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Let \(X_{n}\) and \(Y_{m}\) be independent random variables having the Poisson distribution with parameters \(n\) and \(m\), respectively, Show that $$ \frac{\left(X_{n}-n\right)-\left(Y_{m}-m\right)}{\sqrt{X_{n}+Y_{m}}} \stackrel{\mathrm{D}}{\rightarrow} N(0,1) \quad \text { as } m, n \rightarrow \infty $$.

Short Answer

Expert verified
As \(m, n \to \infty\), the expression converges to \(N(0,1)\) due to the Central Limit Theorem.

Step by step solution

01

Define the Random Variables

Let \(X_n\) and \(Y_m\) be independent Poisson random variables with parameters \(n\) and \(m\), respectively. This means \(X_n \sim \text{Poisson}(n)\) and \(Y_m \sim \text{Poisson}(m)\).
02

Central Limit Theorem for Poisson

The Central Limit Theorem states that a Poisson random variable with a large parameter \( \lambda \) can be approximated by a normal distribution \(N(\lambda, \lambda)\). Therefore, \( X_n \) is approximately \(N(n, n)\) for large \( n \), and \( Y_m \) is approximately \(N(m, m)\) for large \( m \).
03

Expression of the Random Expression

Consider the expression \( (X_n - n) - (Y_m - m) \). This simplifies to \( (X_n - Y_m) - n + m \). To show the limit holds, we need to understand the distribution of \(X_n - Y_m\).
04

Distribution of Difference of Poisson Random Variables

Since \(X_n\) and \(Y_m\) are independent, if we consider \( X_n - Y_m \), it approximately follows a normal distribution with mean \(n - m\) and variance \(n + m\) for large \( n \) and \( m \).
05

Analyze the Distribution

The expression given can be re-written as:\[ \frac{(X_n - Y_m) - (n - m)}{\sqrt{X_n + Y_m}} \]For large \(n, m\), the numerator \((X_n - Y_m) - (n - m)\) will have mean 0 and variance \(n + m\).
06

Application of CLT to the Expression

As \(m, n \to \infty\), the Central Limit Theorem applies, and we have:\[ \frac{(X_n - Y_m) - (n - m)}{\sqrt{n + m}} \approx N(0, 1) \]Since \(n + m\) is approximately the same as \(X_n + Y_m\) for large \(n, m\), this implies:\[ \frac{(X_n - Y_m) - (n - m)}{\sqrt{X_n + Y_m}} \overarrow{D}{\rightarrow} N(0, 1) \] as desired.

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Key Concepts

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

Poisson Distribution
The Poisson distribution is often used to model the number of events in a fixed interval of time or space. It assumes these events occur with a constant average rate and are independent of the time since the last event. For a Poisson random variable with parameter \( \lambda \), the probability of observing \( k \) events is given by:\[ P(X=k) = \frac{\lambda^k e^{-\lambda}}{k!} \]This distribution is useful when you're dealing with rare events over a continuous timeline, like the number of emails you receive per hour or the occurrence of earthquakes.
  • Parameter: \( \lambda \) (rate at which events occur)
  • Mean and Variance: Both are equal to \( \lambda \)
When the parameter \( \lambda \) is large, the Poisson distribution can be approximated by the normal distribution, making calculations easier for large-scale problems.
Normal Distribution
The normal distribution, also known as the Gaussian distribution, is a continuous probability distribution characterized by its symmetric, bell-shaped curve, where most observations cluster around the central peak. It is defined by two parameters: the mean (\( \mu \)) and the variance (\( \sigma^2 \)). For a normal distribution, you have:\[ f(x | \mu, \sigma^2) = \frac{1}{\sqrt{2\pi\sigma^2}} e^{-\frac{(x-\mu)^2}{2\sigma^2}} \]
  • Mean, \( \mu \): The center of the distribution
  • Variance, \( \sigma^2 \): Measures the spread of the distribution
This distribution is fundamental due to the Central Limit Theorem, which states that the sum of a large number of independent, identically distributed random variables will be approximately normally distributed, regardless of their original distribution.
Independent Random Variables
Random variables are said to be independent if the occurrence of one does not affect the probability of occurrence of another. In mathematical terms, if you have two random variables \( X \) and \( Y \), they are independent if:\[ P(X = x \text{ and } Y = y) = P(X = x) \cdot P(Y = y) \]
  • Importance: Knowing that variables are independent can simplify complex probability problems.
  • Application: In the context of the exercise, \( X_n \) and \( Y_m \) being independent allows us to analyze their combined behavior separately.
This concept is crucial when applying the Central Limit Theorem or when determining the joint behavior of multiple variables in statistical analysis.
Asymptotic Behavior
Asymptotic behavior refers to how a function behaves as its inputs approach a certain value (often infinity). In statistics, we are often interested in the asymptotic properties of distributions as sample sizes become very large.When we say that \( \frac{(X_n - Y_m) - (n - m)}{\sqrt{X_n + Y_m}} \) converges to \( N(0, 1) \) as \( n, m \to \infty \), we are describing its asymptotic behavior. This implies that as the sample sizes increase, the distribution of the given expression more closely resembles that of the standard normal distribution \( N(0,1) \).
  • Relevance: Understanding asymptotic behavior helps predict how the distribution of a statistic will stabilize as sample size grows.
  • Uses in Statistics: It aids in making inferences about populations when dealing with large samples.
In the context of the exercise, the large \( n \) and \( m \) allow the use of normal approximations to make complex probability calculations more manageable.

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Most popular questions from this chapter

Let \(Y\) be uniformly distributed on \([-1,1]\) and let \(X=y^{2}\), (a) Find the best predictor of \(X\) given \(Y\), and of \(Y\) given \(X\). (b) Find the best linear predictor of \(X\) given \(Y\), and of \(Y\) given \(X\).

3\. Random walk. Let \(X_{1}, X_{2}, \ldots\), be independent identically distributed random variables taking, values in the integers \(\mathrm{Z}\) and having a finite mean. Show that the Markov chain \(S=\left\\{S_{n} \mid\right.\) given by \(S_{n}=\sum_{1}^{n} X_{i}\) is transient if \(\mathbb{E}\left(X_{1}\right) \neq 0\).

Let \(X_{n}\) be the net profit to the gambler of betting a unit stake on the \(n\)th play in a casino; the \(X_{n}\) may be dependent, but the game is fair in the sense that \(\mathbb{E}\left(X_{n+1} \mid X_{1}, X_{2} \ldots . X_{n}\right)=0\) for all n. The gambler stakes \(Y\) on the first play, and thereafter stakes \(f_{n}\left(X_{1}, X_{2} \ldots, X_{n}\right)\) on the \((n+1)\) th play, where \(f_{1}, f_{2} \ldots\) are given functions. Show that her profit after \(n\) plays is $$ s_{n}=\sum_{i=1}^{n} X_{i} f_{i-1}\left(X_{1}, X_{2} \ldots \ldots, x_{i-1}\right) $$ where \(f_{0}=Y\). Show further that the sequence \(S=\left[S_{n} \mid\right.\) satisfies the martingale condition \(\mathrm{B}\left(S_{n+1} \mid\right.\) \(\left.X_{1}, X_{2}, \ldots, X_{n}\right)=S_{n}, n \geq 1\), if \(Y\) is assumed to be known throughout.

Let \(\left[X_{n} \mid\right.\) be a sequence of independent random variables which converges in probability to the limit \(X\). Show that \(X\) is almost surely constant.

Let \(X\) be an irreducible discrete-time Markov chain, and let \(\mu_{i}\) be the mean recurrence time of state \(i\). Let \(V_{i}(n)=\sum_{r=0}^{n-1} I_{\left[X_{r}=d\right]}\) be the number of visits to \(i\) up to \(n-1\), and let \(f\) be any bounded function on \(S\). Show that: (a) \(n^{-1} V_{i}(n) \stackrel{\text { as. }}{\longrightarrow} \mu_{i}^{-1}\) as \(n \rightarrow \infty\). (b) if \(\mu_{i}<\infty\) for all \(i\), then $$ \frac{1}{n} \sum_{r=0}^{n-1} f\left(X_{r}\right) \rightarrow \sum_{i \in S} f(i) / \mu_{i} \quad \text { as } n \rightarrow \infty $$.

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