Understanding the proof of sample mean being unbiased estimator of population mean in Simple Random Sampling Without Replacement (SRSWOR)

Question

I was reading about the proof of the sample mean being the unbiased estimator of population mean. Here is the concerned derivation:

Let us consider the simple arithmetic mean $\bar y = \frac{1}{n}\,\sum_{i=1}^{n} y_i$ as an unbiased estimator of population mean $\overline Y = \frac{1}{N}\,\sum_{i=1}^{N} Y_i$.

Simple Random Sampling Without Replacement

Let $t_i = \sum_{i=1}^n y_i\;.$

\begin{align}\mathrm E(\bar y)&= \frac{1}{n}\,\mathrm E{\left(\sum_{i=1}^n y_i\right)}\\&= \frac{1}{n}\,\mathrm E(t_i)\\ &= \frac{1}{n}\color{red}{\left(\frac{1}{N \choose n}\,\sum_{i=1}^{N \choose n}t_i\right)}\\ &= \frac{1}{n}\left(\frac{1}{N \choose n}\,\sum_{i=1}^{N \choose n}\left(\sum y_i\right)\right)\\ &= \frac{1}{n}\left(\frac{1}{N \choose n}\,\color{red}{{N-1\choose n-1}\sum_{i=1}^N y_i}\right)\\ &=\frac{1}{N}\sum_{i=1}^{N}y_i\\ &= \overline Y\;.\end{align}

I couldn't understand the derivation; I really couldn't conceive how the red coloured terms came from nowhere.

Can anyone please help me explain the derivation by showing how the red coloured terms came in the concerned steps?

Answer 1

The author's derivation is pretty strange. The fact that $\bar{y}$ is an unbiased estimate of $\mu$ the population mean when sampling without replacement is true due to linearity of expectation alone:

\begin{align} \text{E}(\bar{y}) &= \text{E} \left ( \frac{1}{n} \sum_{i=1}^{n} y_i \right ) \\ &= \frac{1}{n} \sum_{i=1}^{n} \text{E}(y_i) \\ &= \frac{1}{n} \sum_{i=1}^{n} \mu \\ &= \mu . \end{align}

There's no reason to bother with combinatorics.

Answer 2

Ok this question has been silent for a while, but I was intrigued and researched it. Here is what I came up with:

The term in red on the third line arises as follows:

We can calculate the expectation for a discrete random variable, $X$ using $$\text{E}(X) = \sum_{i=1}^k x_i p_i$$ where we have a finite number of outcomes, $x_1,x_2,\dots,x_k$ that occur with probabilities $p_1, p_2, \dots, p_k$ respectively.

In the case of SRSWOR, we are selecting $n$ (sample size) units out of $N$ (population). The total number of possible samples is $N \choose n$, since the probability of choosing a sample is uniform, the probability of selecting any one of these samples is $\frac{1}{N \choose n}$.

In this example, $x_i=t_i=\sum_{j=1}^n y_i$ and $k = {N \choose n}$ and $p_i = \frac{1}{N \choose n}$. So, we have \begin{align} E[t_i] &= \sum_{i=1}^{N \choose n} \frac{1}{N \choose n} t_i\\ &= \frac{1}{N \choose n} \sum_{i=1}^{N \choose n} \sum_{i=1}^n y_i\\ \end{align}

The term in red on the 5th line, I can only provide a high level intuition for. The short answer is essentially, $$\sum_{i=1}^{N \choose n} \sum_{i=1}^n y_i = {N-1 \choose n-1} \sum_{i=1}^N y_i$$ Note on the right that the summation now goes from 1 to $N$ instead of 1 to $n$.

You do some straightforward (?) math to find $$ \frac1n \frac{1}{N \choose n}{N-1 \choose n-1} = \frac1N$$ And you have your proof. Using combinatorics provides one way to gain intuition regarding key aspects of choosing n samples from a population of N possible samples without replacement (SRSWOR). In this case it gives us a way to determine whether the sample arithmetic mean is an unbiased estimator of the population mean at a deeper level.