Let $ \{Y_t,0 \leq t \leq T\} $ be a Gaussian process and $ f(.) $ is a measurable function. How can I represent the $n$th moment of $ f(Y_t) $ which is $ E[f(Y_t)^n] $?
Is it $$ E[f(Y_t)^n] = \int_{\mathbb{R}_n} \frac{f(y_1) ... f(y_n)}{(2\pi)^{z/2}\; |\Sigma|^{1/2}} \exp\left(\frac{1}{2} (\mathbf{y}-\mu)^\intercal \Sigma^{-1}\; (\mathbf{y}-\mu)\right) \mathbf{dy} $$ where $ \mathbf{y} = (y_1,...,y_n) $, $ \mathbf{dy} = dy_1 \times ... \times dy_n $, and $ \mu = (\mu_1,...,\mu_n) $ or $$ E[f(Y_t)^n] = \int_{-\infty}^\infty \frac{f(y)^n}{\sigma \sqrt{2\pi}} \exp\left(-\frac{(y-\mu)^2}{2\sigma^2}\right) dy $$ where $ \mu = E[Y_t] $ and $ \sigma^2 = \operatorname{Var}(Y_t) $.
Please help me. If the answer is one of the equations then why is the reason behind the answer, if the answer is both of them, then why and when to use each equations. This confuses me a lot as Gaussian process is a multivariate normal but when we only take $ Y_t $ then it would be just normal.