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Chapter 5: Problem 12

For any function \(f(x)\) the expected value is \(\mathrm{E}[f]=\sum p_{i} f\left(x_{i}\right)\) or \(\int p(x) f(x) d x\) (discrete or continuous probability). The function can be \(x\) or \((x-m)^{2}\) or \(x^{2} .\) If the mean is \(\mathrm{E}[x]=m\) and the variance is \(\mathrm{E}\left[(x-m)^{2}\right]=\sigma^{2}\) what is \(\mathbf{E}\left[x^{2}\right] ?\)

Short Answer

Expert verified
\( \mathrm{E}[x^2] = \sigma^2 + m^2 \).

Step by step solution

01

Understand the Problem

We are given a few functions concerning the expected value, including the mean and variance. The goal is to determine the expected value of the square of a random variable, \[ \mathrm{E}\left[x^2\right] \].
02

Use the Definition of Variance

The variance is given as \( \mathrm{E}\left[(x - m)^2\right] = \sigma^2 \). By expanding the squared term, we have:\[ (x - m)^2 = x^2 - 2mx + m^2. \]
03

Apply Linearity of Expectation

Using the linearity of expectation on the expanded form of variance, we get:\[ \mathrm{E}\left[(x - m)^2\right] = \mathrm{E}[x^2] - 2m\mathrm{E}[x] + m^2. \]
04

Substitute Known Values

Substitute the expressions for mean and variance into the equation:\[ \sigma^2 = \mathrm{E}[x^2] - 2m \times m + m^2 = \mathrm{E}[x^2] - 2m^2 + m^2. \]
05

Simplify the Equation

Simplify the right-hand side:\[ \sigma^2 = \mathrm{E}[x^2] - m^2. \]
06

Solve for E[x^2]

Now, solve the equation for \( \mathrm{E}[x^2] \):\[ \mathrm{E}[x^2] = \sigma^2 + m^2. \]

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Key Concepts

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

Variance
Variance is a statistical concept that measures how much a set of numbers is spread out. Think of it as the average of the squared differences from the mean. If the variance is high, the numbers are more spread out from their average value. Conversely, a low variance indicates that the numbers are closer to the mean.
  • In mathematical terms, for a random variable, variance is defined as: \( \sigma^2 = \mathrm{E}[(x - m)^2] \), where \( m \) is the mean of the data set.
  • The variance provides insight into the variability of data, which is essential in probability and statistics to determine the reliability of data set outcomes.
Understanding variance is crucial as it helps in assessing the consistency of data. In other words, variance quantifies the uncertainty in predicting future outcomes based on the given data.
Linearity of Expectation
The linearity of expectation is a property of expected values that makes calculations involving expectations easier to handle. This property states that for any two random variables \( X \) and \( Y \), the expected value of their sum is the sum of their expected values, i.e., \( \mathrm{E}[X + Y] = \mathrm{E}[X] + \mathrm{E}[Y] \).
  • This property holds true regardless of whether \( X \) and \( Y \) are independent or dependent.
  • It's particularly useful for simplifying the computations of expected values when dealing with linear combinations of random variables.
Applying this property allows us to break down complex expressions into simpler parts, making them much more manageable. That's what we did with the expression \( \mathrm{E}[(x - m)^2] \) to find \( \mathrm{E}[x^2] \). It helped in expanding and isolating part of the equation to solve for the unknown.
Discrete and Continuous Probability
Probability comes in two main types: discrete and continuous. These are used to describe two different kinds of outcomes.
  • Discrete Probability: This involves scenarios where the outcomes can be counted, often finite. Examples include rolling dice or flipping a coin. For discrete probability, probabilities are often calculated using sums.
  • Continuous Probability: This involves outcomes that can take any value within a range, like the exact height of students in a class. Here, probabilities are calculated using integrals.
In the context of expected values, these definitions matter because they determine whether we sum probabilities or integrate over a range. This also affects the function we use to compute the expected value, whether it's using \( \sum \) for discrete or \( \int \) for continuous situations. Knowing the distinction helps in applying the correct mathematical methods.
Random Variable
A random variable is a fundamental concept in probability and statistics, representing a numerical outcome of a random phenomenon.
  • There are two types: Discrete and Continuous random variables. Discrete random variables take on a finite or countable number of values, while continuous random variables can take any value within a given range.
  • Random variables allow for quantifying the probabilities of different outcomes in various experiments, like flipping a coin or measuring rainfall.
It's important to note that the expected value of a random variable is a key aspect of understanding its behavior. This value provides the average outcome you would expect if you could repeat the experiment infinite times.

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Most popular questions from this chapter

When the covariance matrix for outputs \(\boldsymbol{X}\) is \(\boldsymbol{V}\), the covariance matrix for outputs \(Z=A X\) is \(A V A^{\mathrm{T}}\). Explain this neat formula by linearity: $$ Z=\mathrm{E}\left[(A X-\overline{A X})(A X-\overline{A X})^{\mathrm{T}}\right]=A \mathrm{E}\left[(X-\bar{X})(X-\bar{X})^{\mathrm{T}}\right] A^{\mathrm{T}} . $$

Two measurements of the same variable \(x\) give two equations \(x=b_{1}\) and \(x=b_{2}\). Suppose the means are zero and the variances are \(\sigma_{1}^{2}\) and \(\sigma_{2}^{2}\), with independent errors: \(V\) is diagonal with entries \(\sigma_{1}^{2}\) and \(\sigma_{2}^{2} .\) Write the two equations as \(A \boldsymbol{x}=\boldsymbol{b}\) ( \(A\) is 2 by 1 ). As in the text Example 1 , find this best estimate \(\widehat{x}\) based on \(b_{1}\) and \(b_{2}\) : $$ \widehat{\boldsymbol{x}}=\frac{b_{1} / \sigma_{1}^{2}+b_{2} / \sigma_{2}^{2}}{1 / \sigma_{1}^{2}+1 / \sigma_{2}^{2}} \quad \mathrm{E}\left[\widehat{\boldsymbol{x}} \widehat{\boldsymbol{x}}^{\mathrm{T}}\right]=\left(\frac{1}{\sigma_{1}^{2}}+\frac{1}{\sigma_{2}^{2}}\right)^{-1} $$

Why is \(\left\|B Z B^{-1}\right\| \leq 1\) impossible for any \(B\) but \(\left\|C Y C^{-1}\right\| \leq 1\) is possible ? $$ Z=\left[\begin{array}{ll} 1 & 1 \\ 0 & 1 \end{array}\right] \quad Y=\left[\begin{array}{cc} 1 & 1 \\ 0 & -1 \end{array}\right] $$

Suppose you have \(N=4\) samples \(157,312,696,602\) in Problem 5 . What are the first digits \(x_{1}\) to \(x_{4}\) of the squares? What is the sample mean \(\mu\) ? What is the sample variance \(S^{2}\) ? Remember to divide by \(N-1=3\) and not \(N=4\).

Find the probability generating function \(G(z)\) for Poisson's \(p_{n}=e^{-\lambda} \lambda^{n} / n !\)
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