/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 9 Let \(X\) be a random variable h... [FREE SOLUTION] | 91Ó°ÊÓ

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Let \(X\) be a random variable having expected value \(\mu\) and variance \(\sigma^{2}\). Find the expected value and variance of $$ Y=\frac{X-\mu}{\sigma} $$

Short Answer

Expert verified
The expected value of \(Y = \frac{X - \mu}{\sigma}\) is 0, and the variance of \(Y\) is 1.

Step by step solution

01

Write the expressions for expected value and variance of Y

We are given that \(Y = \frac{X - \mu}{\sigma}\). We want to find: 1. The expected value of \(Y\), \(E[Y]\). 2. The variance of \(Y\), \(Var[Y]\).
02

Find the expected value of Y

By definition, the expected value of \(Y\), \(E[Y]\) is given by: $$ E[Y] = E\left[\frac{X - \mu}{\sigma}\right], $$ where \(E[...]\) represents the expected value of the expression inside the brackets. Now, using the properties of expected value, we can rewrite this expression as: $$ E[Y] = \frac{1}{\sigma} E[X - \mu]. $$ We know that \(E[X] = \mu\), so \(E[X - \mu] = \mu - \mu = 0\). Therefore, $$ E[Y] = \frac{1}{\sigma} \times 0 = 0. $$ So, the expected value of \(Y\) is 0.
03

Find the variance of Y

By definition, the variance of Y, \(Var[Y]\) is given by: $$ Var[Y] = E\left[\left(\frac{X - \mu}{\sigma}\right)^2\right] - \left(E\left[\frac{X-\mu}{\sigma}\right]\right)^2, $$ where \(Var[...]\) represents the variance of the expression inside the brackets and \(E[...]\) represents the expected value. Using the properties of expected value, we can rewrite the expression as: $$ Var[Y] = \frac{1}{\sigma^2} E[(X - \mu)^2] - 0^2, $$ as we found the expected value of \(Y\) to be 0 in Step 2. Now, we know that the variance of \(X\) is \(\sigma^2\) and by definition, \(Var[X] = \sigma^2 = E[(X - \mu)^2]\). Therefore, we can substitute this: $$ Var[Y] = \frac{1}{\sigma^2} \times \sigma^2, $$ which simplifies to: $$ Var[Y] = 1. $$ So, the variance of \(Y\) is 1. Our final results are: 1. The expected value of \(Y\) is 0. 2. The variance of \(Y\) is 1.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Expected Value
The expected value is a fundamental concept in probability theory.
It represents the average or the mean value of a random variable over an infinite number of trials.
In other words, it's what you can expect to get on average if you could repeat an experiment forever.For a random variable like \( X \), the expected value is written as \( E[X] \).
This value is often referred to by the Greek letter \( \mu \), which signifies the population mean.
Expected value can be thought of as a weighted average, where each outcome of \( X \) is weighted by its probability.Here, when we want to calculate the expected value of \( Y = \frac{X - \mu}{\sigma} \), we are interested in the average of \( Y \) over many trials.
Using linearity of expectation, a property of the expected value, the calculation simplifies.
The expected value of a mean-centered random variable becomes zero because \( \mu \) offsets \( X \), resulting in:
  • \( E[Y] = 0 \)
Variance
Variance measures how spread out the values of a random variable are around the mean.
It gives a numerical value to the notion of distribution width.For a random variable \( X \), variance is denoted by \( Var(X) \) or \( \sigma^2 \).
It is calculated by taking the average of the squared differences from the mean.
The formula is \( Var(X) = E[(X - \mu)^2] \).In our exercise, when dealing with \( Y = \frac{X - \mu}{\sigma} \), we transform \( X \) by centering and scaling, hence affecting its variance.
By properties of variance, when all data in a set are multiplied by a constant, the variance is multiplied by the square of that constant.
Here, \( \/ \frac{1}{\sigma} \/ \) is the constant, which therefore leads to:
  • \( Var(Y) = 1 \)
Random Variable
A random variable is essentially a variable that represents possible outcomes of a random process.
It can be understood as a real-valued function that is defined over a sample space.Random variables can be discrete, taking specific values like integers, or continuous, where they take any value within a range.
In probability theory, they are used to quantify and describe random phenomena.In our example, \( X \) is a random variable with a specific expected value \( \mu \) and variance \( \sigma^2 \).
Transforming \( X \) into \( Y = \frac{X - \mu}{\sigma} \) standardizes the random variable.
This standardization makes the new random variable \( Y \) have an expected value of 0 and a variance of 1.
Thus, it aligns with the standard normal distribution properties.
Standardization
Standardization is a statistical method that transforms data to have a mean of zero and a standard deviation of one.
This process is extremely useful when comparing different datasets or working with distributions that have different scales.By standardizing a random variable \( X \), we convert it into a standardized random variable \( Y \).
This is achieved using the formula \( Y = \frac{X - \mu}{\sigma} \), which centers \( X \) by subtracting the mean \( \mu \) and scales it by dividing by the standard deviation \( \sigma \).Standardization allows data to be compared on the same scale, facilitating easier analysis and interpretation.
In our exercise, this results in the standardized random variable \( Y \) having:
  • An expected value of 0.
  • A variance of 1.
This transformation aligns the data with the standard normal distribution, enabling more straightforward application of statistical methods and further calculations.

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