**Definition 1**: If a continuous random variable *x* has frequency function *f*(*x*) then the **expected value** of *g*(*x*) is

**Property 1**: If *g* and *h* are independent then

Proof: Similar to the proof of Property 1b of Expectation

**Definition 2**: If a random variable *x* has frequency function *f*(*x*) then the *n***th moment** *M _{n}*(

*x*

_{0}) of

*f*(

*x*)

**about**

*x*

_{0}is

We also use the following symbols for the *n***th moment around the origin**, i.e. where *x*_{0} = 0

The **mean** is the first moment about the origin.

We use the following symbols for the *n***th moment around the mean**

The **variance** is the second moment about the mean

**Definition 3**: The **moment generating function** of a discrete random variable *x* with frequency function *f*(*x*) is a function of a dummy variable *θ* given by

The moment generating function for a continuous random variable is

**Property 2**: If the moment generating function of *x* for frequency function *f*(*x*) converges for each *k*, then

Proof: We provide the proof where *x* is discrete random variable. The continuous case is similar.

it follows that

Thus, the *k*+1^{th} term in the power series expansion of the moment generating function is

The result now follows by induction on *k*.

**Theorem 1**: A distribution function is completely determined by its moment generating function. I.e. two distribution functions with the same moment generating function are equal.

**Corollary 1**: If *x* is a random variable which depends on *n* with frequency function *f _{n}*(

*x*) and y is another random variable with frequency function

*g*(

*x*), then if

**Definition 4**: If *x* is a discrete random variable with frequency function *f*(*x*), then the **moment generating function** of *g*(*x*), is

The equivalent for a continuous random variable *x* is

**Properties 3**: Where *c* is a constant and *g*(*x*) is any function for which *M _{g}*

_{(}

_{x}_{)}(

*θ*) exists

Proof: We prove the property where *x* is a discrete random variable. The situation is similar where *x* is continuous.

**Property 4**: The moment generating function of the sum of *n* independent variables is equal to the product of the moment generating functions of the individual variables; i.e.

Proof: Since the *x _{i}* are independent, so are the , and so by Property 2

**Theorem 2** (**Change of variables technique**): If y = *h*(*x*) is an increasing function and *f*(*x*) is the frequency function of *x*, then the frequency function *g*(y) of y is given by

Proof: Let *G*(y) be the cumulative distribution function of y, let *h*^{-1} be the inverse function of *h* and let *u* = *h*^{-1} (*t*). Then

Now by changing variable names we have

where *x* = *h*^{-1} (y) and so y = *h(x*).

**Corollary 2**: If y = *h*(*x*) is an decreasing function and *f*(*x*) is the frequency function of *x*, then the frequency function *g*(y) of y is given by

**Corollary 3**: If z = *t*(*x, *y) is an increasing function of y keeping *x *fixed and *f*(*x, *y) is the joint frequency function of *x *and y and *h*(*x, *z) is the joint frequency function of *x* and *z*, then

Proof: If *z* = *t*(*x*, y) is an increasing function of y, keeping *x* fixed, and *g*(y|*x*) is the frequency function of y|*x* and *k*(*z*|*x*) is the frequency function of *z*|*x*), then by the theorem

Now let *f*(*x*, y) be the joint frequency function of *x* and y. Then *f*(*x*, y) = *f*(*x*) · *g*(y|*x*). Similarly if *h*(*x, z*) is the joint frequency function of *x* and *z*, we have* h*(*x, z*) = *h*(*x*) · *k*(*z*|*x*). Thus

Since both *f* and *h* are the pdf for *x*, *f*(*x*) = *h*(*x*), and so we have

**Corollary 4**: If z = *t*(*x, *y) is an decreasing function of y keeping *x *fixed and *f*(*x, *y) is the joint frequency function of *x *and y and *h*(*x, *z) is the joint frequency function of *x* and *z*, then

**Example 1**: Suppose *x* has pdf *f*(*x*) = *e*^{-x} where *x* ≥ 0, and y =. Find the pdf *g* of y

Since is an increasing function, where *x* = y^{2}, we get

**Example 2**: Suppose *x* has pdf *f*(*x*) = *e*^{-x} for *x* > 0 and y has pdf *g*(*x*) = *e*^{-y} for y > 0, and suppose that *x* and y are independently distributed. Define *z* = y/*x*. What is the pdf for *z*?

From Corollary 3, for fixed *x* > 0, *z* = y/*x* is increasing (since y > 0), and so we have

But since *x* and y are independently distributed,

Let *w* = *x*(1+*z*). Then *x* = *w*/(1+*z*). and d*x* = d*w*/(1+*z*). It now follows that

**Example 3**: Suppose *x* has standard normal distribution *N*(0, 1). What is the pdf of the random variable *z* = |*x*| and what is the mean of this distribution?

By Definition 1 of Basic Characteristics of the Normal Distribution, the pdf of *x* is (with *μ* = 0 and *σ* = 1)

and so the probability distribution function is

Now |*x*| < *a* is equivalent to –*a < x < a*, and so we have the following formula for *z*’s distribution function *G*(*z*):

Since the pdf *g*(*z*) is the derivative of *G*(*z*), it follows that

We next use the following using the substitution

Since *z* ≥ 0, we have

**Property** 5: If *x* ~ *N*(0, *σ*), then the mean of |*x*| is

Proof: Let z = |*x*/*σ*|. Thus *z* ~ *N*(0, 1), and so as we saw in Example 3, *E*[*z*] = . But *z* = |*x*|/*σ*, and so |*x*| = *σz*, from which it follows that *E*[|*x*|] = .