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Assuming $f:\mathbb R^N \to \mathbb R^N$ is a differentiable function, and $X \in \mathbb R^N$ a random variable with expectation $\mu$ and finite co-variance matrix $\Sigma$.

Does the following hold true, given the linearity of expectation?

$\mathop{\mathbb E} \left[ \sum_i^N f(X)_i \right ] = \sum_i^N \mathop{\mathbb{E}}\left[ f(X)_i \right]$

PedanticCat
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    Assuming everything here is finite, such as $N$ and $\mathop{\mathbb{E}}\left[ f(X)_i \right]$, I would expect it to hold true no matter what the other properties of $X$ and $f$ are – Henry Apr 22 '21 at 14:01
  • An issue arises if say individual $\mathop{\mathbb{E}}\left[ f(X)_1 \right]$ and $\mathop{\mathbb{E}}\left[ f(X)_2 \right]$ are not finite, but $\mathop{\mathbb{E}}\left[ f(X)_1 + f(X)_2\right]$ is finite – Henry Apr 22 '21 at 14:17

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I will use expectations under the image measure $\mu_X(X \in B),B \in \mathcal{B}(\mathbb{R})$ (i.e. the probability distribution of $X$). Differentiability is not a needed requirement. If $f(X)=[f_1(X),f_2(X),...,f_N(X)]$, then let's just consider the finite sequence $(f_n)_{n=1,2,...N}$ of functions $f_n:\mathbb{R}^N\to \mathbb{R}$. If $(f_n)_{n=1,2,...N} \in \mathcal{L}^1(\mu_X)$, that is if they have finite expectation, then by linearity $$\sum_{n=1}^N\mathbb{E}_X[f_n(X)]=\sum_{n=1}^N\int_{\mathbb{R}^N}f_n(x)\mu_X(X \in dx)=\int_{\mathbb{R}^N}\sum_{n=1}^Nf_n(x)\mu_X(X \in dx)=\mathbb{E}_X\bigg[\sum_{n=1}^Nf_n(X)\bigg]<\infty$$ and $$\sum_{n=1}^Nf_n(X) \in \mathcal{L}^1(\mu_X)$$ This is because the set of $\mu_X$-integrable functions $\mathcal{L}^1(\mu_X)$ is a vector space.

Snoop
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