So for example, if I had to find $E(\sum_1^n\log(X_i))$ would this be equal to $\sum_1^nE(\log(X_i))$ and then do I proceed from there?
By the way, $X_1, X_2, ...,X_n $ is a random sample from the distribution with probability density function $ f(x)=\theta x^{\theta-1}$, $0<x<1, \theta>0$, just in case anyone wanted to know!
That's right. The reason is that expectation is linear:
$$E[a_1X_1 + ... + a_nX_n] = a_1E[X_1] + ... + a_nE[X_n]$$
This holds for any random variables $X_1, ..., X_n$ (They don't have to be independent or identically distributed) and any finite constants $a_1, ..., a_n$, if all the expectations exist
However, if we consider an infinite sum of random variables and constants, linearity might not hold.
Next:
$$E[\log(X_i)] = \int_{\mathbb R} \log(x_i)f_{X_i}(x_i)dx_i $$
$$ = \int_{\mathbb R} \log(x_i) \theta x_i^{\theta-1}dx_i $$
$$ = \int_{0}^{1} \log(x_i) \theta x_i^{\theta-1}dx_i $$
Do you know how to evaluate
$$\int \log(x_i) x_i^{\theta-1}dx_i$$
?
I found a case where it works and a case where it doesn't.
X <- matrix(rep((1:9/9),3), 3,3)
X
[,1] [,2] [,3]
[1,] 0.1111111 0.4444444 0.7777778
[2,] 0.2222222 0.5555556 0.8888889
[3,] 0.3333333 0.6666667 1.0000000
mean(colSums(log(X)))
[1] -2.324398
sum(colMeans(log(X)))
[1] -2.324398
Here is my attempt to show mathematically. I am going to assume n is the length of each sample X in this case.
\begin{align} E\bigg(\sum^n_1 \log(X_i)\bigg) &= \frac{\sum^n_1 \log(X_1) + \sum^n_1 \log(X_2) + ... \sum^n_1 \log(X_n)}{n} \\[10pt] \sum^n_1 E\big(\log(X_i)\big) &= \sum^n_1 \frac{\sum^n_1 \log(X_1)}{n} + \frac{\sum^n_1 \log(X_2)}{n} + ... \frac{\sum^n_1 \log(X_n)}{n} \\[10pt] &= \frac{\sum^n_1 \log(X_1) + \sum^n_1 \log(X_2) + ... \sum^n_1 \log(X_n)}{n} \end{align}
If the size of $X$ is different they would not be equal.
> X <- matrix(rep((1:12/12),4), 4,3)
> X
[,1] [,2] [,3]
[1,] 0.08333333 0.4166667 0.7500000
[2,] 0.16666667 0.5000000 0.8333333
[3,] 0.25000000 0.5833333 0.9166667
[4,] 0.33333333 0.6666667 1.0000000
> sum(colMeans(log(X)))
[1] -2.457916
> mean(colSums(log(X)))
[1] -3.277222
\begin{align} \sum^n_1 E(\log(X_i)) &= \sum^n_1 \frac{\sum^n_1 \log(X_1)}{k} + \frac{\sum^n_1 \log(X_2)}{k} + ... \frac{\sum^n_1 \log(X_n)}{k} \\[10pt] &= \frac{\sum^n_1 \log(X_1) + \sum^n_1 \log(X_2) + ... \sum^n_1 \log(X_n)}{k} \end{align} where $k$ is the length of each $X$ and $k \ne n$.
Assuming you have samples $X_1^{(j)},...,X_N^{(j)}$ of each variable $X^{(j)}$, and there are $p$ variables:
$$ E\left[ \sum_{j=1}^p \log(X^{(j)})\right] \approx (1/N) \sum_{i=1}^N \left \{ \sum_{j=1}^p \log(X_i^{(j)}) \right \} $$
