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Chapter 7: Problem 24

Let \(X\) be a standard normal random variable. Find the m.g.f. and density of \(X^{2}\).

Short Answer

Expert verified
The m.g.f. of \(X^2\) is \((1 - 2t)^{-1/2}\); its pdf is \(f(x) = \frac{1}{\sqrt{2\pi x}} e^{-x/2}\).

Step by step solution

01

Understand the Problem

We need to find the moment generating function (m.g.f.) and the probability density function (pdf) of the random variable \(X^2\), where \(X\) is a standard normal random variable with mean 0 and variance 1.
02

Identify the Distribution of \(X^2\)

Since \(X\) is a standard normal random variable, \(X^2\) follows a Chi-squared distribution with 1 degree of freedom, denoted as \(\chi^2_1\).
03

Find the Moment Generating Function of \(X^2\)

The m.g.f. of a \(\chi^2_k\) distribution is \((1 - 2t)^{-k/2}\) for \(t < 1/2\). Here \(k = 1\), so the m.g.f. of \(X^2\) is \((1 - 2t)^{-1/2}\).
04

Find the Probability Density Function of \(X^2\)

The pdf of a \(\chi^2_1\) distribution is \(f(x) = \frac{1}{\sqrt{2\pi x}} e^{-x/2}\) for \(x > 0\). This is derived from the density function of \(X\), which is normal.

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Key Concepts

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

Understanding Moment Generating Function (MGF)
A moment generating function, or m.g.f., is a crucial concept in probability theory. It is a tool used for summarizing the probability distribution of a random variable.
The m.g.f. provides a way to understand all moments (such as mean, variance, skewness, etc.) of a distribution.
Let's break it down further to make it simple.
  • The m.g.f. of a random variable X is defined as \( M_X(t) = E[e^{tX}] \), where \( E \) denotes the expected value.
  • The function is called "moment generating" because its derivatives, evaluated at \( t = 0 \), give us the moments of the distribution.
  • For a positive integer \( n \), the \( n^{th} \) moment is found by taking the \( n^{th} \) derivative of the m.g.f., then evaluating it at \( t = 0 \).
In the case of the Chi-squared distribution with 1 degree of freedom (since \( X^2 \) when \( X \) is standard normal is \( \chi^2_1 \)), the m.g.f. is given by \( (1 - 2t)^{-1/2} \). This formula shows us how the distribution of \( X^2 \) "behaves" in terms of its probability distribution, using the moment generating function.
Exploring the Probability Density Function (PDF)
The probability density function, or pdf, is another fundamental concept in probability. It defines how the probabilities are distributed over the values of a random variable.
Understanding the pdf is key in evaluating the likelihood of different outcomes.
  • The pdf gives us the probability that a continuous random variable falls within a particular range of values. It is denoted as \( f(x) \).
  • For any value \( x \), \( f(x) \) should be non-negative, and the area under the pdf curve over all possible values equals 1.
  • The pdf for a \( \chi^2_1 \) distribution is given by \( f(x) = \frac{1}{\sqrt{2\pi x}} e^{-x/2} \) for \( x > 0 \). This tells us how values of \( X^2 \) are distributed when \( X \) is a standard normal variable.
By understanding the pdf, you can determine the likelihood of different values for \( X^2 \) based on its distribution. This concludes how small or large values are expected depending on its shape.
Grasping the Standard Normal Distribution
The standard normal distribution is a special type of normal distribution, and it plays a fundamental role in statistics.
This distribution is characterized by a mean of 0 and a standard deviation of 1.
  • It is symmetric around the mean, which in this case is zero. This symmetry implies that data near the mean are more frequent in occurrence than data far from the mean.
  • The standard normal distribution uses a bell curve, which is described mathematically by the equation \( f(x) = \frac{1}{\sqrt{2\pi}} e^{-x^2/2} \).
  • Within this context, any normal random variable can be transformed into a standard normal random variable \( Z \) using the transformation \( Z = \frac{X - \mu}{\sigma} \), where \( \mu \) is the mean and \( \sigma \) is the standard deviation of the original variable.
Understanding the standard normal distribution is key for statistical analysis, as it serves as the foundation for many statistical methods and tests found in advanced statistics. This distribution helps pave the way for further concepts, like hypothesis testing and confidence intervals.

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Most popular questions from this chapter

Suppose a machine's lifetime \(T\) has hazard rate \(\lambda \sqrt{t}\), where \(\lambda>0\). Find \(\mathbf{P}(T>t)\).

Suppose that \(X\) has distribution function $$ F(x)=1-\exp \left(-\int_{0}^{x} g(u) d u\right) $$ for some function \(g(.)\). Show that this is possible if and only if \(g(u) \geq 0\) and \(\int_{0}^{\infty} g(u) d u=\infty\).

You have two independent random variables, each uniform on \((0,1)\). Explain how you would use them to obtain a random variable \(X\) with density $$ f(x)=\frac{3}{5}\left(1+x+\frac{1}{2} x^{2}\right) \quad \text { for } 0 \leq x \leq 1 . $$

What is the distribution function of the random variable having the beta density with \(a=b=\frac{1}{2}\) ?

(a) Let \(X\) have the standard normal density. Show that \(|X|\) has distribution function \(F=2 \Phi(x)-1\), for \(x>0\). (b) Let \(X\) have distribution \(F(x)\). What is the distribution of \(|X|\) ? What is the density of \(|X|\) if it exists?
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