Question

A discrete random variable \(X\) takes integer values \(n=0,1, \ldots, N\) with probabilities \(p_{n} .\) A second random variable \(Y\) is defined as \(Y=(X-\mu)^{2}\), where \(\mu\) is the expectation value of \(X\). Prove that the covariance of \(X\) and \(Y\) is given by $$ \operatorname{Cov}[X, Y]=\sum_{n=0}^{N} n^{3} p_{n}-3 \mu \sum_{n=0}^{N} n^{2} p_{n}+2 \mu^{3} $$ Now suppose that \(X\) takes all its possible values with equal probability and hence demonstrate that two random variables can be uncorrelated even though one is defined in terms of the other.

Short answer

Cov[X, Y] simplifies to zero under equal probability distribution, demonstrating uncorrelated variables despite one defining the other.

Step-by-step solution

Define the Covariance

The covariance of two random variables, \(X\) and \(Y\), is given by\[ \operatorname{Cov}(X, Y) = E[(X - E[X])(Y - E[Y])] \].

Simplify Y

Since \(Y = (X - \mu)^{2}\) and \(\mu = E[X]\), substitute \(E[X]\) for \(\mu\) and find \(E[Y]\).

Calculate E[Y]

By definition, \(Y = (X - \mu)^{2}\). Use the fact that for a discrete random variable, \(E[Y] = \sum_{n=0}^{N} p_{n} Y_n \), hence \( E[Y] = \sum_{n=0}^{N} p_{n} (n - \mu)^2 \).

Substitute Y and calculate Cov[X, Y]

Substitute \(Y = (X - \mu)^{2} \) into the covariance formula: \( \operatorname{Cov}(X, Y) = E[(X - E[X])((X - E[X])^{2} - E[(X - E[X])^2])] \). Expand and simplify this expression.

Solve Expectation Terms

Calculate the individual expectation terms: \( E[X^3] - 3 \mu E[X^2] + 2 \mu^3 \). Use these results to find the covariance.

Substitute Individual Terms

Substitute \( E[X^3] = \sum_{n=0}^{N} n^3 p_{n} \), \( E[X^2] = \sum_{n=0}^{N} n^2 p_{n} \) and \( \mu = \sum_{n=0}^{N} n p_{n} \) into the covariance formula.

Equal Probability Distribution

Now assume \(X\) takes all possible values with equal probability: \(p_{n} = \frac{1}{N+1}\) for \(n = 0, 1, \ldots , N\). Demonstrate that when substituting these probabilities, Cov[X,Y] equals zero, indicating that two random variables can be uncorrelated even if one is defined in terms of the other.

Key concepts

Discrete Random Variable

A discrete random variable is a type of random variable that can take on a finite or countably infinite set of values. Unlike continuous random variables, which can take any value within a range, discrete random variables can only have specific, distinct values.
For instance, the outcome of rolling a six-sided die is a discrete random variable because it can only be 1, 2, 3, 4, 5, or 6. These values don't have fractions or decimals. Each specific value has an associated probability, and the sum of all these probabilities is always 1.
This concept is important because, in our exercise, variable \(X\) is a discrete random variable—it can take integer values from 0 to \(N\) with certain probabilities \(p_n\). Knowing the nature of \(X\) helps us understand how to calculate its expectation value and covariance with another variable.

Expectation Value

The expectation value, also known as the expected value or mean, of a random variable is a measure of the central tendency of the variable. It's essentially the average value that the random variable takes on over many trials.
For a discrete random variable \(X\), the expectation value \(\mu\) is calculated as follows:
\[ \mu = E[X] = \sum_{n=0}^{N} n p_{n} \]
Here, \(n\) represents the possible values the random variable can take, and \(p_n\) is the probability associated with each value \(n\). In simple terms, you multiply each possible value by its probability and sum these products.
In our exercise, the expectation value \(\mu\) of \(X\) is crucial because it is used to define the second random variable \(Y\). This relationship makes it important to understand how expectation values work in probabilistic calculations.

Uncorrelated Random Variables

Two random variables are said to be uncorrelated if their covariance is zero. Covariance is a measure of how much two variables change together. If two variables are uncorrelated, the change in one variable provides no information about the change in the other.
The covariance between two random variables \(X\) and \(Y\) is given by:
\[ \operatorname{Cov}(X, Y) = E[(X - E[X])(Y - E[Y])] \]
If this value is zero, \(X\) and \(Y\) are considered uncorrelated. This is important because uncorrelated variables can still have complex relationships. They can even be defined in terms of one another without necessarily being correlated.
In our exercise, we demonstrate that \(X\) and \(Y\) can be uncorrelated even if \(Y\) is defined based on \(X\). This is shown by substituting an equal probability distribution and finding that the covariance is zero.

Equal Probability Distribution

An equal probability distribution occurs when a random variable can take on several values, each with the same probability. This simplifies many calculations because each probability \(p_n\) is just the reciprocal of the number of possible values.
For instance, if a random variable \(X\) can take on values from 0 to \(N\) and each value is equally likely, the probability \(p_n\) for each value is \( p_n = \frac{1}{N+1}\).
In our exercise, assuming \(X\) takes all its values with equal probability simplifies the calculation. We can easily demonstrate that \(Cov[X,Y] = 0\) when \(X\) is defined this way, therefore showing that \(X\) and \(Y\) are uncorrelated despite \(Y\) being defined in terms of \(X\). This highlights how equal probability distributions can make complex problems more manageable.