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91Ó°ÊÓ

If \(Y\) has an exponential distribution with mean \(\theta\), show that \(U=2 Y / \theta\) has a \(\chi^{2}\) distribution with 2 df.

Short Answer

Expert verified
\(U\) follows a \(\chi^2(2)\) distribution.

Step by step solution

01

Understanding the Problem

We are given a random variable \(Y\) which is exponentially distributed with mean \(\theta\). We need to show that \(U = \frac{2Y}{\theta}\) has a \(\chi^2\) distribution with 2 degrees of freedom.
02

Convert Exponential to Chi-Square Distribution

Note that if \(Y\) is an exponential distribution with mean \(\theta\), then the rate parameter is \(\lambda = \frac{1}{\theta}\). The transformation \(U = \frac{2Y}{\theta}\) can be seen as scaling \(Y\) by \(\frac{2}{\theta}\).
03

Determine the Equivalent Distribution

Using the transformation \(Y = \frac{U \theta}{2}\), identify that if \(Y\) follows an exponential distribution with the rate parameter \(\lambda\), that is \( \text{Exponential}(1/\theta)\), then \(U\) must be related to a \(\chi^2\) distribution. Recall that \(2\times \text{Exponential}(\lambda)\) is \(\chi^2\) with 2 df.
04

Confirm the Chi-Square Distribution

The transformation \(U = \frac{2Y}{\theta}\) means \(U \sim \chi^2(2)\), because scaling an exponential random variable with rate \(\lambda = 1/\theta\) by 2 results in a \(\chi^2\) distribution with 2 degrees of freedom. Therefore, \(U\) can be identified as the chi-square random variable.

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Key Concepts

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

Chi-Square Distribution
The Chi-Square distribution is a vital concept in statistics, especially when dealing with hypothesis testing and variance estimation. This distribution is defined only for non-negative random variables and is often used when we need to analyze the variation within a dataset. The Chi-Square distribution with \( k \) degrees of freedom is the distribution of a sum of the squares of \( k \) independent standard normal random variables. To further simplify:
  • It is skewed to the right, but the skewness diminishes with increasing degrees of freedom.
  • The mean of a chi-square distribution \( \chi^2(k) \) is equal to its degrees of freedom, \( k \).
  • The mode of a chi-square distribution with \( k \) degrees of freedom is \( k - 2 \) (when \( k > 2 \)).
In practical usage, it's essential to know that chi-square tests (like the Chi-Square goodness-of-fit test or Chi-Square test for independence) rely on this powerful distribution. In the context of the problem, when an exponential distribution is scaled accordingly, it transforms into a chi-square distribution with specific degrees of freedom.
Transformation of Variables
Transformation of variables is a critical technique in statistics, enabling us to convert a non-standard distribution into a more familiar one for easier analysis. The transformation typically involves applying a mathematical function to a variable to alter its distribution while preserving its statistical properties. In our specific problem:
  • We are transforming an exponentially distributed variable \( Y \) into another variable \( U = \frac{2Y}{\theta} \).
  • This transformation entails scaling the variable \( Y \) by the factor \( \frac{2}{\theta} \).
Through this process, otherwise complex distributions can often present a form that aligns with known distributions, allowing statisticians to leverage existing theories and tools for analysis. In this exercise, the transformation aligns with the properties of the chi-square distribution, aiding in confirming that the transformed variable fits a well-understood distribution.
Degrees of Freedom
Degrees of freedom, commonly abbreviated as df, represent the number of independent values that can vary in a statistical calculation. In simple terms, they are the number of independent observations that are free to vary while estimating some statistical parameters. This concept affects the shape of many critical probability distributions, including the chi-square distribution. For the Chi-Square distribution:
  • The degrees of freedom determine how the distribution looks. More degrees of freedom indicate a more symmetric and less skewed distribution.
  • In our problem, the transformation yields a chi-square distribution with 2 degrees of freedom, implying that the distribution involves two independent components.
Degrees of freedom are crucial when conducting various types of statistical tests; they help define the specific chi-square distribution needed for a particular analysis. Understanding this concept allows statisticians to correctly determine the distribution parameters necessary for accurate hypothesis testing.

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Most popular questions from this chapter

a. Find \(t_{05}\) for a \(t\) -distributed random variable with 5 df. b. Refer to part (a). What is \(P\left(T^{2}>t_{.05}^{2}\right) ?\) c. Find \(F_{10}\) for an \(F\) -distributed random variable with 1 numerator degree of freedom and 5 denominator degrees of freedom. d. Compare the value of \(F_{10}\) found in part \((c)\) with the value of \(t_{.05}^{2}\) from parts \((a)\) and \((b)\). e. In Exercise 7.33 , you will show that if \(T\) has a \(t\) distribution with \(\nu\) df, then \(U=T^{2}\) has an \(F\) distribution with 1 numerator degree of freedom and \(\nu\) denominator degrees of freedom. How does this explain the relationship between the values of \(F_{.10}\) (1 num. df, 5 denom df) and \(t_{05}^{2}(5 \mathrm{df})\) that you observed in part \((d)?\)

The downtime per day for a computing facility has mean 4 hours and standard deviation. .8 hour. a. Suppose that we want to compute probabilities about the average daily downtime for a period of 30 days.i. What assumptions must be true to use the result of Theorem 7.4 to obtain a valid approximation for probabilities about the average daily downtime? i. Under the assumptions described in part (i), what is the approximate probability that the average daily downtime for a period of 30 days is between 1 and 5 hours? b. Under the assumptions described in part (a), what is the approximate probability that the total downtime for a period of 30 days is less than 115 hours?

If \(Y\) is a random variable that has an \(F\) distribution with \(\nu_{1}\) numerator and \(\nu_{2}\) denominator degrees of freedom, show that \(U=1 / Y\) has an \(F\) distribution with \(\nu_{2}\) numerator and \(\nu_{1}\) denominator degrees of freedom.

Let \(Y_{1}, Y_{2}, \ldots, Y_{5}\) be a random sample of size 5 from a normal population with mean 0 and variance 1 and let \(\bar{Y}=(1 / 5) \sum_{i=1}^{5} Y_{i} .\) Let \(Y_{6}\) be another independent observation from the same population. What is the distribution of a. \(W=\sum_{i=1}^{5} Y_{i}^{2} ?\) Why? b. \(U=\sum_{i=1}^{5}\left(Y_{i}-\bar{Y}\right)^{2} ?\) Why? c. \(\sum_{i=1}^{5}\left(Y_{i}-\bar{Y}\right)^{2}+Y_{6}^{2} ?\) Why?

a. If \(U\) has a \(\chi^{2}\) distribution with \(\nu\) df, find \(E(U)\) and \(V(U)\). b. Using the results of Theorem 7.3 , find \(E\left(S^{2}\right)\) and \(V\left(S^{2}\right)\) when \(Y_{1}, Y_{2}, \ldots, Y_{n}\) is a random sample from a normal distribution with mean \(\mu\) and variance \(\sigma^{2}\).

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