Mean of the bootstrap sample vs statistic of the sample Say I have a sample and the bootstrap sample from this sample for a stastitic $\chi$ (e.g. the mean). As we all know, this bootstrap sample estimates the sampling distribution of the estimator of the statistic.
Now, is the mean of this bootstrap sample a better estimate  of the population statistic than the statistic of the original sample? Under what conditions would that be the case?
 A: Let's generalize, so as to focus on the crux of the matter.  I will spell out the tiniest details so as to leave no doubts.  The analysis requires only the following:


*

*The arithmetic mean of a set of numbers $z_1, \ldots, z_m$ is defined to be 
$$\frac{1}{m}\left(z_1 + \cdots + z_m\right).$$

*Expectation is a linear operator.  That is, when $Z_i, i=1,\ldots,m$ are random variables and $\alpha_i$ are numbers, then the expectation of a linear combination is the linear combination of the expectations,
$$\mathbb{E}\left(\alpha_1 Z_1 + \cdots + \alpha_m Z_m\right) = \alpha_1 \mathbb{E}(Z_1) + \cdots + \alpha_m\mathbb{E}(Z_m).$$
Let $B$ be a sample $(B_1, \ldots, B_k)$ obtained from a dataset $x = (x_1, \ldots, x_n)$ by taking $k$ elements uniformly from $x$ with replacement.  Let $m(B)$ be the arithmetic mean of $B$.  This is a random variable.  Then
$$\mathbb{E}(m(B)) = \mathbb{E}\left(\frac{1}{k}\left(B_1+\cdots+B_k\right)\right) = \frac{1}{k}\left(\mathbb{E}(B_1) + \cdots + \mathbb{E}(B_k)\right)$$
follows by linearity of expectation.  Since the elements of $B$ are all obtained in the same fashion, they all have the same expectation, $b$ say:
$$\mathbb{E}(B_1) = \cdots = \mathbb{E}(B_k) = b.$$
This simplifies the foregoing to
$$\mathbb{E}(m(B)) = \frac{1}{k}\left(b + b + \cdots + b\right) = \frac{1}{k}\left(k b\right) = b.$$
By definition, the expectation is the probability-weighted sum of values.  Since each value of $X$ is assumed to have an equal chance of $1/n$ of being selected,
$$\mathbb{E}(m(B)) = b = \mathbb{E}(B_1) = \frac{1}{n}x_1 + \cdots + \frac{1}{n}x_n = \frac{1}{n}\left(x_1 + \cdots + x_n\right) = \bar x,$$
the arithmetic mean of the data.
To answer the question, if one uses the data mean $\bar x$ to estimate the population mean, then the bootstrap mean (which is the case $k=n$) also equals $\bar x$, and therefore is identical as an estimator of the population mean.

For statistics that are not linear functions of the data, the same result does not necessarily hold.  However, it would be wrong simply to substitute the bootstrap mean for the statistic's value on the data: that is not how bootstrapping works.  Instead, by comparing the bootstrap mean to the data statistic we obtain information about the bias of the statistic.  This can be used to adjust the original statistic to remove the bias.  As such, the bias-corrected estimate thereby becomes an algebraic combination of the original statistic and the bootstrap mean.  For more information, look up "BCa" (bias-corrected and accelerated bootstrap) and "ABC".  Wikipedia provides some references.
A: Since the bootstrap distribution associated with an iid sample $X_1,\ldots,X_n$ is defined as$$\hat{F}_n(x) = \frac{1}{n}\sum_{i=1}^n\mathbb{I}_{X_i\le x}\qquad X_i\stackrel{\text{iid}}{\sim}F(x)\,,$$
the mean of the bootstrap distribution $\hat{F}_n$ (conditional on the iid sample $X_1,\ldots,X_n$) is$$\mathbb{E}_{\hat{F}_n}[X]=\frac{1}{n}\sum_{i=1}^n X_i=\bar{X}_ n$$
When you (if you have to) implement a simulation version of this expectation, i.e., compute an average of $B$ random draws from $\hat{F}_n$, $$\hat{\mathbb{E}}_{\hat{F}_n}[X]=\frac{1}{B} \sum_{b=1}^B X^*_b \qquad X^*_i\stackrel{\text{iid}}{\sim}\hat F_n(x)\,,$$there is some Monte Carlo variability in this approximation of $\mathbb{E}_{\hat{F}_n}[X]$, but its mean (the expectation of the empirical average, conditional on the original sample $X_1,\ldots,X_n$) and its limit when the number $B$ of bootstrap simulations grows to infinity are both exactly $\bar{X}_ n$.
