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Chapter 1: Problem 10

Suppose that the two variables \(x\) and \(z\) are statistically independent. Show that the mean and variance of their sum satisfies $$ \begin{aligned} \mathbb{E}[x+z] &=\mathbb{E}[x]+\mathbb{E}[z] \\ \operatorname{var}[x+z] &=\operatorname{var}[x]+\operatorname{var}[z] \end{aligned} $$

Short Answer

Expert verified
The expectations and variances of independent variables sum separately: \(\mathbb{E}[x+z] = \mathbb{E}[x] + \mathbb{E}[z]\) and \(\operatorname{var}[x+z] = \operatorname{var}[x] + \operatorname{var}[z]\).

Step by step solution

01

Understand the Problem

We are tasked with proving two properties involving the expectation and variance of two independent random variables, \(x\) and \(z\). Specifically, we need to show that the expectation of their sum equals the sum of their expectations and that the variance of their sum equals the sum of their variances.
02

Apply the Expectation Property

For independent variables, the expectation of the sum is the sum of the expectations. Mathematically, \(\mathbb{E}[x+z] = \mathbb{E}[x] + \mathbb{E}[z]\). This holds because the expectation is a linear operator, meaning it distributes over addition.
03

State Variance Definition

The variance of a random variable is defined as the expectation of the squared deviation from its mean: \(\operatorname{var}[x] = \mathbb{E}[(x - \mathbb{E}[x])^2]\).
04

Independent Variables and Variance

For independent random variables, the variance of their sum is the sum of their variances: \(\operatorname{var}[x+z] = \operatorname{var}[x] + \operatorname{var}[z]\). This holds because the covariance term \(\operatorname{cov}(x, z)\), which is \(\operatorname{var}[x+z] = \operatorname{var}[x] + \operatorname{var}[z] + 2\operatorname{cov}(x, z)\), is zero for independent variables.
05

Summarize the Proof

Since we have shown \(\mathbb{E}[x+z] = \mathbb{E}[x] + \mathbb{E}[z]\) using the linearity of expectation, and \(\operatorname{var}[x+z] = \operatorname{var}[x] + \operatorname{var}[z]\) due to their independence, both statements are proven true.

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Key Concepts

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

Expectation
Expectation, often symbolized as \( \mathbb{E} \), is a fundamental concept in probability that averages out the different outcomes of a random variable, weighted by their probabilities. Think of it like the long-term average if you were to repeat an experiment many times.
This is sometimes called the "expected value" or "mean" of a random variable. When dealing with two independent variables, like \( x \) and \( z \), the expectation of their sum has a neat property: the expectation of their sum equals the sum of their expectations. This is because the expectation is a linear operator, allowing us to distribute over addition.
  • For example, if \( \mathbb{E}[x] = 3 \) and \( \mathbb{E}[z] = 5 \), then \( \mathbb{E}[x + z] = 3 + 5 = 8 \).
  • This principle holds true no matter how complex the variables are, as long as they remain independent.

This idea is intuitive and foundational when dealing with expected outcomes, providing a straightforward way to predict the behavior of combined random variables.
Variance
Variance, often expressed as \( \operatorname{var}(X) \) for a random variable \( X \), quantifies the spread of possible outcomes around the expected value. It's calculated by averaging the squared differences from the mean. A higher variance indicates more spread out data, while a lower variance indicates data is more clustered around the mean.
For two independent variables, like \( x \) and \( z \), the variance of their sum is simply the sum of their variances. This can be written as:
  • \( \operatorname{var}(x+z) = \operatorname{var}(x) + \operatorname{var}(z) \)
  • This occurs because, for independent variables, the covariance term鈥擻( 2\operatorname{cov}(x, z) \)鈥攊s zero.
When dealing with complex systems, variance helps us understand the uncertainty or risk involved by providing a measure of how much results can vary.
Random Variables
A random variable is a variable that can take on different values, each with an associated probability. Random variables are a foundational concept in probability and statistics, allowing us to model uncertain outcomes like rolling a die or choosing a card from a deck.
Random variables can be discrete, like the number of heads in ten coin flips, or continuous, like the height of randomly chosen individuals.
  • Independence is a key feature. If two random variables are independent, the occurrence of one doesn't affect the other. This lack of influence simplifies many calculations.
  • For example, if we have a dice roll \( x \) and a coin flip \( z \), the result from the dice roll doesn't change the probability of landing heads or tails.
By understanding and applying the properties of random variables, we can predict and quantify outcomes in uncertain conditions. This is essential for statistical analysis and informs decision-making in many fields.

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Most popular questions from this chapter

The gamma function is defined by $$ \Gamma(x) \equiv \int_{0}^{\infty} u^{x-1} e^{-u} \mathrm{~d} u $$ Using integration by parts, prove the relation \(\Gamma(x+1)=x \Gamma(x)\). Show also that \(\Gamma(1)=1\) and hence that \(\Gamma(x+1)=x !\) when \(x\) is an integer.

Suppose that we have three coloured boxes \(r\) (red), \(b\) (blue), and \(g\) (green). Box \(r\) contains 3 apples, 4 oranges, and 3 limes, box \(b\) contains 1 apple, 1 orange, and 0 limes, and box \(g\) contains 3 apples, 3 oranges, and 4 limes. If a box is chosen at random with probabilities \(p(r)=0.2, p(b)=0.2, p(g)=0.6\), and a piece of fruit is removed from the box (with equal probability of selecting any of the items in the box), then what is the probability of selecting an apple? If we observe that the selected fruit is in fact an orange, what is the probability that it came from the green box?

Given a loss matrix with elements \(L_{k j}\), the expected risk is minimized if, for each \(\mathbf{x}\), we choose the class that minimizes (1.81). Verify that, when the loss matrix is given by \(L_{k j}=1-I_{k j}\), where \(I_{k j}\) are the elements of the identity matrix, this reduces to the criterion of choosing the class having the largest posterior probability. What is the interpretation of this form of loss matrix?

Consider two variables \(\mathbf{x}\) and \(\mathbf{y}\) having joint distribution \(p(\mathbf{x}, \mathbf{y})\). Show that the differential entropy of this pair of variables satisfies $$ \mathrm{H}[\mathbf{x}, \mathbf{y}] \leqslant \mathrm{H}[\mathbf{x}]+\mathrm{H}[\mathbf{y}] $$ with equality if, and only if, \(\mathbf{x}\) and \(\mathbf{y}\) are statistically independent.

In this exercise, we prove the normalization condition (1.48) for the univariate Gaussian. To do this consider, the integral $$ I=\int_{-\infty}^{\infty} \exp \left(-\frac{1}{2 \sigma^{2}} x^{2}\right) \mathrm{d} x $$ which we can evaluate by first writing its square in the form $$ I^{2}=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \exp \left(-\frac{1}{2 \sigma^{2}} x^{2}-\frac{1}{2 \sigma^{2}} y^{2}\right) \mathrm{d} x \mathrm{~d} y $$ Now make the transformation from Cartesian coordinates \((x, y)\) to polar coordinates \((r, \theta)\) and then substitute \(u=r^{2}\). Show that, by performing the integrals over \(\theta\) and \(u\), and then taking the square root of both sides, we obtain $$ I=\left(2 \pi \sigma^{2}\right)^{1 / 2} $$ Finally, use this result to show that the Gaussian distribution \(\mathcal{N}\left(x \mid \mu, \sigma^{2}\right)\) is normalized.
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