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Legitimate probability density functions

by , PhD

This lecture discusses two properties characterizing probability density functions (pdfs). Not only any pdf satisfies these two properties, but also any function that satisfies these two properties is a legitimate pdf.

Table of Contents

Properties of probability density functions

The following proposition formally describes the two properties.

Proposition Let X be a continuous random variable. Its probability density function, denoted by [eq1], satisfies the following two properties:

  1. Non-negativity: [eq2] for any $xin U{211d} $;

  2. Integral over R equals 1: [eq3].

Proof

Remember that, by the definition of a pdf, [eq4] is such that[eq5]for any interval $left[ a,b
ight] $. Probabilities cannot be negative, therefore [eq6] and[eq7]for any interval $left[ a,b
ight] $. But the above integral can be non-negative for all intervals $left[ a,b
ight] $ only if the integrand function itself is non-negative, that is, if [eq2] for all x. This proves property 1 above (non-negativity).

Furthermore, the probability of a sure thing must be equal to 1. Since [eq9] is a sure thing, then[eq10]which proves property 2 above (integral over R equals 1).

Identification of legitimate probability density functions

Any pdf must satisfy property 1 and 2 above. It can be demonstrated that also the converse holds: any function enjoying these properties is a pdf.

Proposition Let [eq4] be a function satisfying the following two properties:

  1. Non-negativity: [eq2] for any $xin U{211d} $;

  2. Integral over R equals 1: [eq3].

Then, there exists a continuous random variable X whose pdf is [eq14].

This proposition gives us a powerful method for constructing probability density functions. Take any non-negative function g(x) (non-negative means that [eq15] for any $xin U{211d} $). If the integral[eq16]exists and is finite and strictly positive, then define[eq17]I is strictly positive, thus [eq4] is non-negative and it satisfies property 1. It also satisfies Property 2 because[eq19]Thus, any non-negative function g(x) can be used to build a pdf if its integral over R exists and is finite and strictly positive.

Example Define a function g(x) as follows:[eq20]How do we construct a pdf from g(x)? First, we need to verify that g(x) is non-negative. But this is true because $x^{2}$ is always non-negative. Then, we need to verify that the integral of g(x) over R exists and is finite and strictly positive:[eq21]Having verified that I exists and is finite and strictly positive, we can define[eq22]By the above proposition, [eq4] is a legitimate pdf.

Solved exercises

Below you can find some exercises with explained solutions.

Exercise 1

Consider the following function:[eq24]

where [eq25]. Prove that [eq26] is a legitimate probability density function.

Solution

Since $lambda >0$ and the exponential function is strictly positive, [eq27] for any $xin U{211d} $, so the non-negativity property is satisfied. The integral property is also satisfied because[eq28]

Exercise 2

Consider the following function:[eq29]

where $l,uin U{211d} $ and $l<u$. Prove that [eq4] is a legitimate probability density function.

Solution

$l<u$ implies $frac{1}{u-l}>0$, so [eq2] for any $xin U{211d} $ and the non-negativity property is satisfied. The integral property is also satisfied because[eq32]

Exercise 3

Consider the following function:[eq33]where $nin U{2115} $ and [eq34] is the Gamma function. Prove that [eq4] is a legitimate probability density function.

Solution

Remember the definition of Gamma function:[eq36][eq37] is obviously strictly positive for any $z$, since [eq38] is strictly positive and $x^{z-1}$ is strictly positive on the interval of integration (except at 0 where it is 0). Therefore, [eq4] satisfies the non-negativity property because the four factors in the product[eq40]are all non-negative on the interval [eq41].

The integral property is also satisfied because[eq42]

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