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## 3.2 Properties of the Sample Mean

A more precise way to express consistency of an estimator $\hat\mu$ for a parameter $\mu$ is

$P(|\hat{\mu} - \mu|<\epsilon) \xrightarrow[n \rightarrow \infty]{p} 1 \quad \text{for any}\quad\epsilon>0.$

This expression says that the probability of observing a deviation from the true value $\mu$ that is smaller than some arbitrary $\epsilon > 0$ converges to $1$ as $n$ grows. Consistency does not require unbiasedness.

To examine properties of the sample mean as an estimator for the corresponding population mean, consider the following R example.

We generate a population pop consisting of observations $Y_i$, $i=1,\dots,10000$ that origin from a normal distribution with mean $\mu = 10$ and variance $\sigma^2 = 1$.

To investigate the behavior of the estimator $\hat{\mu} = \bar{Y}$ we can draw random samples from this population and calculate $\bar{Y}$ for each of them. This is easily done by making use of the function replicate(). The argument expr is evaluated n times. In this case we draw samples of sizes $n=5$ and $n=25$, compute the sample means and repeat this exactly $N=25000$ times.

For comparison purposes we store results for the estimator $Y_1$, the first observation in a sample for a sample of size $5$, separately.

# generate a fictious population
pop <- rnorm(10000, 10, 1)

# sample from the population and estimate the mean
est1 <- replicate(expr = mean(sample(x = pop, size = 5)), n = 25000)

est2 <- replicate(expr = mean(sample(x = pop, size = 25)), n = 25000)

fo <- replicate(expr = sample(x = pop, size = 5), n = 25000)

Check that est1 and est2 are vectors of length $25000$:

# check if object type is vector
is.vector(est1)
##  TRUE
is.vector(est2)
##  TRUE
# check length
length(est1)
##  25000
length(est2)
##  25000

The code chunk below produces a plot of the sampling distributions of the estimators $\bar{Y}$ and $Y_1$ on the basis of the $25000$ samples in each case. We also plot the density function of the $\mathcal{N}(10,1)$ distribution.

# plot density estimate Y_1
plot(density(fo),
col = 'green',
lwd = 2,
ylim = c(0, 2),
xlab = 'estimates',
main = 'Sampling Distributions of Unbiased Estimators')

# add density estimate for the distribution of the sample mean with n=5 to the plot
lines(density(est1),
col = 'steelblue',
lwd = 2,
bty = 'l')

# add density estimate for the distribution of the sample mean with n=25 to the plot
lines(density(est2),
col = 'red2',
lwd = 2)

# add a vertical line at the true parameter
abline(v = 10, lty = 2)

# add N(10,1) density to the plot
curve(dnorm(x, mean = 10),
lwd = 2,
lty = 2,

# add a legend
legend("topleft",
legend = c("N(10,1)",
expression(Y),
expression(bar(Y) ~ n == 5),
expression(bar(Y) ~ n == 25)
),
lty = c(2, 1, 1, 1),
col = c('black','green', 'steelblue', 'red2'),
lwd = 2) First, all sampling distributions (represented by the solid lines) are centered around $\mu = 10$. This is evidence for the unbiasedness of $Y_1$, $\overline{Y}_{5}$ and $\overline{Y}_{25}$. Of course, the theoretical density $\mathcal{N}(10,1)$ is centered at $10$, too.

Next, have a look at the spread of the sampling distributions. Several things are noteworthy:

• The sampling distribution of $Y_1$ (green curve) tracks the density of the $\mathcal{N}(10,1)$ distribution (black dashed line) pretty closely. In fact, the sampling distribution of $Y_1$ is the $\mathcal{N}(10,1)$ distribution. This is less surprising if you keep in mind that the $Y_1$ estimator does nothing but reporting an observation that is randomly selected from a population with $\mathcal{N}(10,1)$ distribution. Hence, $Y_1 \sim \mathcal{N}(10,1)$. Note that this result does not depend on the sample size $n$: the sampling distribution of $Y_1$ is always the population distribution, no matter how large the sample is. $Y_1$ is a good a estimate of $\mu_Y$, but we can do better.

• Both sampling distributions of $\overline{Y}$ show less dispersion than the sampling distribution of $Y_1$. This means that $\overline{Y}$ has a lower variance than $Y_1$. In view of Key Concepts 3.2 and 3.3, we find that $\overline{Y}$ is a more efficient estimator than $Y_1$. In fact, this holds for all $n>1$.

• $\overline{Y}$ shows a behavior illustrating consistency (see Key Concept 3.2). The blue and the red densities are much more concentrated around $\mu=10$ than the green one. As the number of observations is increased from $1$ to $5$, the sampling distribution tightens around the true parameter. Increasing the sample size to $25$, this effect becomes more apparent. This implies that the probability of obtaining estimates that are close to the true value increases with $n$.

We encourage you to go ahead and modify the code. Try out different values for the sample size and see how the sampling distribution of $\overline{Y}$ changes!

#### $\overline{Y}$ is the Least Squares Estimator of $\mu_Y$

Assume you have some observations $Y_1,\dots,Y_n$ on $Y \sim \mathcal{N}(10,1)$ (which is unknown) and would like to find an estimator $m$ that predicts the observations as well as possible. By good we mean to choose $m$ such that the total squared deviation between the predicted value and the observed values is small. Mathematically, this means we want to find an $m$ that minimizes

$\begin{equation} \sum_{i=1}^n (Y_i - m)^2. \tag{3.1} \end{equation}$

Think of $Y_i - m$ as the mistake made when predicting $Y_i$ by $m$. We could also minimize the sum of absolute deviations from $m$ but minimizing the sum of squared deviations is mathematically more convenient (and will lead to a different result). That is why the estimator we are looking for is called the least squares estimator. $m = \overline{Y}$, the sample mean, is this estimator.

We can show this by generating a random sample and plotting (3.1) as a function of $m$.

# define the function and vectorize it
sqm <- function(m) {
sum((y-m)^2)
}
sqm <- Vectorize(sqm)

# draw random sample and compute the mean
y <- rnorm(100, 10, 1)
mean(y)
##  10.00543
# plot the objective function
curve(sqm(x),
from = -50,
to = 70,
xlab = "m",
ylab = "sqm(m)")

# add vertical line at mean(y)
abline(v = mean(y),
lty = 2,
col = "darkred")

# add annotation at mean(y)
text(x = mean(y),
y = 0,
labels = paste(round(mean(y), 2))) Notice that (3.1) is a quadratic function so that there is only one minimum. The plot shows that this minimum lies exactly at the sample mean of the sample data.

Some R functions can only interact with functions that take a vector as an input and evaluate the function body on every entry of the vector, for example curve(). We call such functions vectorized functions and it is often a good idea to write vectorized functions yourself, although this is cumbersome in some cases. Having a vectorized function in R is never a drawback since these functions work on both single values and vectors.

Let us look at the function sqm(), which is non-vectorized:

sqm <- function(m) {
sum((y-m)^2) #body of the function
}

Providing, e.g., c(1,2,3) as the argument m would cause an error since then the operation y-m is invalid: the vectors y and m are of incompatible dimensions. This is why we cannot use sqm() in conjunction with curve().

Here Vectorize() comes into play. It generates a vectorized version of a non-vectorized function.

#### Why Random Sampling is Important

So far, we assumed (sometimes implicitly) that the observed data $Y_1, \dots, Y_n$ are the result of a sampling process that satisfies the assumption of simple random sampling. This assumption often is fulfilled when estimating a population mean using $\overline{Y}$. If this is not the case, estimates may be biased.

Let us fall back to pop, the fictive population of $10000$ observations and compute the population mean $\mu_{\texttt{pop}}$:

# compute the population mean of pop
mean(pop)
##  9.992604

Next we sample $10$ observations from pop with sample() and estimate $\mu_{\texttt{pop}}$ using $\overline{Y}$ repeatedly. However, now we use a sampling scheme that deviates from simple random sampling: instead of ensuring that each member of the population has the same chance to end up in a sample, we assign a higher probability of being sampled to the $2500$ smallest observations of the population by setting the argument prop to a suitable vector of probability weights:

# simulate outcomes for the sample mean when the i.i.d. assumption fails
est3 <-  replicate(n = 25000,
expr = mean(sample(x = sort(pop),
size = 10,
prob = c(rep(4, 2500), rep(1, 7500)))))

# compute the sample mean of the outcomes
mean(est3)
##  9.443454

Next we plot the sampling distribution of $\overline{Y}$ for this non-i.i.d. case and compare it to the sampling distribution when the i.i.d. assumption holds.

# sampling distribution of sample mean, i.i.d. holds, n=25
plot(density(est2),
col = 'steelblue',
lwd = 2,
xlim = c(8, 11),
xlab = 'Estimates',
main = 'When the i.i.d. Assumption Fails')

# sampling distribution of sample mean, i.i.d. fails, n=25
lines(density(est3),
col = 'red2',
lwd = 2)

# add a legend
legend("topleft",
legend = c(expression(bar(Y)[n == 25]~", i.i.d. fails"),
expression(bar(Y)[n == 25]~", i.i.d. holds")
),
lty = c(1, 1),
col = c('red2', 'steelblue'),
lwd = 2) Here, the failure of the i.i.d. assumption implies that, on average, we underestimate $\mu_Y$ using $\overline{Y}$: the corresponding distribution of $\overline{Y}$ is shifted to the left. In other words, $\overline{Y}$ is a biased estimator for $\mu_Y$ if the i.i.d. assumption does not hold.