# Using integrals to prove that the mean of the sampling distribution is the population mean

Let the random variables $X_1, X_2, \dots X_n$ denote a random sample from a population.

The sample mean of these random variables is: $\overline{X}=\frac{1}{n}\sum\limits_{i=1}^{n}X_i$

I would like to show that the mean of the sampling distribution of the sample mean is $\mu$, the population mean.

Here's what I have done:

\begin{align} E(\overline{X}) &= \int\limits_{\overline{X}} \bar{x} f(\bar{x})\,\, d\bar{x} \\ &=\int\limits_{\overline{X}} \left(\frac{1}{n} \sum\limits_{i=1}^n X_i \right) f(\bar{x}) \, d\bar{x} \end{align}

From here, I am not sure what to do anymore but anyway I end up with:

$$\begin{array} {cc} &=& \frac{1}{n} \left( \int\limits_{\overline{X}}X_1f(\bar{x}) \, d\bar{x} + \int\limits_{\overline{X}}X_2f(\bar{x}) \, d\bar{x}+ \dots + \int\limits_{\overline{X}}X_nf(\bar{x}) \, d\bar{x}\right) \end{array}$$

Now, I don't know how to complete this as I am unsure how to interpret the last equation. Somehow, the $\int\limits_{\overline{X}}X_if(\bar{x}) \, d\bar{x}$ is suppose to equal to $\mu$ but I don't see how that can be true.

I know the answer will be $\mu$ because of here but I would like to arrive at the answer using integrals instead.

• @AlecosPapadopoulos Thanks for the link. I can't follow it very well yet as I have not covered the topics you touched on. Is trying to arrive at the answer using integrals really more complicated than relying on the properties of expected values? Oct 6, 2013 at 22:09
• It is more complicated, although not really more complex. You cannot avoid the joint density and the multiple integrals, since the sample mean is a function of many random variables (and you do not assume that the variables are independent in order to prove unbiasedness of the sample mean). Oct 6, 2013 at 22:12
• Okay, noted. Then I will have to revisit this question later on in the future. But I would appreciate it if you can post a complete answer. That way, I can accept it and mark this question as answered. Oct 6, 2013 at 22:26

No problem giving a complete answer. Let's see

As already stated, the sample mean is a function of many random variables, and so the symbol $E$ refers to the expected value with respect to their joint distribution. Denoting $\mathbf X$ the multivariate vector of the $n$ r.v.'s, their joint density can be written as $f_{\mathbf X}(\mathbf x)= f_{X_1,...,X_n}(x_1,...,x_n)$ and their joint support $D = S_{X_1} \times ...\times S_{X_n}$

The sample mean is a function of this multivariate vector, $\bar X = \frac 1n \sum_{i=1}^{n}X_i = g(\mathbf X)$. Using the Law of Unconcscious Statistician We have

$$E[\frac 1n \sum_{i=1}^{n}X_i] = \int_D g(\mathbf x)f_{\mathbf X}(\mathbf x)d\mathbf x$$.

Under convergence regularity conditions we can decompose the multidimensional integral into an n-iterative integral:

$$E[\frac 1n \sum_{i=1}^{n}X_i] = \int_{S_{X_n}}...\int_{S_{X_1}}\left[\frac 1n \sum_{i=1}^{n}x_i\right]f_{X_1,...,X_n}(x_1,...,x_n)dx_1...dx_n$$

and using the linearity of integrals we can decompose into

$$= \frac 1n\int_{S_{X_n}}...\int_{S_{X_1}}x_1f_{X_1,...,X_n}(x_1,...,x_n)dx_1...dx_n \; + ...\\ ...+\frac 1n\int_{S_{X_n}}...\int_{S_{X_1}}x_nf_{X_1,...,X_n}(x_1,...,x_n)dx_1...dx_n$$

For each n-iterative integral we can re-arrange the order of integration so that, in each, the outer integration is with respect to the variable that is outside the joint density. Namely,

$$\frac 1n\int_{S_{X_n}}...\int_{S_{X_1}}x_1f_{X_1,...,X_n}(x_1,...,x_n)dx_1...dx_n = \\\frac 1n\int_{S_{X_1}}x_1\int_{S_{X_n}}...\int_{S_{X_2}}f_{X_1,...,X_n}(x_1,...,x_n)dx_2...dx_ndx_1$$

and in general

$$\frac 1n\int_{S_{X_n}}...\int_{S_{X_j}}...\int_{S_{X_1}}x_jf_{X_1,...,X_n}(x_1,...,x_n)dx_1...dx_j...dx_n =$$ $$=\frac 1n\int_{S_{X_j}}x_j\int_{S_{X_n}}...\int_{S_{X_{j-1}}}\int_{S_{X_{j+1}}}...\int_{S_{X_1}}f_{X_1,...,X_n}(x_1,...,x_n)dx_1...dx_{j-1}dx_{j+1}......dx_ndx_j$$

As we calculate one-by-one the integral in each n-iterative integral (starting from the inside), we "integrate out" a variable and we obtain in each step the "joint-marginal" distribution of the other variables. Each n-iterative integral therefore will end up as $\frac 1n\int_{S_{X_j}}x_jf_{X_j}(x_j)dx_j$.

Bringing it all together we arrive at

$$E[\frac 1n\sum_{i=1}^{n} X_i ] = \frac 1n\int_{S_{X_1}}x_1f_{X_1}(x_1)dx_1 +...+\frac 1n\int_{S_{X_n}}x_nf_{X_n}(x_n)dx_n$$

But now each simple integral is the expected value of each random variable separately, so

$$= E[\frac 1n\sum_{i=1}^{n} X_i ] = \frac 1nE(X_1) + ...+\frac 1nE(X_n) = \frac 1nE(X) + ...+\frac 1nE(X)$$ $$= \frac 1n nE(X) = E(X)$$

You should not confuse the argument to the probability density function with the random variable. Often one uses a lower-case letter for the former and a capital letter for the latter. For example, if the density function for the random variable $X$ is $f$, then one can speak of $\int_{-\infty}^\infty xf(x)\,dx$, and the lower-case "$x$" is not the random variable $X$. It is analogous to such things as "$P(X\le x)$": the "$X$" and the "$x$" mean two different things. One can also write $f(3)$, and it doesn't mean the density function of a random variable called "$3$"; it means the value at the number $3$, of the random variable $X$.

Thus one can write $f_X(4)$ and $f_Y(4)$ and they're the values at $4$ of the density functions of two different random variables $X$ and $Y$.

If you're writing $$\mathbb E(\bar X) = \int_{-\infty}^\infty \bar x f(\bar x)\,d\bar x,$$ then that means the same thing as $$\mathbb E(\bar X) = \int_{-\infty}^\infty x f_{\bar X}(x)\,dx.$$ You can't put the random variable $\bar X$ in place of the bound variable $\bar x$.

To understand this, it may help to realize that $$\sum_{j=1}^3 (i^2 j^3) \text{ and } \sum_{k=1}^3 (i^2 k^3)$$ both mean $$i^2 1^3 + i^2 2^3 + i^2 3^3,$$ and thus both depend on the value of the "free variable" $i$ but not on any value of the "bound variable" that is either $j$ or $k$. There's no $j$ or $k$ for them to depend on. One can freely change the name of the bound variable from $j$ to $k$ without changing the value of the expression.

The same thing applies to $$\mathbb E(X) = \int_{\mathbb R} x f_X(x)\,dx = \int_{\mathbb R} w f_X(w)\,dw.$$ One can freely change the name of the bound variable from $x$ to $w$. But the capital $X$ still refers to the same random variable.

If you intend $f$ to be the density function of $\bar X$, then in your second displayed line in your question, $f$ is still the density function of the random variable $\bar X$.

If I wanted to do this by using the density function of the sample mean $\bar X$, I'd actually need to show how that function depends on the density function of $X_1$. That would be an $n$-gold convolution.

I wouldn't do it that way if I could help it. Instead I'd use the linearity of expectation, thus: $$\mathbb E\left(\frac{X_1+\cdots+X_n}{n}\right) = \frac1n\mathbb E(X_1+\cdots+X_n) = \frac1n((\mathbb E X_1)+\cdots+(\mathbb E X_n))$$

There is of course the problem of how to prove the linearity of expectation. Maybe that's posted here somewhere as a separate question already.