Proving the Expectation of a Standard Normal Random Variable: Simplified Techniques and Integral Evaluations

In this article, we will delve into the method of proving the expectation of a standard normal random variable using both variable substitution and linear combinations. By eliminating the need to write integrals explicitly, we simplify the proof and provide a clearer understanding of the underlying concepts.

Introduction to Standard Normal Distribution

The standard normal distribution is a fundamental concept in probability theory and statistics, characterized by a normal distribution with a mean (μ) of 0 and a standard deviation (σ) of 1. Denoted as (X sim N(0,1)), the probability density function (PDF) of a standard normal random variable is represented as:

PDF(X) ( frac{1}{sqrt{2pi}} e^{-X^2 / 2} )

Variable Substitution and Linear Combinations

The problem at hand is to prove the expectation of a function involving the standard normal random variable (X). Specifically, we need to evaluate:

( mathbb{E}left[frac{1}{1 e^X}right] )

Given that (X) is a standard normal random variable, we can exploit its symmetry. Since (X) and (-X) have the same distribution, we can write:

( mathbb{E}left[frac{1}{1 e^X}right] mathbb{E}left[frac{1}{1 e^{-X}}right] )

Using the property that the expectation is a linear operator, we have:

( 2mathbb{E}left[frac{1}{1 e^X}right] mathbb{E}left[frac{1}{1 e^X}right] mathbb{E}left[frac{1}{1 e^{-X}}right] mathbb{E}left[frac{1 e^X}{1 e^X}right] 1 )

This simplifies to:

( 2mathbb{E}left[frac{1}{1 e^X}right] 1 )

Thus, we conclude:

( mathbb{E}left[frac{1}{1 e^X}right] frac{1}{2} )

Integral Evaluation Strategy

To further support this result, we can conduct an integral evaluation directly. Starting with the expectation of (frac{1}{1 e^X}), we write:

( mathbb{E}left[frac{1}{1 e^X}right] int_{-infty}^{infty} frac{1}{1 e^x} cdot frac{1}{sqrt{2pi}} e^{-x^2/2} , dx )

By making the substitution (x -t), we transform the integral:

( mathbb{E}left[frac{1}{1 e^X}right] frac{1}{sqrt{2pi}} int_{-infty}^{infty} frac{1}{1 e^{-t}} e^{-t^2/2} , (-dt) frac{1}{sqrt{2pi}} int_{-infty}^{infty} frac{e^t}{e^t 1} e^{-t^2/2} , dt )

Adding the original integral to this transformed version:

( 2mathbb{E}left[frac{1}{1 e^X}right] frac{1}{sqrt{2pi}} int_{-infty}^{infty} left( frac{1}{1 e^x} frac{e^x}{1 e^x} right) e^{-x^2/2} , dx )

This simplifies to:

( 2mathbb{E}left[frac{1}{1 e^X}right] frac{1}{sqrt{2pi}} int_{-infty}^{infty} e^{-x^2/2} , dx )

The integral on the right-hand side is the integral of the standard normal PDF, which is equal to 1. Thus, we have:

( 2mathbb{E}left[frac{1}{1 e^X}right] 1 )

Therefore, the final result is:

( mathbb{E}left[frac{1}{1 e^X}right] frac{1}{2} )

Conclusion

In conclusion, we have successfully proven that the expectation of the function (frac{1}{1 e^X}) for a standard normal random variable (X) is (frac{1}{2}) using both variable substitution and integral evaluation techniques. This result highlights the symmetrical properties and linearity of the expectation operator in the context of the standard normal distribution.