SPEA V706 · Math Camp · Day 4
populations vs samples
estimands, estimators, parameters
Sampling distributions
Reading: Hansen, PSE, chapters 6–7
These two ideas are equivalent. But the generative process analogy is often less confusing.
If you are puzzling over what the population is in a given problem, ask yourself how to see it like a bingo ball machine.
If X is a random variable with CDF F_X, we can think data collection as a method of generating realizations from F_X. - Turn the crank on the bingo ball to generate 1 draw from F_X. - Repeat this process to generate multiple draws.
A sample is a set of observations drawn from some population CDF.
Use i = 1...n to index individual observations on draws from F_X.
\{X_1, \dots, X_n\} are the individual draws.
A random sample is a set n “independent and identically distributed (iid)” observations.
Independent: the value of observation i is statistically independent of the value of observation j for all i \neq j.
Identically distributed: every draw comes from the same F_X
An estimand is the population quantity we want to learn, say E[Y]. It is a fixed number, a feature of F.
An estimator is a recipe that turns a sample into a guess at the estimand
The estimate is the value of the estimator for a particular sample.
Thus the estimate is a random variable: if you draw a new sample, you get a different number.
Our estimand is the expected value of Y, denoted E[Y].
Suppose we have a random sample of size n from F_Y() with observations \{Y_1, \dots, Y_n\}.
The sample mean is an intuitive estimator of E[Y]:
\bar Y_n = \frac{1}{n} \sum_{i=1}^{n} Y_i
\bar Y_n is a random variable, so it has a distribution.
That distribution is the sampling distribution of the estimator.
Repeated Random Sampling: Imagine drawing a fresh sample of size n, computing \bar Y_n, and writing the number down. Then repeat this process many times.
The histogram of those sample avarages is the sampling distribution of the the mean
It is not the distribution of Y. It is the distribution of a summary of a sample of Y’s.
The same population, sampled at n = 5, n = 30, and n = 200, with 5000 samples at each size
All three pile up on the same true mean.
Nothing about the population changed here.
The bias of an estimator is how far off it is on average across repeated samples:
\operatorname{Bias}(\hat\theta) = E[\hat\theta] - \theta
The sample mean is unbiased for E[Y]. E[\bar Y_n] = \frac{1}{n} \sum_{i=1}^{n} E[Y_i] = \frac{1}{n}\,n\,\mu = \mu
On average, over many samples, the sample mean lands right on the population expected value.
Unbiasedness says nothing about any specific sample. Your particular estimate can still be badly wrong.
Start with V2 to pull the 1/n out, then V5 to split the variance of the sum. Independence kills every cross term:
\begin{align*} \operatorname{Var}(\bar Y_n) &= \operatorname{Var}\left(\frac{1}{n}\sum_{i=1}^n Y_i\right) \\ &= \frac{1}{n^2}\operatorname{Var}\left(\sum_{i=1}^n Y_i\right) \\ &= \frac{1}{n^2}\sum_{i=1}^n \operatorname{Var}(Y_i) \\ &= \frac{1}{n^2}\,n\,\sigma^2 \;=\; \frac{\sigma^2}{n} \end{align*}
The standard error is the standard deviation of the sampling distribution:
\operatorname{se}(\bar Y_n) = \sqrt{\operatorname{Var}(\bar Y_n)} = \frac{\sigma}{\sqrt{n}}
So far, we have been thinking about the sampling distribution of \bar Y_n for a fixed n across repeated samples.
A different thought experiment is to imagine drawing a single sample, but letting n grow.
The sample mean converges to the population mean as the sample grows:
\bar Y_n \;\xrightarrow{\;p\;}\; E[Y] \qquad \text{as } n \to \infty
The sampling variance \sigma^2/n collapses to zero as n grows
The LLN says \bar Y_n lands on \mu. The CLT tells us the shape of the distribution around it:
\sqrt{n}\,\frac{\bar Y_n - \mu}{\sigma} \;\xrightarrow{\;d\;}\; N(0, 1)
Under weak conditions on the underlying random variable, standardized sample means are distributed approximately Normal as n grows
pacman::p_load(data.table, ggplot2)
set.seed(20260814)
# 5000 sample means of size n = 30, from a
# right-skewed Exponential(1) population
sims <- data.table(
xbar = replicate(
5000, mean(rexp(30, rate = 1))))
# the LLN in one number: close to the
# population mean of 1
sims[, mean(xbar)]
# the CLT in one picture: bell-shaped
p <- ggplot(sims, aes(xbar)) +
geom_histogram(bins = 30,
fill = "#000080")
p # draw the histogram