Let $\{X_t\}$ be a strict-stationary process and $w \in [0,1]$ be some constant. I wonder if there is a way to calculate (or lower bound) the expected product
$$\mathbb{E} \left[ \prod_{t=1}^T (1+w X_t) \right]$$
for some integer $T>1$.
It is well-known that if $X_t$ are independent, then the expectation above can be reduced to the product of the expected value of individual term; i.e., $\mathbb{E}[1+wX_i]$. Since strict stationarity implies that the $X_i$ has the same mean, say $\mu$, we can further simplify the term to obtain
$$\mathbb{E}[1+wX_i] = 1+w\mu.
$$
However, if we drop the independence, then I fail to see how this might be calculated (perhaps, the only way that I can perceive is via joint probability distribution of $X_1,X_2,\dots,X_T$ to calculate the $\int \int \cdots \int \prod_{t=1}^T(1+wx_t) dF(x_1,x_2,\dots,x_T)$…) Any comment is appreciated.
Best Answer
Please check if the below works:
From the definition of strictly stationary, we have: $F_{X}(x_{t_1+\tau} ,\ldots, x_{t_n+\tau}) = F_{X}(x_{t_1},\ldots, x_{t_n}) \quad \text{for all } \tau,t_1, \ldots, t_n \in \mathbb{R} \text{ and for all } n \in \mathbb{N}$. Specifically, setting $n =1, \tau = i$ and letting $t$ be some arbitrary starting point, $F_{X}(x_{t+i})=F_{X}(x_{t}) \quad \forall i$.
Thus $\{X_t\}$ are identically distributed. Then $\mathbb{E}\left[\prod\limits_{t=1}^T(1+wX_t)\right] = \mathbb{E}\left[(1+wX)^T\right]$ where $T$ is fixed and $X$ is the common distribution of $\{X_1,\dots,X_T\}$.
Now we can compute this using the binomial expansion as $\mathbb{E}\left[(1+wX)^T\right]$ \begin{align*} \mathbb{E}\left[(1+wX)^T\right] &= \mathbb{E}\left[1+\binom{T}{1}wX + \binom{T}{2}(wX)^2+\dots+\binom{T}{T}(wX)^T\right]\\ &=1+\binom{T}{1}w\mathbb{E}[X] + \binom{T}{2}w^2\mathbb{E}[X^2]+\dots+\binom{T}{T}w^T\mathbb{E}[X^T] \end{align*}
which can be computed, if one knows each of the moments of the random variable $X$.