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Legitimate probability density functions

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.

Properties of probability density functions

The following proposition formally describes the two properties.

Proposition Let X be an absolutely 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 an absolutely continuous random variable X whose pdf is [eq4].

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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