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If \(W \sim \chi_{v}^{2}\), show that \(\mathrm{E}(W)=v, \operatorname{var}(W)=2 v\) and \((W-v) / \sqrt{2 v} \stackrel{D}{\longrightarrow} N(0,1)\) as \(v \rightarrow\) \(\infty\)

Short Answer

Expert verified
\(E(W)=v\), \(\operatorname{var}(W)=2v\), \(\frac{(W-v)}{\sqrt{2v}} \to N(0,1)\) as \(v \to \infty\).

Step by step solution

01

Define the Chi-Square Distribution

The random variable \( W \sim \chi_{v}^{2} \) indicates that it follows a chi-square distribution with \( v \) degrees of freedom. This is a special case of the gamma distribution with shape parameter \( k = \frac{v}{2} \) and scale parameter \( \theta = 2 \).
02

Determine the Expectation

For a chi-square distribution with \( v \) degrees of freedom, the expected value (mean) is given by \( \mathrm{E}(W) = v \). This follows from its relationship with the gamma distribution, where the mean is \( k\theta = \frac{v}{2} \times 2 = v \).
03

Determine the Variance

The variance of a chi-square distribution with \( v \) degrees of freedom is \( \operatorname{var}(W) = 2v \). This is derived from the fact that the variance of the gamma distribution is \( k\theta^{2} = \frac{v}{2} \times 2^{2} = 2v \).
04

Show the Distribution Converges to Normal

To demonstrate that \((W-v) / \sqrt{2v} \stackrel{D}{\longrightarrow} N(0, 1)\) as \( v \rightarrow \infty \), we apply the central limit theorem. Given the mean and variance established, by the theorem: \[ \frac{W-v}{\sqrt{2v}} \to N(0,1) \text{ as } v \to \infty. \]

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Key Concepts

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

Expectation in Statistics
Expectation, often referred to as the expected value, is a fundamental concept in statistics that provides a measure of the center of a distribution. Essentially, it can be viewed as the average or mean of a random variable if we were to repeat an experiment numerous times and calculate the average outcome. For a chi-square distribution with \( v \) degrees of freedom, the expectation is represented as \( \mathrm{E}(W) = v \). This means that if we draw many samples from a chi-square distribution with \( v \) degrees of freedom, the average of these samples will tend to be close to \( v \).

Consider a simpler analogy: think of expectation as the balance point of a seesaw where the weights are the probabilities of outcomes. In the context of the chi-square distribution, this balance point is equivalent to its degrees of freedom \( v \). This ties back to the gamma distribution, which underlies the chi-square distribution, providing it a solid mathematical grounding. Understanding expectation helps us make predictions and perform statistical analyses efficiently.
Variance in Statistics
Variance in statistics measures the spread or dispersion of a set of random variable outcomes. Simply put, it quantifies how much the values of a random variable differ from the expected value (mean). For the chi-square distribution, variance is essential because it provides insight into the precision of the distribution.

For a chi-square distribution with \( v \) degrees of freedom, the variance is given by \( \operatorname{var}(W) = 2v \). This means that the outcomes can deviate by a range determined by \( 2v \) from its mean, \( v \).

Think of variance like the extent to which students' test scores differ in a classroom. If all students have similar scores, variance is low. If there’s a wide range of scores, variance is high. High variance indicates that the chi-square values are spread out over a wider range, while a low variance would suggest values are tightly clustered around the mean.
Central Limit Theorem
The Central Limit Theorem (CLT) is a cornerstone concept in statistics, describing how the sum (or average) of a large number of independent random variables, regardless of their distribution, will tend to follow a normal distribution. This is incredibly useful because it means we can apply normal distribution properties to a wide range of real-world situations.

In the context of the chi-square distribution with \( v \) degrees of freedom, CLT is applied to show that as \( v \) becomes very large, the distribution of \( \frac{W-v}{\sqrt{2v}} \) approaches the standard normal distribution \( N(0, 1) \).

This convergence to a normal distribution is critical for performing hypothesis tests and constructing confidence intervals in statistics when sample sizes are large. It assures us that, regardless of the original distribution's shape, the average or sum of a large number of observations will still be predictably normal.

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Most popular questions from this chapter

If \(R\) is binomial with denominator \(m\) and probability \(\pi\), show that $$ \frac{R / m-\pi}{\\{\pi(1-\pi) / m\\}^{1 / 2}} \stackrel{D}{\longrightarrow} Z \sim N(0,1) $$ and that the limits of a \((1-2 \alpha)\) confidence interval for \(\pi\) are the solutions to $$ R^{2}-\left(2 m R+m z_{\alpha}^{2}\right) \pi+m\left(m+z_{\alpha}^{2}\right) \pi^{2}=0 $$ Give expressions for them. In a sample with \(m=100\) and 20 positive responses, the \(0.95\) confidence interval is \((0.13,0.29)\). As this interval either does or does not contain the true \(\pi\), what is the meaning of the \(0.95 ?\)

Suppose that \(Y_{1}, \ldots, Y_{4}\) are independent normal variables, each with variance \(\sigma^{2}\), but with means \(\mu+\alpha+\beta+\gamma, \mu+\alpha-\beta-\gamma, \mu-\alpha+\beta-\gamma, \mu-\alpha-\beta+\gamma\). Let \(Z^{\mathrm{T}}=\frac{1}{4}\left(Y_{1}+Y_{2}+Y_{3}+Y_{4}, Y_{1}+Y_{2}-Y_{3}-Y_{4}, Y_{1}-Y_{2}+Y_{3}-Y_{4}, Y_{1}-Y_{2}-Y_{3}+Y_{4}\right)\) Calculate the mean vector and covariance matrix of \(Z\), and give the joint distribution of \(Z_{1}\) and \(V=Z_{2}^{2}+Z_{3}^{2}+Z_{4}^{2}\) when \(\alpha=\beta=\gamma=0\). What is then the distribution of \(Z_{1} /(V / 3)^{1 / 2} ?\)

Show how to use inversion to generate Bernoulli random variables. If \(0<\pi<1\), what distribution has \(\sum_{j=1}^{m} I\left(U_{j} \leq \pi\right) ?\)

Conditional on \(M=m, Y_{1}, \ldots, Y_{n}\) is a random sample from the \(N\left(m, \sigma^{2}\right)\) distribution. Find the unconditional joint distribution of \(Y_{1}, \ldots, Y_{n}\) when \(M\) has the \(N\left(\mu, \tau^{2}\right)\) distribution. Use induction to show that the covariance matrix \(\Omega\) has determinant \(\sigma^{2 n-2}\left(\sigma^{2}+n \tau^{2}\right)\), and show that \(\Omega^{-1}\) has diagonal elements \(\left\\{\sigma^{2}+(n-1) \tau^{2}\right) /\left\\{\sigma^{2}\left(\sigma^{2}+n \tau^{2}\right)\right\\}\) and offdiagonal elements \(-\tau^{2} /\left\\{\sigma^{2}\left(\sigma^{2}+n \tau^{2}\right)\right\\}\)

Let \(f(t)\) denote the probability density function of \(T \sim t_{v}\). (a) Use \(f(t)\) to check that \(\mathrm{E}(T)=0, \operatorname{var}(T)=v /(v-2)\), provided \(v>1,2\) respectively. (b) By considering \(\log f(t)\), show that as \(v \rightarrow \infty, f(t) \rightarrow \phi(t)\).

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