\alpha s)}]\\ = e^{-\alpha t} e^{-\alpha s} \beta \min (\exp (2\alpha t) , \exp (2\alpha s)) = \beta \min (e^{\alpha (t-s)}, e^{\alpha (s-t)}) = \beta e^{-\alpha \vert t - s\vert}$$ Your answer is correct.

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