# 24.4 - Two or More Parameters

24.4 - Two or More Parameters

In each of the examples we considered so far in this lesson, there is one and only one parameter. What happens if a probability distribution has two parameters, $$\theta_1$$ and $$\theta_2$$, say, for which we want to find sufficient statistics, $$Y_1$$ and $$Y_2$$? Fortunately, the definitions of sufficiency can easily be extended to accommodate two (or more) parameters. Let's start by extending the Factorization Theorem.

Definition (Factorization Theorem)

Let $$X_1, X_2, \ldots, X_n$$ denote random variables with a joint p.d.f. (or joint p.m.f.):

$$f(x_1,x_2, ... ,x_n; \theta_1, \theta_2)$$

which depends on the parameters $$\theta_1$$ and $$\theta_2$$. Then, the statistics $$Y_1=u_1(X_1, X_2, ... , X_n)$$ and $$Y_2=u_2(X_1, X_2, ... , X_n)$$ are joint sufficient statistics for $$\theta_1$$ and $$\theta_2$$ if and only if:

$$f(x_1, x_2, ... , x_n;\theta_1, \theta_2) =\phi\left[u_1(x_1, ... , x_n), u_2(x_1, ... , x_n);\theta_1, \theta_2 \right] h(x_1, ... , x_n)$$

where:

• $$\phi$$ is a function that depends on the data $$(x_1, x_2, ... , x_n)$$ only through the functions $$u_1(x_1, x_2, ... , x_n)$$ and $$u_2(x_1, x_2, ... , x_n)$$, and
• the function $$h(x_1, ... , x_n)$$ does not depend on either of the parameters $$\theta_1$$ or $$\theta_2$$.

Let's try the extended theorem out for size on an example.

## Example 24-6

Let $$X_1, X_2, \ldots, X_n$$ denote a random sample from a normal distribution $$N(\theta_1, \theta_2$$. That is, $$\theta_1$$ denotes the mean $$\mu$$ and $$\theta_2$$ denotes the variance $$\sigma^2$$. Use the Factorization Theorem to find joint sufficient statistics for $$\theta_1$$ and $$\theta_2$$.

Because $$X_1, X_2, \ldots, X_n$$ is a random sample, the joint probability density function of $$X_1, X_2, \ldots, X_n$$ is, by independence:

$$f(x_1, x_2, ... , x_n;\theta_1, \theta_2) = f(x_1;\theta_1, \theta_2) \times f(x_2;\theta_1, \theta_2) \times ... \times f(x_n;\theta_1, \theta_2) \times$$

Inserting what we know to be the probability density function of a normal random variable with mean $$\theta_1$$ and variance $$\theta_2$$, the joint p.d.f. is:

$$f(x_1, x_2, ... , x_n;\theta_1, \theta_2) = \dfrac{1}{\sqrt{2\pi\theta_2}} \text{exp} \left[-\dfrac{1}{2}\dfrac{(x_1-\theta_1)^2}{\theta_2} \right] \times ... \times = \dfrac{1}{\sqrt{2\pi\theta_2}} \text{exp} \left[-\dfrac{1}{2}\dfrac{(x_n-\theta_1)^2}{\theta_2} \right]$$

Simplifying by collecting like terms, we get:

$$f(x_1, x_2, ... , x_n;\theta_1, \theta_2) = \left(\dfrac{1}{\sqrt{2\pi\theta_2}}\right)^n \text{exp} \left[-\dfrac{1}{2}\dfrac{\sum_{i=1}^{n}(x_i-\theta_1)^2}{\theta_2} \right]$$

Rewriting the first factor, and squaring the quantity in parentheses, and distributing the summation, in the second factor, we get:

$$f(x_1, x_2, ... , x_n;\theta_1, \theta_2) = \text{exp} \left[\text{log}\left(\dfrac{1}{\sqrt{2\pi\theta_2}}\right)^n\right] \text{exp} \left[-\dfrac{1}{2\theta_2}\left\{ \sum_{i=1}^{n}x_{i}^{2} -2\theta_1\sum_{i=1}^{n}x_{i} +\sum_{i=1}^{n}\theta_{1}^{2} \right\}\right]$$

Simplifying yet more, we get:

$$f(x_1, x_2, ... , x_n;\theta_1, \theta_2) = \text{exp} \left[ -\dfrac{1}{2\theta_2}\sum_{i=1}^{n}x_{i}^{2}+\dfrac{\theta_1}{\theta_2}\sum_{i=1}^{n}x_{i} -\dfrac{n\theta_{1}^{2}}{2\theta_2}-n\text{log}\sqrt{2\pi\theta_2} \right]$$

Look at that! We have factored the joint p.d.f. into two functions, one ($$\phi$$) being only a function of the statistics $$Y_1=\sum_{i=1}^{n}X^{2}_{i}$$ and $$Y_2=\sum_{i=1}^{n}X_i$$, and the other (h) not depending on the parameters $$\theta_1$$ and $$\theta_2$$:

Therefore, the Factorization Theorem tells us that $$Y_1=\sum_{i=1}^{n}X^{2}_{i}$$ and $$Y_2=\sum_{i=1}^{n}X_i$$ are joint sufficient statistics for $$\theta_1$$ and $$\theta_2$$. And, the one-to-one functions of $$Y_1$$ and $$Y_2$$, namely:

$$\bar{X} =\dfrac{Y_2}{n}=\dfrac{1}{n}\sum_{i=1}^{n}X_i$$

and

$$S_2=\dfrac{Y_1-(Y_{2}^{2}/n)}{n-1}=\dfrac{1}{n-1} \left[\sum_{i=1}^{n}X_{i}^{2}-n\bar{X}^2 \right]$$

are also joint sufficient statistics for $$\theta_1$$ and $$\theta_2$$. Aha! We have just shown that the intuitive estimators of $$\mu$$ and $$\sigma^2$$ are also sufficient estimators. That is, the data contain no more information than the estimators $$\bar{X}$$ and $$S^2$$ do about the parameters $$\mu$$ and $$\sigma^2$$! That seems like a good thing!

We have just extended the Factorization Theorem. Now, the Exponential Criterion can also be extended to accommodate two (or more) parameters. It is stated here without proof.

## Exponential Criterion

Let $$X_1, X_2, \ldots, X_n$$ be a random sample from a distribution with a p.d.f. or p.m.f. of the exponential form:

$$f(x;\theta_1,\theta_2)=\text{exp}\left[K_1(x)p_1(\theta_1,\theta_2)+K_2(x)p_2(\theta_1,\theta_2)+S(x) +q(\theta_1,\theta_2) \right]$$

with a support that does not depend on the parameters $$\theta_1$$ and $$\theta_2$$. Then, the statistics $$Y_1=\sum_{i=1}^{n}K_1(X_i)$$ and $$Y_2=\sum_{i=1}^{n}K_2(X_i)$$ are jointly sufficient for $$\theta_1$$ and $$\theta_2$$.

Let's try applying the extended exponential criterion to our previous example.

## Example 24-6 continuted

Let $$X_1, X_2, \ldots, X_n$$ denote a random sample from a normal distribution $$N(\theta_1, \theta_2)$$. That is, $$\theta_1$$ denotes the mean $$\mu$$ and $$\theta_2$$ denotes the variance $$\sigma^2$$. Use the Exponential Criterion to find joint sufficient statistics for $$\theta_1$$ and $$\theta_2$$.

The probability density function of a normal random variable with mean $$\theta_1$$ and variance $$\theta_2$$ can be written in exponential form as:
Therefore, the statistics $$Y_1=\sum_{i=1}^{n}X^{2}_{i}$$ and $$Y_2=\sum_{i=1}^{n}X_i$$ are joint sufficient statistics for $$\theta_1$$ and $$\theta_2$$.