Question

Let $X$ and $Y$ be random variables such that $E\left(X^k\right)$ and $E\left(Y^k\right)\ne 0$ exist for $k=1,2,3,\dots ..$ If the ratio $X/Y$ and its denominator $Y$ are independent, prove that $E[(X/Y{)}^k]=E\left(X^k\right)/E\left(Y^k\right),k=1,2,3,\dots$ Hint: Write $E\left(X^k\right)=E[Y^k(X/Y{)}^k]$.

Short answer

The result is proven by rewriting the expectation of $X^k$, applying the property of independent variables, simplifying the expression and cancelling out a common term from both sides of equation.

Step-by-step solution

Rewrite the expectation

Following the hint provided, represent the expectation $E[X^k]$ as $E[Y^k.(X/Y{)}^k]$. This provides us a starting point to simplify and compare with the provided expression.

Move the expectation inside

Since the ratio $X/Y$ and its denominator $Y$ are independent, according to the property of independent random variables, we can move the Expectation function inside the brackets to get: $E[Y^k.E[(X/Y{)}^k]]$. This lets us break down the complex expression into simpler components.

Simplify the expression

The inside expression now changes to $E[(X/Y{)}^k]$, the expectation of the ratio $X/Y$ raised to the power $k$. However, we know that the expectation of $Y^k$ is non-zero from the problem statement. So we can rewrite this as $E[Y^k]\ast E[(X/Y{)}^k]$, where $\ast$ denotes multiplication.

Proving the equality

Observe that we can cancel out one $E[Y^k]$ from both sides of the equation. This leads to $E[(X/Y{)}^k]=E[X^k]/E[Y^k]$, which is the conclusion we aimed to prove.

Key concepts

Expectation

The concept of expectation, often referred to as the expected value, is a fundamental idea in probability and statistics. It provides a measure of the central tendency of a random variable's possible values, weighed by their probabilities. Think of it as the weighted average of all possible outcomes.

For a discrete random variable, expectation is calculated by the formula:

• $$E(X)=\sum _i{x}_{i}P(x_i)$$

Where

• $x_i$ are the possible values the random variable can take, and
• $P(x_i)$ is the probability of each event occurring.

For continuous random variables, the expectation is determined using an integral:

• $$E(X)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}xf(x)dx$$

Where

• $f(x)$ is the probability density function of the random variable.

Notably, expectations have linearity properties. For any random variables $X$ and $Y$, and constants $a$ and $b$, the expected value operates as:

• $E(aX+bY)=aE(X)+bE(Y)$

This property becomes particularly useful when dealing with independent random variables.

Random Variables

Random variables are central to the study of probability. A random variable can be thought of as a function that assigns a numerical value to each outcome in a sample space of a random process. They are classified mainly into two types:

• Discrete Random Variables: These take on a countable number of possible values. Examples include the result of rolling a die or the number of heads when flipping coins.
• Continuous Random Variables: These can take on an infinite set of values within a given interval. Examples are the exact time taken for an activity or the measurement of heights.

Understanding random variables is crucial because they allow us to model and analyze real-world phenomena probabilistically. When we describe a random variable, we typically use several statistical measures, including:

• Probability distribution: Describes how probabilities are spread over the values that the random variable can take.
• Expectation: As discussed, this gives the average or mean value of the random variable.
• Variance: Indicates how much the values of the random variable deviate from the mean.

Knowing how random variables behave and interact, especially when they are independent, helps in predicting outcomes and making decisions based on probabilistic models.

Ratio of Random Variables

When dealing with random variables, we often need to assess how one random variable behaves relative to another, which leads us to consider the ratio of random variables. This concept becomes interesting and sometimes complex because the ratio involves division and possibly creating new kinds of distributions.

For instance, if $X$ and $Y$ are two random variables, their ratio $X/Y$ creates another random variable. The key challenge with a ratio is ensuring that the denominator does not become zero, as division by zero is undefined.

When $X$ and $Y$ are independent, like in our exercise, some simplifications are possible. Independence implies that the behavior of one variable does not influence the other, allowing us to simplify expectations involving their ratios easily. As illustrated in the problem, if $X/Y$ and $Y$ are independent, we can derive equivalence relations for expectations of powers of their ratios:

• $$E[(X/Y{)}^k]=\frac{E\left(X^k\right)}{E\left(Y^k\right)}$$

This relation demonstrates that when the ratio $X/Y$ and $Y$ are independent, it simplifies the calculation to individual expectations making the analytical process much easier.