Understanding the Probability Density Function and Statistical Measures of a Random Variable

Probability density functions (PDFs) are fundamental in statistical analysis, particularly in continuous probability distributions. One such distribution is the exponential distribution, which is often used in reliability theory, queuing theory, and other stochastic models. This article will explore the concept of the probability density function (PDF), specifically in the context of a random variable X with a given PDF, and will then delve into how to find the mean and variance of another random variable Y derived from X.

The Probability Density Function (PDF) of a Random Variable X

Given the probability density function (PDF) of a random variable X as:

fx  k * exp^(-x/3) for x  0

The PDF must satisfy the properties of a valid probability density function, one of which is that the integral of the PDF over the entire domain must equal 1. Thus, we have:

∫[0 to ∞] k * exp^(-x/3) dx  1

To find the value of k, we need to solve this integral. This is an example of an improper integral over a semi-infinite interval. Let’s proceed with the integration:

Set up the integral for k:

∫[0 to ∞] k * exp^(-x/3) dx  1

Integrate the function:

Let u  -x/3, then du  -1/3 dx

∫[0 to ∞] k * exp^u * (-3) du  1

-3k * ∫[0 to ∞] exp^u du  1

Evaluate the integral:

-3k * [-3 exp^u] from 0 to ∞  1

-3k * [-3 exp^(-x/3)] from 0 to ∞  1

-3k * [0 - (-3)]  1

-3k * 3  1

-9k  1

k  -1/9

Check if k is consistent:

However, k must be positive because the PDF must be non-negative. The correct form for k should be:

k  3/9  1/3

Mean and Variance of Y

Given that Y 5 - 2X, we need to find the mean (expected value) and variance of Y using the properties of expected value and variance for linear transformations of random variables. Let’s start with the mean:

Find the mean of X:

Mean of X, E[X]  ∫[0 to ∞] x * (1/3 * exp^(-x/3)) dx

Let u  -x/3, then du  -1/3 dx

Mean of X  1/3 * ∫[0 to ∞] x * exp^(-x/3) dx

Mean of X  1/3 * ∫[0 to ∞] 3u * exp^u * (-3) du

Mean of X  - ∫[0 to ∞] u * exp^u du

Mean of X  - [u * exp^u] from 0 to ∞   ∫[0 to ∞] exp^u du

Mean of X  0   [exp^u] from 0 to ∞

Mean of X  0   [0 - 1]  1

Find the mean of Y:

Mean of Y, E[Y]  E[5 - 2X]

E[Y]  5 - 2 * E[X]

E[Y]  5 - 2 * 1

E[Y]  3

Find the variance of X:

Variance of X, Var[X]  E[X^2] - (E[X])^2

E[X^2]  ∫[0 to ∞] x^2 * (1/3 * exp^(-x/3)) dx

Let u  -x/3, then du  -1/3 dx

E[X^2]  1/3 * ∫[0 to ∞] x^2 * exp^(-x/3) dx

E[X^2]  1/3 * ∫[0 to ∞] (3u)^2 * exp^u * (-3) du

E[X^2]  - ∫[0 to ∞] 9u^2 * exp^u du

E[X^2]  - [9u^2 * exp^u] from 0 to ∞   2 * ∫[0 to ∞] u * exp^u du

E[X^2]  0   18 * [exp^u] from 0 to ∞

E[X^2]  0   18 * [0 - 1]  9

Var[X]  9 - 1^2  8

Find the variance of Y:

Variance of Y, Var[Y]  Var[5 - 2X]

Var[Y]  Var[-2X]

Var[Y]  (-2)^2 * Var[X]

Var[Y]  4 * 8  32

In this context, we have found that the mean of Y is 3 and the variance of Y is 32, given the transformation Y 5 - 2X and the probability density function fX (x) 1/3 * exp^(-x/3).

Conclusion

This analysis showcases the importance of understanding probability density functions and their properties in statistical analysis. The key concepts we explored include the proper calculation of k in a given PDF, the mean and variance of a transformed random variable, and the use of integration to find these statistical measures.

Keywords

probability density function statistical measures random variable transformation