Bias and Variance

Jan 25, 2020
11 min read
Jun 30, 2020 21:42 EDT
The bias, variance and mean squared error of an estimator. The efficiency is used to compare two estimators.
To get started, we’re going to introduce some basic concepts.

Basic concepts

Def: A parameter is a numerical characteristic of a population distribution, and is often unknown. A Statistic is a numerical summary of the sample, and it should depend on the sample only and does not involve unknown parameters.

If we have a model $Y_1, \cdots, Y_n \overset{i.i.d.}{\sim} Bern(p)$, here $p$ is unknown and is a parameter. $Y_1$ is a statistic of the sample, and so is $Y_1 + \cdots + Y_n$. However, $\sum_{i=1}^n (Y_i - p)^2$ is not a statistic because the unknown parameter $p$ is involved.

More rigorously speaking, a quantity $T$ is called a statistic if it can be expressed as a function of the sample

$$ T = f(Y_1, \cdots, Y_n) $$

where $f$ is known. The sample size $n$ is always treated as known. A statistic is a fixed rule to calculate a quantity based on the sample. The same sample values yields the same statistic values.

Def: Point estimation is a single value estimate of a parameter $\theta$ based on the sample. Often, an estimate of $\theta$ is called an estimator and is denoted $\hat{\theta}$. An estimator is a statistic as it doesn’t involve anything unknown. Because of this, an estimator is also a function of the sample.

Example: suppose we have an i.i.d. sample $Y_1, \cdots, Y_n$ following the same distribution with mean $\mu$ and variance $\sigma^2$. Are the following estimators?

EstimatorKnown $\mu$Unknown $\mu$
$T_1 = \frac{1}{n} \sum_{i=1}^n (Y_i - \bar{Y})^2$YesYes
$T_2 = \frac{1}{n} \sum_{i=1}^n (Y_i - \mu)^2$YesNo

A commonly used type of estimator is called an interval estimate. In general, this is an interval constructed based on the sample, which hopefully contains the true parameter with certain quantified accuracy. The lower and upper limits $L$ and $U$ do not involve unknown parameters.

Example: confidence intervals (a frequentist method where we assume the unknown parameters are fixed) and credible intervals (a Bayesian method where the unknown parameters are random).


Some notations before we start:

  • $\theta$: unknown parameter
    • Taking values within a set $\Theta$ called the parameter space, which is often a subset of the real line $\mathbb{R}$.
    • E.g. $p$ in $Bern(p)$ has a parameter space of $\Theta = [0, 1]$.
  • $\hat{\theta}$: an estimator of $\theta$. It has to be a known function of sample $(Y_1, \cdots, Y_n)$.

Now, a very basic question is how do we evaluate an estimator? Suppose we have an observed (realized) sample $(y_1, \cdots, y_n)$. We can plug these values into two estimators

$$ \begin{gather*} \hat{\theta}_1 = f_1(Y_1, \cdots, Y_n) \\
\hat{\theta}_2 = f_2(Y_1, \cdots, Y_n) \end{gather*} $$

and get two point estimates. Let’s say we know the true value of $\theta$ and the value of $\hat{\theta}_1$ was closer to it. Is $\hat{\theta}_1$ better than $\hat{\theta}_2$? The answer’s no because we cannot evaluate an estimator based on a single realized estimate. However, $\hat{\theta}$ is a function of random sample $(Y_1, \cdots, Y_n)$. Some realizations may make $\hat{\theta}$ closer to $\theta$, while others may not.

To address the randomness, an idea is to repeat realizations of sample $(Y_1, \cdots, Y_n)$ and plot a histogram of the realized estimator. Comparing the histograms of different estimators may give us some insight on which is a better one.

Bias

Def: The bias of estimator $\hat{\theta}$ of $\theta$ is defined as

$$ Bias(\hat{\theta}; \theta) = E[\hat{\theta}] - \theta, $$

where the expectation $E$ is taken by assuming that $\theta$ is the true parameter ($E_{\theta}$).

Remark: the bias is a function of the parameter $\theta \in \Theta$. We’ll come back to this when introducing the concept of unbiasedness.

Uniform distribution example

$(Y_1, \cdots, Y_n) \overset{i.i.d.}{\sim} Unif(0, \theta)$. We want to estimate $\theta$. The proposed estimator is $\bar{Y}_n$. The bias of our estimator is

$$ \begin{aligned} Bias(\bar{Y}_n; \theta) &= E[\bar{Y}_n] - \theta \\
&= E\left[\frac{1}{n} (Y_1 + \cdots + Y_n)\right] - \theta \\
&= \frac{1}{n} E[Y_1 + \cdots + Y_n] - \theta \\
&= \frac{1}{n} \sum_{i=1}^n E[Y_i] - \theta \\
&= \frac{1}{n} \cdot n \cdot \frac{\theta}{2} - \theta \\
&= -\frac{\theta}{2} \end{aligned} $$

Note that $E[\bar{Y}_n] = E[Y_i]$. This holds true when the samples are i.i.d. Note also that the bias is a function of $\theta$. Our parameter space is $\Theta = (0, \infty)$.

Def: An estimator $\hat{\theta}$ of $\theta$ is said to be unbiased if $Bias(\hat{\theta}; \theta) = 0$, i.e. $E[\hat{\theta}] = \theta$ for any $\theta \in \Theta$. The “for any” part is important because $\theta$ is unknown.

Unbiasedness example

Suppose $(Y_1, \cdots, Y_n)$ is i.i.d. and $E[Y_i] = \theta \in \Theta = \mathbb{R}$. $\bar{Y}_n$ is an estimator of $\theta$. We always have $E[\bar{Y}_n] = \theta$, so the bias of this estimator is $0$ for all $\theta \in \mathbb{R}$. According to the definition, the estimator $\bar{Y}_n$ is unbiased for $\theta$.

Unbiasedness is a desirable property. It says, averagely speaking, the estimator captures the true parameter no matter where the latter is located.

Biased example

$(Y_1, \cdots, Y_n) \overset{i.i.d.}{\sim} Unif(0, \theta)$. The proposed estimator $\hat{\theta}_1 = 1$. Suppose the true $\theta = 1$. Then the bias would be

$$ Bias(\hat{\theta}_1; \theta) = E[\hat{\theta}_1] - \theta = 1 - 1 = 0. $$

This is wrong because $\theta$ is unknown and we can’t just assume its value. What we know is that $\Theta = (0, \infty)$. Suppose $\theta \in \Theta$,

$$ Bias(\hat{\theta}_1; \theta) = E[\hat{\theta}_1] - \theta = 1 - \theta. $$

So the conclusion is $\hat{\theta}_1$ is biased. Here we emphasize the importance of the “for all” language in the definition of unbiasedness.

Another proposed estimator is $\hat{\theta}_2 = 2\bar{Y}_n$. Since we know $E[\bar{Y}_n] = \frac{\theta}{2}$,

$$ E[\hat{\theta}_2] = 2E[\bar{Y}_n] = \theta. $$

Thus, $Bias(\hat{\theta}_2; \theta) \equiv 0$ and $\hat{\theta}_2$ is unbiased.

Proposition: suppose $\hat{\theta}$ is an unbiased estimator for $\theta$, and $\hat{\eta}$ is an unbiased estimator for $\eta$.

  1. Let $a$ be a known non-random scalar. $a\hat{\theta}$ would be unbiased for $a\theta$.
  2. $\hat{\theta} + a$ would be unbiased for $\theta + a$.
  3. $a\hat{\theta} + b\hat{\eta}$ would be unbiased for $a\theta + b\eta$.

In conclusion, unbiasedness is preserved under linear transformations. However, it’s often not preserved under nonlinear transformations.

Nonlinear transformation example

Suppose $\hat{\theta}$ is an unbiased estimator for $\theta$. Is $\hat{\theta}^2$ an unbiased estimator of $\theta^2$?

$$ E\left[\hat\theta^2\right] = Var(\hat\theta) + \left( E\left[\hat\theta\right] \right)^2 = Var(\hat\theta) + \theta^2 $$

Unless $ Var(\hat\theta) = 0$, we always have a bias. $ Var(\hat\theta) = 0$ implies $\hat\theta$ is a constant, which is very rare.

Proposition: $Y_1, \cdots, Y_n$ ($n \geq 2$) are i.i.d. random samples with population variance $\sigma^2$. The estimator $$ \hat{\sigma}_n^2 = \frac{1}{n-1}\sum_{i=1}^n (Y_i - \bar{Y}_n)^2 $$

is an unbiased estimator of $\sigma^2$. The proof is given as follows.

$$ \begin{aligned} S &= \sum_{i=1}^n \left(Y_i - \bar{Y}\right)^2 \\
&= \sum_{i=1}^n \left(Y_i^2 - 2Y_i\bar{Y} + \bar{Y}^2\right) \\
&= \sum_{i=1}^n Y_i^2 - 2\bar{Y}\underbrace{\sum_{i=1}^n Y_i}_{n\bar{Y}} + n\bar{Y}^2 \\
&= \sum_{i=1}^n Y_i^2 - n\bar{Y}^2 = A - B \end{aligned} $$

Recall that $E\left[ Y_i^2\right] = \sigma^2 + \mu^2$ where $\mu = E[Y_i]$.

$$ \begin{aligned} E[A] &= \sum_{i=1}^n E[Y_i^2] = n\sigma^2 + n\mu^2 \\
Var\left(\bar{Y}\right) &= Var\left( \frac{Y_1 + \cdots + Y_n}{n} \right) \\
&= \frac{1}{n^2}\left( Var(Y_1) + \cdots + Var(Y_n) \right) \\
&= \frac{1}{n^2}n\sigma^2 = \frac{\sigma^2}{n} \\
E\left[ \bar{Y}^2 \right] &= Var(\bar{Y}) + E[\bar{Y}]^2 = \frac{\sigma^2}{n} + \mu^2 \\
\Rightarrow E[B] &= \sigma^2 + n\mu^2 \\
E\left[ \frac{S}{n-1} \right] &= \frac{1}{n-1}(E[A] - E[B]) \\
&= \sigma^2 \end{aligned} $$

Variance

The importance of the variance is best explained with an example. Suppose $Y_1, \cdots, Y_n$ are i.i.d. with mean $\mu$ and variance $\sigma^2$. Previously we’ve used $\bar{Y}$ to estimate $\mu$. Why didn’t we use $Y_1$ if $E[Y_1] = \mu$? Well,

$$ Var(\bar{Y}) = \frac{\sigma^2}{n} \text{ vs. }Var(Y_1) = \sigma^2 $$

We can easily see that when $n > 2$, $Var(\bar{Y}) < Var(Y_1)$.

Def: The variance of estimator $\hat\theta$ of $\theta$ is

$$ Var(\hat\theta; \theta) = Var(\hat\theta), $$

where the variance is computed by assuming that $\theta$ is the true parameter. It is also a function of $\theta$. The square root of the variance is denoted

$$ S.E.(\hat\theta) = \sqrt{Var(\hat\theta; \theta)} $$

In this and later chapters, “standard error” and “standard deviation” are used interchangeably.

Bernoulli example

$Y_1, \cdots, Y_n \overset{i.i.d.}{\sim} Bern(p)$. We know that $E[Y_i] = p$ and $Var(Y_i) = p(1-p)$. Let $X_n = Y_1 + \cdots + Y_n$. The estimator

$$ \hat{p} = \bar{Y} = \frac{X_n}{n} $$

is unbiased for $p$. Find the variance of $\hat{p}$.

$$ Var(\hat{p}) = Var(\bar{Y}) = \frac{Var(Y_1)}{n} = \frac{p(1-p)}{n} $$

Linear combination example

$\{X_1, \cdots, X_{n_1}\}$ are i.i.d. with $E[X_i] = \mu_1$ and $Var(X_i) = \sigma_1^2$. $\{Y_1, \cdots, Y_{n_2}\}$ are i.i.d. with $E[Y_i] = \mu_2$ and $Var(Y_i) = \sigma_2^2$. We assume that ${X_i}$ and ${Y_i}$ are independent of each other. We want to estimate $\theta = \mu_1 - \mu_2$.

Our proposed estimator is $\hat\theta = \bar{X} - \bar{Y}$. We know that $\bar{X}$ is unbiased for $\mu_1$ and $\bar{Y}$ is unbiased for $\mu_2$. Their linear combination would also be unbiased.

We’re also interested in the variability of this estimator. We know that

$$ Var(A + B) = Var(A) + Var(B) + 2Cov(A, B), $$

so the variance of $\hat\theta$ would be

$$ \begin{aligned} Var(\hat\theta) &= Var(\bar{X} - \bar{Y}) \\
&= Var(\bar{X}) + Var(-\bar{Y}) + 2Cov(\bar{X}, \bar{Y}) \\
&= \frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2} + 0 \\
&= \frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2} \end{aligned} $$

So the standard error of $\hat\theta$ is

$$ S.E.(\hat\theta) = \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}} $$

Uniform distribution example

$Y_1, \cdots, Y_n \overset{i.i.d.}{\sim} Unif(0, \theta)$. The two proposed estimators are $\hat{\theta}_1 = 1$ and $\hat{\theta}_2 = 2\bar{Y}$. Find the variance of the two estimators.

$$ \begin{gather*} Var(\hat\theta_1) = 0 \\
Var(\hat\theta_2) = 4Var(\bar{Y}) = \frac{4Var(Y_1)}{n} = \frac{4\sigma^2}{12n} = \frac{\theta^2}{3n} \end{gather*} $$

We’ve shown earlier that $\hat\theta_1$ is biased, but here in terms of variability it’s actually perfect. How do we determine which one is the better estimator?

Mean squared error

Def: We need a measure of goodness for estimators combining both bias and variance. The MSE is defined as

$$ MSE(\hat\theta; \theta) = E\left[(\hat\theta - \theta)^2\right] $$

where the expectation is taken by assuming that $\theta$ is the true parameter.

Theorem: $MSE(\hat\theta; \theta) = Bias(\hat\theta; \theta)^2 + Var(\hat\theta; \theta)$. This decomposition is derived as follows.

$$ \begin{aligned} MSE(\hat\theta) &= E\left[ \left((\hat\theta - E[\hat\theta]) + Bias(\hat\theta)\right)^2 \right] \\
&= E\left[ (\hat\theta - E[\hat\theta])^2 \right] + 2E\left[ Bias(\hat\theta)(\hat\theta - E[\hat\theta]) \right] + E\left[ Bias(\hat\theta)^2 \right] \\
&= Var(\hat\theta) + 2Bias(\hat\theta)\underbrace{E\left[\hat\theta - E[\hat\theta]\right]}_{0} + Bias(\hat\theta)^2 \\
&= Var(\hat\theta) + Bias(\hat\theta)^2 \end{aligned} $$

If $\hat\theta$ is unbiased,

$$ MSE(\hat\theta; \theta) = Var(\hat\theta; \theta) $$

Notation: for simplicity, we can write $Bias(\hat\theta)$, $Var(\hat\theta)$ and $MSE(\hat\theta)$ if it’s clear from the context what’s the assumed true parameter $\theta$.

A question here is why do we take the square? Why not $E\left[|\hat\theta - \theta|\right]$?

  • The main reason is just mathematical convenience.
  • Another reason is “risk aversion”. The square penalizes large deviations more than the absolute value.

However, the other choices of “penalization functions” are not useless. These are further studied in robust statistics. Another note here is there’s almost always a trade-off between the variance and the bias. We’ll discuss this in detail later.

Calculation example

Suppose $Y_i \overset{i.i.d.}{\sim} Unif(0, \theta)$. Three estimators are proposed:

  1. $\hat\theta_1 = 1$,
  2. $\hat\theta_2 = 2\bar{Y}_n$, and
  3. $\hat\theta_3 = \max(Y_1, \cdots, Y_n)$.

Calculate the MSE of each estimator.

$$ \begin{gather*} MSE(\hat\theta_1) = Bias(\hat\theta_1)^2 + Var(\hat\theta_1) = (1 - \theta)^2 + 0 \\
MSE(\hat\theta_2) = 0^2 + \frac{\theta^2}{3n} = \frac{\theta^2}{3n} \quad \cdots \rightarrow 0 \text{ as } n \rightarrow \infty \end{gather*} $$

In addition to $\theta$, the sample size also enters in the $MSE$ for $\hat\theta_2$. when $n$ is large, the $MSE$ becomes really small. Usually we fix $n$ at some value.

For $\hat\theta_3$, we’re calculating the 1st order statistic, which is often denoted $Y_{(1)}$. The first step is to compute the PDF of $\hat\theta_3$, which is also the hardest part of the problem. We know for a fact that

$$ \max(Y_1, \cdots, Y_n) \leq x \Leftrightarrow Y_1 \leq x, Y_2 \leq x, \cdots, Y_n \leq x $$

and this is very useful in getting the CDF. $$ \begin{aligned} P(\hat\theta_3 \leq x) &= P\Big(\max(Y_1, \cdots, Y_n) \leq x \Big), \quad 0 < x < \theta \\
&= P(Y_1 \leq x, \cdots, Y_n \leq x) \\
&= P({Y_1 \leq x} \cap \cdots \cap {Y_n \leq x}) \\
&= P(Y_1 \leq x) \cdot P(Y_2 \leq x) \cdots P(Y_n \leq x) \quad \cdots \text{independence} \\
&= P(Y_1 \leq x)^n \\
&= \left(\frac{x}{\theta}\right)^n \end{aligned} $$

By differentiation, the PDF is

$$ f(x) = \frac{d}{dx}\left(\frac{x}{\theta}\right)^n =\frac{nx^{n-1}}{\theta^n}, \quad 0 < x < \theta $$

Now we can find the bias and the variance.

$$ \begin{aligned} E[\hat\theta_3] &= \int_0^\theta xf(x)dx \\
&= \int_0^\theta x \cdot \frac{nx^{n-1}}{\theta^n} dx \\
&= \frac{n}{\theta^n} \cdot \frac{x^{n+1}}{n+1} \bigg|_0^\theta \\
&= \frac{n}{n+1}\theta \\
Bias(\hat\theta_3) &= -\frac{1}{n+1}\theta \end{aligned} $$

Remark: although $\hat\theta_3$ is biased, bias $\rightarrow 0$ as $n \rightarrow \infty$. We call this asymtotically unbiased.

To find $Var(\hat\theta_3)$, we need to find $E\left[ \hat\theta_3^2 \right]$ first. $$ \begin{aligned} E\left[ \hat\theta_3^2 \right] &= \int_0^\theta x^2 \cdot \frac{nx^{n-1}}{\theta^n} dx \\
&= \frac{n}{\theta^n} \int_0^\theta x^{n+1}dx \\
&= \frac{n\theta^2}{n+2} \end{aligned} $$

Now we have

$$ \begin{aligned} Var(\hat\theta_3) &= E\left[ \hat\theta_3^2 \right] - E[\hat\theta_3]^2 \\
&= \frac{n\theta^2}{n+2} - \frac{n^2\theta^2}{(n+1)^2} \\
&= \frac{n\theta^2}{(n+1)^2(n+2)} \end{aligned} $$

And now we can find the MSE: $$ \begin{aligned} MSE(\hat\theta_3) &= Bias(\hat\theta_3)^2 + Var(\hat\theta_3) \\
&= \frac{\theta^2}{(n+1)^2} + \frac{n\theta^2}{(n+1)^2(n+2)} \\
&= \frac{2\theta^2}{(n+1)(n+2)} \end{aligned} $$

We can also find that

$$ \frac{MSE(\hat\theta_3)}{MSE(\hat\theta_2)} = \frac{6n}{(n+1)(n+2)} < 1 \text{ if } n > 2, $$

which means $MSE(\hat\theta_3) < MSE(\hat\theta_2)$ when $n > 2$.

Efficiency

Def: Given two estimators $\hat\theta_1$ and $\hat\theta_2$ of the same parameter based on the sample random sample, the efficiency of $\hat\theta_1$ relative to $\hat\theta_2$ is defined as $$ eff(\hat\theta_1, \hat\theta_2) = \frac{MSE(\hat\theta_2)}{MSE(\hat\theta_1)} $$

This works for both biased and unbiased estimators, although we usually look at the efficiency for unbiased estimators.


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