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Suppose that \(X\) and \(Y\) are discrete random variables, and that \(\phi(X)\) and \(\psi(X)\) are two functions of \(X\) satisfying $$ \mathbb{E}(\phi(X) g(X))=\mathbb{E}(\psi(X) g(X))=\mathbb{E}(Y g(X)) $$ for any function \(g\) for which all the expectations exist. Show that \(\phi(X)\) and \(\psi(X)\) are almost surely equal, in that \(\mathrm{P}(\phi(X)=\psi(X))=1\).

Short Answer

Expert verified
\( \phi(X) \) and \( \psi(X) \) are almost surely equal, i.e., \( \mathrm{P}(\phi(X) = \psi(X)) = 1 \).

Step by step solution

01

Understand the Given Condition

We know that the expectations \( \mathbb{E}(\phi(X) g(X)) = \mathbb{E}(Y g(X)) \) and \( \mathbb{E}(\psi(X) g(X)) = \mathbb{E}(Y g(X)) \) hold for any function \( g \). This implies that \( \mathbb{E}(\phi(X) g(X)) = \mathbb{E}(\psi(X) g(X)) \) for any choice of \( g \).
02

Consider the Difference Function

Define a new function \( h(X) = \phi(X) - \psi(X) \). By the properties given, we have \( \mathbb{E}(h(X) g(X)) = \mathbb{E}((\phi(X) - \psi(X)) g(X)) = 0 \) for any function \( g \) for which the expectation exists.
03

Choose a Specific Function for g(X)

Take \( g(X) = \mathbf{1}_{\{h(X) > 0\}} \), where \( \mathbf{1} \) is an indicator function that is 1 if \( h(X) > 0 \) and 0 otherwise. Thus, we have \( \mathbb{E}((\phi(X) - \psi(X)) \mathbf{1}_{\{h(X) > 0\}}) = 0 \).
04

Analyze the Implication of Zero Expectation

Since \( \mathbb{E}((\phi(X) - \psi(X)) \mathbf{1}_{\{h(X) > 0\}}) = 0 \), the quantity \( (\phi(X) - \psi(X)) \mathbf{1}_{\{h(X) > 0\}} = 0 \) almost surely. This implies \( \mathbf{1}_{\{h(X) > 0\}} = 0 \) almost surely, meaning \( \mathrm{P}(h(X) > 0) = 0 \).
05

Repeat for h(X) < 0

Similarly, if we choose \( g(X) = \mathbf{1}_{\{h(X) < 0\}} \), we get \( \mathbb{E}((\phi(X) - \psi(X)) \mathbf{1}_{\{h(X) < 0\}}) = 0 \), implying \( \mathrm{P}(h(X) < 0) = 0 \).
06

Conclude the Equality of \\phi and \\psi

Since \( \mathrm{P}(h(X) > 0) = 0 \) and \( \mathrm{P}(h(X) < 0) = 0 \), we can deduce that \( \mathrm{P}(h(X) = 0) = 1 \). Hence, \( \mathrm{P}(\phi(X) = \psi(X)) = 1 \).

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Key Concepts

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

Discrete Random Variables
In probability theory, a discrete random variable is a type of variable that can only take distinct, separate values. For example, if you roll a die, the result can be any integer from 1 to 6. Discrete random variables are contrasted with continuous random variables, which can take any value within a range. Understanding discrete random variables is crucial for dealing with various statistical models, particularly those involving countable outcomes. **Characteristics of Discrete Random Variables:**
  • Countable outcomes: They can take on a finite or countably infinite number of possible values.
  • Probability Mass Function (PMF): For a discrete random variable, probabilities are assigned through a PMF, which gives the probability that the variable equals a specific value.
  • Cumulative Distribution Function (CDF): The CDF of a discrete random variable gives the probability that the variable is less than or equal to a specific value.
Studying discrete random variables allows learners to apply these concepts to practical problems, such as determining probabilities in games, predicting outcomes based on known data, and making decisions under uncertainty.
Almost Sure Equality
The concept of *almost sure equality* is a fascinating part of probability theory. It describes a situation where two random variables are equal with probability 1, even if they are not equal in every scenario. The term 'almost sure' indicates an overwhelming probability but allows for exceptions in some negligible cases. **Understanding Almost Sure Equality:**
  • Probability 1: When we say two variables are almost surely equal, it means the probability that they are not equal is zero.
  • Negligible cases: There could be a set of outcomes of probability zero where the two variables differ, but this does not impact the overall conclusion of equality.
  • Application: Almost sure equality is particularly useful in theoretical probability and stochastic processes where exact equality cannot be guaranteed at every single point.
In this context, the exercise provides the basis for understanding how conditions like symmetric expectations can lead to the conclusion that two functions of a random variable must be almost surely equal.
Expectation
Expectation, often called expected value, is a fundamental concept in probability and statistics. It refers to the average or mean value of a random variable over a large number of experiments or trials. **Calculating Expectation:**For a discrete random variable, the expectation is calculated using the formula:\[\mathbb{E}(X) = \sum_{i} x_i P(X = x_i)\]where:
  • \(X\) is the random variable,
  • \(x_i\) are the possible values it can take,
  • \(P(X = x_i)\) is the probability that \(X\) equals \(x_i\).
**Significance of Expectation:**
  • Central Tendency: The expectation provides a measure of the central location of the distribution of the variable.
  • Predictive Power: It is used to predict long-term average results if the same experiment is repeated multiple times.
  • Applications: Expectation is extensively used in decision theory, economics, and various applications of statistical inference.
By understanding and calculating expectations, we can harness statistical models to anticipate outcomes effectively and make informed decisions based on probabilistic forecasts.

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Most popular questions from this chapter

A particle performs a random walk on the non-negative intesers as follows. When at the point \(n\) \((>0)\) its next position is uniformly distributed on the set \((0,1,2, \ldots, n+1) .\) When it hits 0 for the first time, it is absorbed. Suppose it starts at the point \(a .\) (a) Find the probability that its position never exceeds \(a\), and prove that, with probability 1 , it is absorbed ultimately- (b) Find the probability that the final step of the walk is from 1 to 0 when \(a=1\). (c) Find the expected number of steps taken before absorption when \(a=1\).

\(N+1\) plates are laid out around a circular dining table, and a hot cake is passed between them in the manner of a symmetric random walk: each time it arrives on a plate, it is tossed to one of the two neighbouring plates, each possibility having probability \(\frac{1}{2}\). The game stops at the moment when the cake has visited every plate at least once. Show that, with the exception of the plate where the eake began, each plate has probability \(1 / N\) of being the last plate visited by the cake.

A coin is tossed repeatedly, heads turning up with probability \(p\) on each toss. Player A wins the game if heads appears at least \(m\) times before tails has appeared \(n\) times; otherwise player \(\mathrm{B}\) wins the game. Find the probability that A wins the game.

In 1710, J. Arbuthnot observed that male births had exceeded female births in London for 82 successive years. Arguing that the two sexes are equally likely, and \(2^{-82}\) is very small, he attributed this run of masculinity to Divine Providence. Let us assume that each birth results in a girl with probability \(p=0.485\), and that the outcomes of different confinements are independent of each other. Ignoring the possibility of twins (and so on), show that the probability that girls outnumber boys in \(2 n\) live births is no greater than \(\left(\begin{array}{c}2 \pi \\ n\end{array}\right) p^{n} q^{n}\\{q /(q-p)\\}\), where \(q=1-p\). Suppose that 20,000 children are bom in each of 82 successive years. Show that the probability that boys outnumber girls every year is at least \(0.99\). You may need Stirling's formula.

Let \(T\) be the time which elapses before a simple random walk is absorbed at either of the absorbing barriers at 0 and \(N\), having started at \(k\) where \(0 \leq k \leq N\). Show that \(P(T<\infty)=1\) and \(\mathbb{E}\left(T^{k}\right)<\infty\) for all \(k \geq 1\).

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