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Given a random variable \(X\), how does the variance of \(c X\) relate to that of \(X\) ?

Short Answer

Expert verified
The variance of \(cX\) is \(c^2\) times the variance of \(X\).

Step by step solution

01

Understanding Variance of a Scaled Random Variable

To determine the relationship between the variances of a scaled random variable, you need to know a basic property of variance. This property states that if a random variable \(X\) has a variance \(\text{Var}(X)\), then multiplying \(X\) by a constant \(c\) results in \(cX\) having a variance of \(c^2 \times \text{Var}(X)\).
02

Applying the Variance Property

Given the above property, we can directly apply it. If \(X\) is a random variable with variance \(\text{Var}(X)\), then the variance of \(cX\) is given by \(\text{Var}(cX) = c^2 \times \text{Var}(X)\).

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Key Concepts

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

Random Variables
Random variables are fundamental concepts in probability and statistics that represent numerical outcomes of a random phenomenon. When dealing with random variables, we typically encounter two types: discrete and continuous.
Discrete random variables may take on a finite or countable sequence of values, like rolling a die where outcomes can be 1 through 6. In contrast, continuous random variables can take on any numerical value within a given range, like measuring the exact weight of an apple.
Random variables are crucial because they allow us to define and analyze probability distributions, which describe how probabilities are spread over the possible outcomes.
Variance Property
Variance is a measure of how much a random variable's outcomes differ from its expected value, known as the mean. The formula for variance is given by equation:\[ \text{Var}(X) = E[(X - \mu)^2] \]where \( \mu \) is the expected value of the random variable \( X \), and \( E \) denotes the expectation.
An essential property of variance is how it responds to transformations of the random variable. If a random variable \( X \) has a variance \( \text{Var}(X) \), then altering \( X \) by multiplying it with a constant \( c \), transforms the variance to \( c^2 \times \text{Var}(X) \).
This property is valuable in statistical analysis, enabling easier calculations when dealing with scaled variables, like financial returns or data units.
Scaling Factor in Variance
The scaling factor's effect on variance highlights the profound impact a constant multiplier has when applied to a random variable. Consider a random variable \( X \) with variance \( \text{Var}(X) \). If we scale \( X \) by a constant \( c \), forming a new random variable \( cX \), the variance of \( cX \) adjusts by the square of the scaling factor, expressed as \( c^2 \times \text{Var}(X) \).
  • This means that scaling amplifies the variability by the square of \( c \).
  • This scaling principle helps in understanding variability in different measurement units or contexts, such as when converting currencies or test scores.
Understanding this concept assists learners in predicting how changes in magnitude affect the spread of a distribution.

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Most popular questions from this chapter

Given an array \(A\) of length \(n\) (chosen from some set that has an underlying ordering), you can select the largest element of the array by first setting \(L=A[1]\) and then comparing \(L\) to the remaining elements of the array, one at a time, replacing \(L\) with \(A[i]\) if \(A[i]\) is larger than \(L\). Assume that the elements of \(A\) are randomly chosen. For \(i>1\), let \(X_{i}=1\) if an element \(i\) of \(A\) is larger than any element of \(A[1: i-1]\). Let \(X_{1}=1 .\) What does \(X_{1}+X_{2}+\cdots+X_{n}\) have to do with the number of times you assign a value to \(L ?\) What is the expected number of times you assign a value to \(L\) ?

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A professor decides that the method proposed for computing the maximum list size is much too complicated. He proposes the following solution: If we let \(X_{i}\) be the size of list \(i\), then what we want to compute is \(E\left(\max _{i}\left(X_{i}\right)\right)\). This means $$ E\left(\max _{i}\left(X_{i}\right)\right)=\max _{i}\left(E\left(X_{i}\right)\right)=\max _{i}(1)=1 . $$ What is the flaw in his solution?

In an independent trials process consisting of six trials with probability p of success, what is the probability that the first three trials are successes and the last three are failures? The probability that the last three trials are successes and the first three are failures? The probability that Trials 1, 3, and 5 are successes and Trials 2, 4, and 6 are failures? What is the probability of three successes and three failures?

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