In the previous lesson, we learned that the expected value of the sample mean \(\bar{X}\) is the population mean *μ*. We also learned that the variance of the sample mean \(\bar{X}\) is *σ*^{2}/*n*, that is, the population variance divided by the sample size *n*. We have not yet determined the probability distribution of the sample mean when, say, the random sample comes from a normal distribution with mean *μ* and variance *σ*^{2}. We are going to tackle that in the next lesson! Before we do that, though, we are going to want to put a few more tools into our toolbox. We already have learned a few techniques for finding the probability distribution of a function of random variables, namely the distribution function technique and the change-of-variable technique. In this lesson, we'll learn yet another technique called the **moment-generating function technique**. We'll use the technique in this lesson to learn, among other things, the distribution of sums of chi-square random variables, Then, in the next lesson, we'll use the technique to find (finally) the probability distribution of the sample mean when the random sample comes from a normal distribution with mean *μ* and variance *σ*^{2}.

- To refresh our memory of the uniqueness property of moment-generating functions.
- To learn how to calculate the moment-generating function of a linear combination of
*n*independent random variables. - To learn how to calculate the moment-generating function of a linear combination of
*n*independent and identically distributed random variables. - To learn the additive property of independent chi-square random variables.
- To use the moment-generating function technique to prove the additive property of independent chi-square random variables.
- To understand the steps involved in each of the proofs in the lesson.
- To be able to apply the methods learned in the lesson to new problems.

Recall that the moment generating function:

\(M_X(t)=E(e^{tX})\)

uniquely defines the distribution of a random variable. That is, if you can show that the moment generating function of \(\bar{X}\) is the same as some known moment-generating function, then \(\bar{X}\)follows the same distribution. So, one strategy to finding the distribution of a function of random variables is:

(1) to find the moment-generating function of the function of random variables

(2) to compare the calculated moment-generating function to known moment-generating functions

(3) if the calculated moment-generating function is the same as some known moment-generating function of *X*, then the function of the random variables follows the same probability distribution as *X*

\(Y=X_1+X_2\)

denote the number of heads in five tosses. What is the probability distribution of *Y*?

**Solution.** We know that:

*X*_{1}is a binomial random variable with*n*= 3 and*p*= ½*X*_{2}is a binomial random variable with*n*= 2 and*p*= ½

Therefore, based on what we know of the moment-generating function of a binomial random variable, the moment-generating function of *X*_{1} is:

\(M_{X_1}(t)=\left(\dfrac{1}{2}+\dfrac{1}{2} e^t\right)^3\)

And, similarly, the moment-generating function of *X*_{2} is:

\(M_{X_2}(t)=\left(\dfrac{1}{2}+\dfrac{1}{2} e^t\right)^2\)

Now, because *X*_{1} and *X*_{2} are independent random variables, the random variable *Y* is the sum of independent random variables. Therefore, the moment-generating function of *Y* is:

\(M_Y(t)=E(e^{tY})=E(e^{t(X_1+X_2)})=E(e^{tX_1} \cdot e^{tX_2} )=E(e^{tX_1}) \cdot E(e^{tX_2} )\)

The first equality comes from the definition of the moment-generating function of the random variable *Y*. The second equality comes from the definition of *Y*. The third equality comes from the properties of exponents. And, the fourth equality comes from the expectation of the product of functions of independent random variables. Now, substituting in the known moment-generating functions of *X*_{1} and *X*_{2}, we get:

\(M_Y(t)=\left(\dfrac{1}{2}+\dfrac{1}{2} e^t\right)^3 \cdot \left(\dfrac{1}{2}+\dfrac{1}{2} e^t\right)^2=\left(\dfrac{1}{2}+\dfrac{1}{2} e^t\right)^5\)

That is, *Y* has the same moment-generating function as a binomial random variable with *n* = 5 and *p* = ½. Therefore, by the uniqueness properties of moment-generating functions, *Y* must be a binomial random variable with *n* = 5 and *p* = ½. (Of course, we already knew that!)

It seems that we could generalize the way in which we calculated, in the above example, the moment-generating function of *Y,* the sum of two independent random variables. Indeed, we can! On the next page!

Theorem. If X_{1}, X_{2}, ... , X_{n}_{ }are n independent random variables with respective moment-generating functions \(M_{X_i}(t)=E(e^{tX_i})\) for i = 1, 2, ... , n, then the moment-generating function of the linear combination: \(Y=\sum\limits_{i=1}^n a_i X_i\) is: \(M_Y(t)=\prod\limits_{i=1}^n M_{X_i}(a_it)\) |

**Proof. **The proof is very similar to the calculation we made in the example on the previous page. That is:

\begin{align}

M_Y(t) &= E[e^{tY}]\\

&= E[e^{t(a_1X_1+a_2X_2+\ldots+a_nX_n)}]\\

&= E[e^{a_1tX_1}]E[e^{a_2tX_2}]\ldots E[e^{a_ntX_n}]\\

&= M_{X_1}(a_1t)M_{X_2}(a_2t)\ldots M_{X_n}(a_nt)\\

&= \prod\limits_{i=1}^n M_{X_i}(a_it)\\

\end{align}

The first equality comes from the definition of the moment-generating function of the random variable *Y*. The second equality comes from the given definition of *Y*. The third equality comes from the properties of exponents, as well as from the expectation of the product of functions of independent random variables. The fourth equality comes from the definition of the moment-generating function of the random variables *X _{i}*, for

While the theorem is useful in its own right, the following corollary is perhaps even more useful when dealing not just with independent random variables, but also random variables that are identically distributed — two characteristics that we get, of course, when we take a random sample.

(1) the moment generating function of the linear combination \(Y=\sum\limits_{i=1}^n X_i\) is \(M_Y(t)=\prod\limits_{i=1}^n M(t)=[M(t)]^n\). (2) the moment generating function of the sample mean \(\bar{X}=\sum\limits_{i=1}^n \left(\dfrac{1}{n}\right) X_i\) is \(M_{\bar{X}}(t)=\prod\limits_{i=1}^n M\left(\dfrac{t}{n}\right)=\left[M\left(\dfrac{t}{n}\right)\right]^n\). |

**Proof.** For (1), use the preceding theorem with *a _{i} *= 1 for

Let *X*_{1}, *X*_{2}, and *X*_{3} denote a random sample of size 3 from a gamma distribution with *α* = 7 and *θ* = 5. Let *Y* be the sum of the three random variables:

\(Y=X_1+X_2+X_3\)

What is the distribution of *Y*?

**Solution.** The moment-generating function of a gamma random variable *X* with *α* = 7 and *θ* = 5 is:

\(M_X(t)=\dfrac{1}{(1-5t)^7}\)

for *t* < 1/5. Therefore, the corollary tells us that the moment-generating function of *Y* is:

\(M_Y(t)=[M_{X_1}(t)]^3=\left(\dfrac{1}{(1-5t)^7}\right)^3=\dfrac{1}{(1-5t)^{21}}\)

for *t* < 1/5, which is the moment-generating function of a gamma random variable with *α* = 21 and *θ* = 5. Therefore, *Y* must follow a gamma distribution with *α* = 21 and *θ* = 5.

What is the distribution of the sample mean \(\bar{X}\)?

**Solution.** Again, the moment-generating function of a gamma random variable *X* with *α* = 7 and *θ* = 5 is:

\(M_X(t)=\dfrac{1}{(1-5t)^7}\)

for *t* < 1/5. Therefore, the corollary tells us that the moment-generating function of \(\bar{X}\) is:

\(M_{\bar{X}}(t)=\left[M_{X_1}\left(\dfrac{t}{3}\right)\right]^3=\left(\dfrac{1}{(1-5(t/3))^7}\right)^3=\dfrac{1}{(1-(5/3)t)^{21}}\)

for *t* < 3/5, which is the moment-generating function of a gamma random variable with *α* = 21 and *θ* = 5/3. Therefore, \(\bar{X}\) must follow a gamma distribution with *α* = 21 and *θ* = 5/3.

We'll now turn our attention towards applying the theorem and corollary of the previous page to the case in which we have a function involving a sum of independent chi-square random variables. The following theorem is often referred to as the "**a****dditive property of independent chi-squares**."

n independent random variables that follow these chi-square distributions: *\(X_1 \sim \chi^2(r_1)\)**\(X_2 \sim \chi^2(r_2)\)*- \(\vdots\)
- \(X_n \sim \chi^2(r_n)\)
Then, the sum of the random variables: \(Y=X_1+X_2+\cdots+X_n\) follows a chi-square distribution with r_{n }degrees of freedom. That is:\(Y\sim \chi^2(r_1+r_2+\cdots+r_n)\) |

**Proof.**

We have shown that *M*_{Y}(*t*) is the moment-generating function of a chi-square random variable with *r*_{1} + *r _{2}* + ... +

\(Y\sim \chi^2(r_1+r_2+\cdots+r_n)\)

as was to be shown.

\(W=Z^2_1+Z^2_2+\cdots+Z^2_n\) follows a χ |

**Proof. **Recall that if *Z _{i }*~

\(W=Z^2_1+Z^2_2+\cdots+Z^2_n \sim \chi^2(1+1+\cdots+1)=\chi^2(n)\)

That is, *W* ~ χ^{2}(*n*), as was to be proved.

\(X_i \sim N(\mu_i,\sigma^2_i)\) for \(W=\sum\limits_{i=1}^n \dfrac{(X_i-\mu_i)^2}{\sigma^2_i} \sim \chi^2(n)\) |

**Proof. **Recall that:

\(Z_i=\dfrac{(X_i-\mu_i)}{\sigma_i} \sim N(0,1)\)

Therefore:

\(W=\sum\limits_{i=1}^n Z^2_i=\sum\limits_{i=1}^n \dfrac{(X_i-\mu_i)^2}{\sigma^2_i} \sim \chi^2(n)\)

as was to be proved.