Why not any random variable is an unbiased estimator of the population mean?

Question

To prove sample mean is an unbiased estimator of the population mean, we use the following step, $\mathbb{E}[\sum \frac{X_i}{n}] = \frac{1}{n} \sum \mathbb{E}[X_i] = \mu$, given that $\mathbb{E}[X_i] = \mu$.

Now my question is if we take say first random variable $X_1$ as an estimate we can go ahead and say that since $\mathbb{E}{[X_1]} = \mu$, $X_1$ is an unbiased estimator. Does that mean $X_1$ or for that matter any value is an unbiased estimator?

I know it is not. I am doing a conceptual mistake in thinking, please let me know where I am going completely wrong.

[Edit] As commented below for iid's every sample is indeed an unbiased estimate of the mean.

Answer 1

It looks like you are making the mistake of confusing data with the random variables.

Let a random variable $X$ have mean $\mu$. Then if $(x_1, \ldots x_n)$ are realizations from $X$, $\frac{1}{n} \sum x_i = \hat{\mu}$ is an unbiased estimator of $\mu$. This is because $\mathbb{E}[\mu - \hat{\mu}] = 0$, a fact that we will use later. In fact if the data is drawn IID from $X$ then each individual sample is an unbiased estimator of the mean; the reason we take many samples is to lower the variance of our estimate.

Now if, as you mentioned, we have multiple random variables, $X_1, \ldots, X_p$, each of which has mean $\mu$, then $\mathbb{E}\left[\frac{1}{p}\sum_{i=1}^p X_i \right] = \mu$ as you suggest. Furthermore, $\mathbb{E}X_1 = \mu$. That is, $\mathbb{E}[\mu - \mathbb{E}X_1]=0$. This implies that the observed realization of $X_1$ is an unbiased estimator of $\mu$. There is nothing startling here but it is distinct, I think, from the question you meant to be asking which was answered in the previous paragraph. This distinction is important and worth understanding well.