Find covariance of a random process

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.

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

• what is $W_{f(t)}$ supposed to mean ? – Ahmad Bazzi May 18 at 10:16
• @Ahmad It's Wiener process – Helen May 18 at 10:20
• Your answer is correct. – Kavi Rama Murthy May 18 at 11:48
• @Kavi Rama Murthy Thank you! – Helen May 18 at 11:52

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.