Cumulative Distribution Function and Expected Value: A Guide to Solving for ( E[Y] ) Given ( Y aX^b )

In this guide, we will explore the cumulative distribution function (CDF) and expected value of a random variable. Specifically, we will solve for ( E[Y] ) where ( Y aX^b ) given the CDF ( F_X(x) 1 - e^{-x} ) for ( x geq 0 ).

Understanding the CDF and Exponential Distribution

The cumulative distribution function ( F_X(x) 1 - e^{-x} ) for ( x geq 0 ) indicates that ( X ) follows a standard exponential distribution with a rate parameter ( lambda 1 ). The probability density function (PDF) of ( X ) is ( f_X(x) e^{-x} ) for ( x geq 0 ). The exponential distribution is often used to model waiting times between events in a Poisson process.

Aims and Objectives

The primary aim is to find the expected value ( E[Y] ) where ( Y aX^b ). To achieve this, we will use the properties of the cumulative distribution function and the linearity of expectation.

Properties and Key Concepts

Cumulative Distribution Function (CDF): The CDF ( F_X(x) ) for a random variable ( X ) is defined as ( F_X(x) P(X leq x) ), which for the standard exponential distribution is given by ( F_X(x) 1 - e^{-x} ).

Exponential Distribution: The exponential distribution is a continuous probability distribution for a non-negative random variable ( X ) that describes the time between events in a Poisson process. Its PDF is ( f_X(x) lambda e^{-lambda x} ), and for ( X ) following a standard exponential distribution, ( lambda 1 ).

Expected Value: The expected value ( E[Y] ) of a random variable ( Y ) is the long-run average value of repetitions of the experiment it represents. It can be calculated using the formula ( E[Y] int_{-infty}^{infty} y , f_Y(y) , dy ) or, for a function of a random variable ( Y g(X) ), using the formula ( E[Y] E[g(X)] int_{-infty}^{infty} g(x) , f_X(x) , dx ).

Solving for ( E[Y] )

Given ( Y aX^b ), we need to find ( E[Y] ). First, let's recall that ( X ) follows a standard exponential distribution with ( f_X(x) e^{-x} ) for ( x geq 0 ).

The expected value of ( Y aX^b ) can be found using the properties of expectation. Specifically, for any function ( g(X) ), the expected value ( E[g(X)] ) can be calculated as:

[E[g(X)] int_{0}^{infty} g(x) , f_X(x) , dx]

Substituting ( g(x) a x^b ) and ( f_X(x) e^{-x} ) into the formula, we get:

[E[Y] E[aX^b] int_{0}^{infty} a x^b , e^{-x} , dx]

This integral can be solved using integration by parts or recognized as a form of the Gamma function ( Gamma(b 1) ). The Gamma function ( Gamma(n) ) is defined as:

[Gamma(n) int_{0}^{infty} t^{n-1} e^{-t} , dt]

For ( g(x) x^b ), we have:

[int_{0}^{infty} x^b , e^{-x} , dx Gamma(b 1)]

Thus, the expected value ( E[Y] ) becomes:

[E[Y] a , Gamma(b 1)]

For ( b 3/4 ), we need to find ( Gamma(3/4 1) ), which is ( Gamma(7/4) ).

Conclusion

Using the properties of the exponential distribution and the Gamma function, we have derived that the expected value ( E[Y] ) for ( Y aX^{3/4} ) is given by:

[E[Y] a , Gamma(7/4)]

This result can be applied to specific values of ( a ) and used in various practical scenarios involving waiting times and other exponential processes.

Related Keywords

Cumulative Distribution Function: A function that gives the probability that a random variable ( X ) is less than or equal to a given value.

Expected Value: The long-run average value of a function of a random variable.

Exponential Distribution: A continuous probability distribution used to model the time between events in a Poisson process.

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