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Show that if \(X_{1}, X_{2}, \ldots, X_{p}\) are independent random variables and \(Y=c_{1} X_{1}+c_{2} X_{2}+\cdots+c_{p} X_{p}\), $$ V(Y)=c_{1}^{2} V\left(X_{1}\right)+c_{2}^{2} V\left(X_{2}\right)+\cdots+c_{p}^{2} V\left(X_{p}\right) $$ You may assume that the random variables are continuous.

Short Answer

Expert verified
The variance of \(Y\) is the sum of \(c_i^2 V(X_i)\) for independent random variables \(X_i\).

Step by step solution

01

Understand Variance of a Random Variable

The variance of a random variable \(X\), denoted by \(V(X)\), measures how much the values of \(X\) deviate from its mean on average. For any constant \(c\) and random variable \(X\), the property \(V(cX) = c^2V(X)\) holds.
02

Expression for Variance of a Linear Combination

The random variable \(Y = c_1 X_1 + c_2 X_2 + \cdots + c_p X_p\) represents a linear combination of independent random variables. We need to find the variance of this linear combination.
03

Apply Variance Properties

Since \(X_1, X_2, \ldots, X_p\) are independent, the variance of the sum is the sum of the variances. Therefore, \[ V(Y) = V(c_1 X_1 + c_2 X_2 + \cdots + c_p X_p) = V(c_1 X_1) + V(c_2 X_2) + \cdots + V(c_p X_p) \] using independence.
04

Use the Constant Variance Rule

Apply the variance property for a constant multiplied by a random variable: \(V(c_i X_i) = c_i^2 V(X_i)\). Substitute into our expression: \[ V(Y) = c_1^2 V(X_1) + c_2^2 V(X_2) + \cdots + c_p^2 V(X_p) \]
05

Conclude the Proof

By applying the properties of variance and the independence of the random variables, we have shown that the variance of \(Y\) is given by the sum of \(c_i^2 V(X_i)\) for each independent \(X_i\). Thus, the statement is proven.

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Key Concepts

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

Independent Random Variables
Understanding independent random variables is crucial for grasping concepts of variance in linear combinations. Independent random variables are those variables whose outcomes do not influence each other. For example, rolling two separate dice can be considered as having independent variables, as the result of one roll doesn't affect the other. In statistical terms, if you have a set of variables \(X_1, X_2, \ldots, X_p\), they are independent if the joint probability of the outcome can be expressed as the product of their individual probabilities. This mathematical property simplifies calculations, especially when you're dealing with variances because it allows easy summation without considering covariances. In our exercise, where we deal with a linear combination of independent variables, understanding their independence is key to effectively calculating the total variance. The independence allows for the variance of the sum to be calculated as the sum of individual variances, because no variable influences any other in this context.
Variance Properties
Variance is a statistical measure that tells us how much the values of a random variable tend to vary. It's often denoted by \(V(X)\), where \(X\) can be any random variable. The variance quantifies the expectation of the squared deviation of a random variable from its mean, providing insights into the data's dispersion. There are key properties of variance that play a significant role when dealing with linear combinations:
  • Variance of a Constant: The variance of a constant is zero since a constant value does not change.
  • Variance of a Scaled Variable: If you multiply a random variable by a constant, the variance is scaled by the square of that constant, i.e., \( V(cX) = c^2 V(X) \).
  • Sum of Independent Variables: If variables are independent, the variance of their sum is equal to the sum of their variances, allowing for straightforward calculations without needing covariances.
These properties simplify finding variance in composite functions involving multiple independent variables. They reduce complex calculations to manageable steps by utilizing the linearity and independence.
Linear Combinations
A linear combination involves combining multiple random variables, each multiplied by a constant, into a new random variable. In the given exercise, we see a linear combination expressed as \(Y = c_1 X_1 + c_2 X_2 + \cdots + c_p X_p\). Here, \(c_1, c_2, \ldots, c_p\) serve as weights applied to their respective variables \(X_1, X_2, \ldots, X_p\).When dealing with linear combinations, especially where your variables are independent, the main task is to calculate the variance of the new composite variable. You're essentially trying to understand how the combined uncertainty from each \(X_i\) contributes to the total uncertainty of \(Y\). Using the variance properties mentioned earlier, because the \(X_i\)s are independent, you calculate the overall variance as follows:
  • Find the variance of each scaled variable: \(V(c_i X_i) = c_i^2 V(X_i)\).
  • Sum these variances: \(V(Y) = c_1^2 V(X_1) + c_2^2 V(X_2) + \cdots + c_p^2 V(X_p)\).
These calculations highlight how the variance from each weighted variable contributes additively to the aggregate variance, simplifying problems into concise solutions.

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Most popular questions from this chapter

An order of 15 printers contains four with a graphics-enhancement feature, five with extra memory, and six with both features. Four printers are selected at random, without replacement, from this set. Let the random variables \(X, Y,\) and \(Z\) denote the number of printers in the sample with graphics enhancement only, extra memory only, and both, respectively. (a) Describe the range of the joint probability distribution of \(X, Y,\) and \(Z\). (b) Is the probability distribution of \(X, Y,\) and \(Z\) multinomial? Why or why not? (c) Determine the conditional probability distribution of \(X\) given that \(Y=2\). Determine the following: (d) \(P(X=1, Y=2, Z=1)\) (e) \(P(X=1, Y=1)\) (f) \(E(X)\) and \(V(X)\) pe (g) \(P(X=1, Y=2 \mid Z=1)\) (h) \(P(X=2 \mid Y=2)\) to (i) Conditional probability distribution of \(X\) given that \(Y=0\) and \(Z=3\).

The lifetimes of six major components in a copier are independent exponential random variables with means of 8000 , \(10,000,10,000,20,000,20,000,\) and 25,000 hours, respectively. (a) What is the probability that the lifetimes of all the components exceed 5000 hours? (b) What is the probability that at least one component lifetime exceeds 25,000 hours?

Test results from an electronic circuit board indicate that \(50 \%\) of board failures are caused by assembly defects, \(30 \%\) are due to electrical components, and \(20 \%\) are due to mechanical defects. Suppose that 10 boards fail independently. Let the random variables \(X, Y,\) and \(Z\) denote the number of assembly, electrical, and mechanical defects among the 10 boards. Calculate the following: (a) \(P(X=5, Y=3, Z=2)\) (b) \(P(X=8)\) (c) \(P(X=8 \mid Y=1)\) (d) \(P(X \geq 8 \mid Y=1)\) (e) \(P(X=7, Y=1 \mid Z=2)\)

The time between surface finish problems in a galvanizing process is exponentially distributed with a mean of 40 hours. A single plant operates three galvanizing lines that are assumed to operate independently. (a) What is the probability that none of the lines experiences a surface finish problem in 40 hours of operation? (b) What is the probability that all three lines experience a surface finish problem between 20 and 40 hours of operation? (c) Why is the joint probability density function not needed to answer the previous questions?

A small-business Web site contains 100 pages and \(60 \%, 30 \%,\) and \(10 \%\) of the pages contain low, moderate, and high graphic content, respectively. A sample of four pages is selected without replacement, and \(X\) and \(Y\) denote the number of pages with moderate and high graphics output in the sample. Determine: (a) \(f_{X Y}(x, y)\) (b) \(f_{X}(x)\) (c) \(E(X)\) (d) \(f_{Y \mid 3}(y)\) (e) \(E(Y \mid X=3)\) (f) \(V(Y \mid X=3)\) (g) Are \(X\) and \(Y\) independent?

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