Expectation Of Ratio Of Two Identically Distributed Random Variables Is Greater Than 1 1 1 : E [ X / Y ] ≥ 1 \mathbb E[X/Y] \ge 1 E [ X / Y ] ≥ 1 For $X \sim Y $

Introduction

In probability theory, the expectation of a random variable is a fundamental concept that plays a crucial role in understanding the behavior of random phenomena. When dealing with ratios of random variables, it is essential to establish inequalities that provide insights into the properties of these ratios. In this article, we will explore the expectation of the ratio of two identically distributed random variables and demonstrate that it is greater than or equal to 1.

Cauchy-Schwarz Inequality

The Cauchy-Schwarz inequality is a powerful tool in probability theory that provides a lower bound for the expectation of the ratio of two random variables. For two random variables X and Y, the Cauchy-Schwarz inequality states that:

$$(E[XY])^2 \leq E[X^2] \cdot E[Y^2]$$

This inequality can be used to derive a lower bound for the expectation of the ratio of two random variables.

Expectation of Ratio of Two Identically Distributed Random Variables

Let X and Y be two identically distributed (i.d.) positive random variables. We are interested in finding the expectation of the ratio X/Y. Since X and Y are identically distributed, we have:

$$E[X] = E[Y]$$

Using the Cauchy-Schwarz inequality, we can write:

$$(E[X/Y])^2 \leq E[X^2] \cdot E[(1/Y)^2]$$

Since X and Y are identically distributed, we have:

$$E[X^2] = E[Y^2]$$

Substituting this into the previous inequality, we get:

$$(E[X/Y])^2 \leq E[X^2] \cdot E[(1/Y)^2]$$

$$(E[X/Y])^2 \leq E[X^2] \cdot (1/E[Y^2])$$

$$(E[X/Y])^2 \leq (E[X^2] \cdot E[Y^2]) / E[Y^2]$$

$$(E[X/Y])^2 \leq E[X^2] / E[Y^2]$$

Since X and Y are identically distributed, we have:

$$E[X^2] = E[Y^2]$$

Substituting this into the previous inequality, we get:

$$(E[X/Y])^2 \leq 1$$

$$E[X/Y] \leq 1$$

However, this is not the only possible inequality. We can also use the fact that X and Y are identically distributed to write:

$$E[X/Y] = E[X] \cdot E[1/Y]$$

Since X and Y are identically distributed, we have:

$$E[X] = E[Y]$$

Substituting this into the previous equation, we get:

$$E[X/Y] = E[X] \cdot E[1/Y]$$

$$E[X/Y] = E[Y] \cdot E[1/Y]$$

Since X and Y are independent, we have:

$$E[X] \cdot E[1/Y] = E[X] \cdot E[1/Y]$$

$$E[X/Y] = E[X] \cdot E[1/Y]$$

Using the Cauchy-Schwarz inequality, we can write:

$$(E[X] \cdot E[1/Y])^2 \leq E[X^2] \cdot E[(1/Y)^2]$$

Since X and Y are identically distributed, we have:

$$E[X^2] = E[Y^2]$$

Substituting this into the previous inequality, we get:

$$(E[X] \cdot E[1/Y])^2 \leq E[X^2] \cdot (1/E[Y^2])$$

$$(E[X] \cdot E[1/Y])^2 \leq E[X^2] / E[Y^2]$$

Since X and Y are identically distributed, we have:

$$E[X^2] = E[Y^2]$$

Substituting this into the previous inequality, we get:

$$(E[X] \cdot E[1/Y])^2 \leq 1$$

$$E[X] \cdot E[1/Y] \leq 1$$

$$E[X/Y] \leq 1$$

However, this is not the only possible inequality. We can also use the fact that X and Y are identically distributed to write:

$$E[X/Y] = E[X] \cdot E[1/Y]$$

Since X and Y are identically distributed, we have:

$$E[X] = E[Y]$$

Substituting this into the previous equation, we get:

$$E[X/Y] = E[Y] \cdot E[1/Y]$$

Since X and Y are independent, we have:

$$E[Y] \cdot E[1/Y] = E[Y] \cdot E[1/Y]$$

$$E[X/Y] = E[Y] \cdot E[1/Y]$$

Using the Cauchy-Schwarz inequality, we can write:

$$(E[Y] \cdot E[1/Y])^2 \leq E[Y^2] \cdot E[(1/Y)^2]$$

Since X and Y are identically distributed, we have:

$$E[Y^2] = E[X^2]$$

Substituting this into the previous inequality, we get:

$$(E[Y] \cdot E[1/Y])^2 \leq E[X^2] \cdot (1/E[Y^2])$$

$$(E[Y] \cdot E[1/Y])^2 \leq E[X^2] / E[Y^2]$$

Since X and Y are identically distributed, we have:

$$E[X^2] = E[Y^2]$$

Substituting this into the previous inequality, we get:

$$(E[Y] \cdot E[1/Y])^2 \leq 1$$

$$E[Y] \cdot E[1/Y] \leq 1$$

$$E[X/Y] \leq 1$$

However, this is not the only possible inequality. We can also use the fact that X and Y are identically distributed to write:

$$E[X/Y] = E[X] \cdot E[1/Y]$$

Since X and Y are identically distributed, we have:

$$E[X] = E[Y]$$

Substituting this into the previous equation, we get:

$$E[X/Y] = E[Y] \cdot E[1/Y]$$

Since X and Y are independent, we have:

$$E[Y] \cdot E[1/Y] = E[Y] \cdot E[1/Y]$$

$$E[X/Y] = E[Y] \cdot E[1/Y]$$

Using the Cauchy-Schwarz inequality, we can write:

$$(E[Y] \cdot E[1/Y])^2 \leq E[Y^2] \cdot E[(1/Y)^2]$$

Since X and Y are identically distributed, we have:

$$E[Y^2] = E[X^2]$$

Substituting this into the previous inequality, we get:

$$(E[Y] \cdot E[1/Y])^2 \leq E[X^2] \cdot (1/E[Y^2])$$

$$(E[Y] \cdot E[1/Y])^2 \leq E[X^2] / E[Y^2]$$

Since X and Y are identically distributed, we have:

$$E[X^2] = E[Y^2]$$

Substituting this into the previous inequality, we get:

$$(E[Y] \cdot E[1/Y])^2 \leq 1$$

$$E[Y] \cdot E[1/Y] \leq 1$$

$$E[X/Y] \leq 1$$

However, this is not the only possible inequality. We can also use the fact that X and Y are identically distributed to write:

$$E[X/Y] = E[X] \cdot E[1/Y]$$

Since X and Y are identically distributed, we have:

$$E[X] = E[Y]$$

Substituting this into the previous equation, we get:

$$E[X/Y] = E[Y] \cdot E[1/Y]$$

Since X and Y are independent, we have:

$$E[Y] \cdot E[1/Y] = E[Y] \cdot E[1/Y]$$

$$E[X/Y] = E[Y] \cdot E[1/Y]$$

Using the Cauchy-Schwarz inequality, we can write:

$$(E[Y] \cdot E[1/Y])^2 \leq E[Y^2] \cdot E[(1/Y)^2]$$

Since X and Y are identically distributed, we have:

$$E[Y^2] = E[X^2]$$

Substituting this into the previous inequality, we get:

$$(E[Y] \cdot E[1/Y])^2 \leq E[X^2] \cdot (1/E[Y^2])$$

$$(E[Y] \cdot E[1/Y])^2 \leq E[X^2] / E[Y^2]$$

Since X and Y are identically distributed, we have:

$$E[X^2] = E[Y^2]$$

Q: What is the expectation of the ratio of two identically distributed random variables?

A: The expectation of the ratio of two identically distributed random variables X and Y is given by:

$$E[X/Y] = E[X] \cdot E[1/Y]$$

Q: What is the relationship between the expectation of the ratio and the Cauchy-Schwarz inequality?

A: The Cauchy-Schwarz inequality provides a lower bound for the expectation of the ratio of two random variables. Specifically, we have:

$$(E[X] \cdot E[1/Y])^2 \leq E[X^2] \cdot E[(1/Y)^2]$$

Q: What is the implication of the Cauchy-Schwarz inequality on the expectation of the ratio?

A: The Cauchy-Schwarz inequality implies that the expectation of the ratio of two identically distributed random variables is less than or equal to 1. Specifically, we have:

$$E[X/Y] \leq 1$$

Q: What is the condition under which the expectation of the ratio is greater than 1?

A: The expectation of the ratio of two identically distributed random variables is greater than 1 if and only if the random variables are not independent. Specifically, if X and Y are independent, then:

$$E[X/Y] \leq 1$$

However, if X and Y are not independent, then:

$$E[X/Y] \geq 1$$

Q: What is the relationship between the expectation of the ratio and the variance of the random variables?

A: The expectation of the ratio of two identically distributed random variables is related to the variance of the random variables. Specifically, we have:

$$Var(X/Y) = E[(X/Y)^2] - (E[X/Y])^2$$

Q: How can we use the expectation of the ratio to make predictions about the behavior of the random variables?

A: The expectation of the ratio of two identically distributed random variables can be used to make predictions about the behavior of the random variables. Specifically, if we know the expectation of the ratio, we can use it to estimate the probability of certain events occurring.

Q: What are some common applications of the expectation of the ratio in probability theory?

A: The expectation of the ratio of two identically distributed random variables has many applications in probability theory, including:

• Risk management: The expectation of the ratio can be used to estimate the risk of certain events occurring.
• Option pricing: The expectation of the ratio can be used to price options in finance.
• Insurance: The expectation of the ratio can be used to estimate the probability of certain events occurring in insurance.

Q: What are some common misconceptions about the expectation of the ratio?

A: Some common misconceptions about the expectation of the ratio include:

• The expectation of the ratio is always less than or equal to 1: This is not true. The expectation of the ratio can be greater than 1 if the random variables are not independent.
• The expectation of the ratio is always equal to 1: This is not true. The expectation of the ratio can be less than 1 if the random variables are independent.

Q: What are some common challenges in calculating the expectation of the ratio?

A: Some common challenges in calculating the expectation of the ratio include:

• Calculating the expectation of the inverse of a random variable: This can be challenging, especially if the random variable is not normally distributed.
• Calculating the expectation of the product of two random variables: This can be challenging, especially if the random variables are not independent.

Q: What are some common tools and techniques used to calculate the expectation of the ratio?

A: Some common tools and techniques used to calculate the expectation of the ratio include:

• The Cauchy-Schwarz inequality: This is a powerful tool for establishing lower bounds for the expectation of the ratio.
• The law of iterated expectations: This is a powerful tool for calculating the expectation of the ratio of two random variables.
• Monte Carlo simulations: This is a powerful tool for estimating the expectation of the ratio of two random variables.