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Chapter 3: Problem 2

Show that if \(\operatorname{var}(X)=0\) then \(X\) is almost surely constant; that is, there exists \(a \in \mathrm{R}\) such that \(\mathbb{P}(X=a)=1\). (First show that if \(\mathrm{E}\left(X^{2}\right)=0\) then \(\mathbb{P}(X=0)=1 .\) )

Short Answer

Expert verified
If \(\operatorname{var}(X) = 0\), then \(X\) is almost surely equal to its mean \(\mu\); thus, \(X\) is a constant almost surely.

Step by step solution

01

Understanding the Problem

We need to show that if the variance of a random variable \(X\) is zero, then \(X\) is almost surely constant. This means there is a real number \(a\) such that the probability that \(X\) equals \(a\) is 1.
02

Recall the Definition of Variance

Recall that the variance of a random variable \(X\) is defined as \(\operatorname{var}(X) = \mathrm{E}[(X - \mu)^2]\), where \(\mu\) is the mean of \(X\). If \(\operatorname{var}(X) = 0\), then \(\mathrm{E}[(X - \mu)^2] = 0\).
03

Apply Expectation on Squared Terms

If \(\mathrm{E}[(X - \mu)^2] = 0\), this implies \(X - \mu\) is almost surely 0 because the expectation of a non-negative random variable being zero indicates that the random variable is zero with probability 1. Hence, \(X = \mu\) almost surely.
04

Special Case of Zero Expectation

In the first sub-result, you are tasked to prove: if \( \mathrm{E}(X^2) = 0 \), then \( \mathbb{P}(X=0) = 1 \). Since \(X^2 \geq 0\), if \( \mathrm{E}(X^2) = 0 \), then \(X^2 = 0\) with probability 1, which means \(X = 0\) with probability 1.

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Key Concepts

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

Almost Sure Convergence
Almost sure convergence refers to a scenario in stochastic processes where a sequence of random variables converges to a specific random variable with probability one. This is a strong form of convergence, indicating that the values of the sequence almost certainly become increasingly close to a particular value or another random variable as the sample size grows.

To put it simply:
  • If you have a random variable sequence \( X_n \), it converges to another random variable \( X \) almost surely if the probability that \( X_n \) converges to \( X \) is 1.
  • This means that as you observe more samples, the sequence essentially matches the behavior of \( X \) with overwhelming certainty.
In our exercise, when demonstrating that if \( \operatorname{var}(X) = 0 \), then \( X \) is almost surely constant, we're applying this concept. Since the variance, which measures how spread out a set of random variable data points are, is zero, there is no variation at all. Consequently, \( X \) must converge to a constant value with probability one.
Expectation
Expectation, often referred to as the mean or expected value, is a foundational concept in probability and statistics. It provides a measure of the central tendency of a random variable. When you compute the expectation of a random variable, it is like calculating a weighted average of all possible outcomes.

Key points about expectation:
  • The expectation of a random variable \( X \), denoted \( \mathrm{E}(X) \), tells you the average outcome if you repeated the process many times.
  • It is calculated as \( \mathrm{E}(X) = \sum (x_i \cdot p(x_i)) \) for discrete variables, where \( x_i \) are possible values and \( p(x_i) \) are their probabilities. For continuous variables, it is an integral.
  • Expectation helps in assessing the probable outcomes in scenarios where outcomes are due to chance.
In our problem, the expectation is used to prove that a variance of zero implies that all observations are fixed at a constant value. The expression \( \mathrm{E}[(X - \mu)^2] = 0 \) suggests that deviations from the mean are non-existent, hence \( X \) is constant.
Mean of Random Variables
The mean of a random variable is a crucial statistic that offers insights into the central value of a probability distribution. Often, the mean and expectation terms are used interchangeably, as they both describe the average outcome for repeated trials or observations.

Understanding the mean of random variables involves:
  • Calculating the average of possible values that a random variable can take.
  • The mean is essentially the point around which the probability mass of a distribution is centered.
For a random variable \( X \), the mean, typically denoted as \( \mu \), is computed in a similar fashion to expectation. In our exercise, demonstrating that a zero variance implies a constant \( X \) heavily relies on understanding the mean. Since \( X \) exhibits no deviation from its mean (as variance is zero), \( X \) consistently equals \( \mu \) 鈥� making \( X \) almost surely constant.

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Most popular questions from this chapter

Let \(\mathbf{X}=\left(X_{1}, X_{2}, \ldots, X_{n}\right)\) be a vector of random variables. The covariance matrix \(\mathbf{V}(\mathbf{X})\) of \(\mathbf{X}\) is defined to be the symmetric \(n\) by \(n\) matrix with entries \(\left(v_{i j}: 1 \leq i, j \leq n\right)\) given by \(v_{i j}=\operatorname{cov}\left(X_{i}, X_{j}\right)\). Show that \(|\mathbf{V}(\mathbf{X})|=0\) if and only if the \(X_{t}\) are linearly dependent with probability one, in that \(\mathbb{P}\left(a_{1} X_{1}+a_{2} X_{2}+\cdots+a_{n} X_{n}=b\right)=1\) for some \(\mathbf{a}\) and \(\dot{b} .(\mid \mathbf{V}]\) denotes the determinant of \(\mathbf{V} .\) )

(a) If \(X\) takes non-negative integer values show that \(\mathbb{E}(X)=\sum_{n=0}^{\infty} \mathrm{P}(X>n)\) (b) An um contains \(b\) blue and \(r\) red balls. Balls are removed at random until the first blue ball is drawn. Show that the expected number drawn is \((b+r+1) /(b+1)\). (c) The balls are replaced and then removed at random until all the remaining balls are of the same colour. Find the expected number remaining in the urn.

(a) Use the inclusion-exclusion formula \((3.4 .2)\) to derive the result of Example (3.4.3), namely: in a random permutation of the first \(n\) integers, the probability that exactly \(r\) retain their original positions is $$ \frac{1}{r !}\left(\frac{1}{2 !}-\frac{1}{3 !}+\cdots+\frac{(-1)^{n-r}}{(n-r) !}\right) $$ (b) Let \(d_{n}\) be the number of derangements of the first \(n\) integers (that is, rearrangements with no integers in their original positions). Show that \(d_{n+1}=n d_{n}+n d_{n-1}\) for \(n \geq 2\). Deduce the result of part (a).

Consider the following fragment of verse entitled 'Note for the scientist'. People who have three daughters try for more, And then its fifty-fifty they'll have four, Those with a son or sons will let things be, Hence all these surplus women, QED. (a) What do you think of the argument? (b) Show that the mean number of children of either sex in a family whose fertile parents have followed this policy equals 1. (You should assume that each delivery yields exactly one child whose sex is equally likely to be male or female.) Discuss.

Suppose you find a warm-hearted bookmaker offering payoff odds of \(\pi(k)\) against the kth horse in an \(n\)-horse race where \(\sum_{k=1}^{n}[\pi(k)+1\\}^{-1}<1\). Show that you can distribute your bets in such a way as to ensure you win.
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