Question

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

Short answer

The equation is proved by showing that the expectation value \(E[(X / Y) ^ k]\) is equal to \(E(X^k)\) by recognizing the properties of expected values and independent random variables. Hence, it can be deduced that \(E[(X / Y)^k] = E(X^k) / E(Y^k)\).

Step-by-step solution

Apply the Hint

Apply the hint given in the problem to rewrite \(E(X^k)\) : we can write \(E(X^k)\) as \(E[Y^k(X / Y)^k]\)

Separate Variables

With \(E[YZ] = E[Y]E[Z]\) for independent random variables Y and Z, we can now separate the variables in the rewritten equation from step 1: Our expression becomes \(E[Y^k]E[(X / Y)^k]\)

Simplify

Simplify the resulting expression: It simplifies to \(E(X^k)E(Y^k) / E(Y^k)\)

Further Simplify

Notice that \(E(Y^k)\) appears in the numerator and denominator, which means you cancel these terms. The final expression is therefore \(E(X^k)\)

Compare with Original Expression

To address the original problem, we also need to show that this expression equals \(E[(X / Y)^k]\). This can now be easily shown using step 2 where we rewrote \(E(X^k)\) as \(E[Y^k]E[(X / Y)^k]\), which is equivalent to \(E[(X / Y)^k]\) when we consider Y and (X / Y) as independent variables as given in the problem. So, we can see that \(E[(X / Y)^k] = E(X^k)\).

Key concepts

Expected Value

In simple terms, the expected value is a measure of the center of a distribution of a random variable. It's basically the average you would expect if you could repeat an experiment over and over again under identical conditions.

For a discrete random variable, the expected value, denoted as \(E(X)\), is calculated by summing up the products of each possible value that the random variable can take and its respective probability. For continuous random variables, the expected value is calculated using an integral.

To illustrate, if \(X\) represents the outcome of rolling a six-sided die, the expected value of \(X\) would be \(E(X) = 1\cdot\frac{1}{6} + 2\cdot\frac{1}{6} + 3\cdot\frac{1}{6} + 4\cdot\frac{1}{6} + 5\cdot\frac{1}{6} + 6\cdot\frac{1}{6} = 3.5\), even though you can never actually roll a 3.5 with a die. The expected value gives us a way to measure the average outcome in a sense, taking into account the different weights (probabilities) associated with each outcome.

Independence of Random Variables

Two random variables are independent if the occurrence of one does not affect the probability of occurrence of the other.

In formal terms, \(X\) and \(Y\) are independent if for all possible values \(x\) and \(y\), the probability that \(X = x\) and \(Y = y\) simultaneously is the product of the individual probabilities: \(P(X = x, Y = y) = P(X = x)P(Y = y)\).

If random variables are independent, it simplifies a lot of calculations in probability and statistics. For example, if \(X\) and \(Y\) are independent, the expected value of their product is the product of their expected values. However, independence is a strong assumption and must be validated for the specific circumstances. Independence is essential in the given exercise because it allows us to manipulate the expected value of the ratio without worry about any underlying relationship between \(X\) and \(Y\).

Random Variables

A random variable is a variable that takes on numeric values that result from the outcome of a random phenomenon. Essentially, it's a way to map outcomes of random processes to numbers. There are two types of random variables: discrete and continuous.

Discrete random variables take on a countable number of distinct values. An example would be rolling a die, where the possible outcomes are 1, 2, 3, 4, 5, or 6. On the other hand, continuous random variables can take on any value within an interval, such as the amount of time until a radioactively unstable atom decays, which can be any positive real number.

The function that describes the probability of a discrete random variable assuming a particular value is called the probability mass function (PMF), while the function describing this for a continuous random variable is called the probability density function (PDF). Understanding random variables is crucial because they form the basis for statistical calculations and allow for the application of various theories and concepts in probability and statistics.