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Chapter 9: Problem 13

Let \(X\) be Cauchy with density \(\frac{1}{\pi\left(1+(x-\alpha)^{2}\right)}\). Show that \(\sigma^{2}(X)\) is not defined, and \(E\left\\{X^{2}\right\\}=\infty\).

Short Answer

Expert verified
Variance is not defined; \( E[X^2] = \infty \).

Step by step solution

01

Understanding the Cauchy Distribution

A Cauchy distribution is characterized by its density function \( f(x) = \frac{1}{\pi(1+(x-\alpha)^2)} \). Unlike the normal distribution, it does not have a defined mean or variance. Our task is to prove that the variance is not defined and the expected value of \( X^2 \) is infinite.
02

Attempt to Calculate Variance

The variance of a random variable \( X \) is given by \( \sigma^2(X) = E[X^2] - (E[X])^2 \). Since Cauchy distribution does not have a defined expected value \( E[X] \), the variance \( \sigma^2(X) \) is intrinsically undefined as it relies on \( (E[X])^2 \).
03

Evaluating \( E[X^2] \) for the Cauchy Distribution

The expected value \( E[X^2] \) is given by the integral \( \int_{-\infty}^{\infty} x^2 f(x) \ dx \). For the Cauchy distribution, this integral diverges. Explicitly, \( \int_{-\infty}^{\infty} \frac{x^2}{\pi(1+(x-\alpha)^2)} \ dx ot= \infty \), which indicates \( E[X^2] = \infty \).
04

Conclusion

Since the mean \( E[X] \) does not exist and \( E[X^2] = \infty \), the variance \( \sigma^2(X) = E[X^2] - (E[X])^2 \) is undefined. Thus, \( E\{X^2\} \) being infinite confirms the undefined nature of variance.

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Key Concepts

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

Variance
Variance is a measure of how much the values of a random variable deviate from the mean, providing insights into the spread of a distribution. For most distributions, variance is given by the formula:\[ \sigma^2(X) = E[X^2] - (E[X])^2 \]However, Cauchy distributions defy this conventional understanding. The Cauchy distribution's variance is undefined. But, why?
  • The variance formula requires a defined expected value \( E[X] \).
  • For Cauchy, \( E[X] \) is not defined due to the nature of its probability density function, which does not result in a converging integral.
Because the expected value does not exist, calculating \( \sigma^2(X) \) becomes impossible, resulting in the statement that variance for a Cauchy distribution is undefined.
Expected Value
Expected value plays a crucial role in probability and statistics by informing us of the average or mean value a random variable takes. Normally, it's calculated as:\[ E[X] = \int_{-\infty}^{\infty} x f(x) \ dx \]For the standard Cauchy distribution, attempting to calculate this integral results in divergence. This means through mathematical evaluation; the integral does not yield a finite value.
  • The probability density function (PDF) of the Cauchy distribution causes the expected value calculation to "blow up"; hence it doesn't exist as a finite number.
  • This breakdown in computation leads to the realization that we cannot determine a "mean" for this distribution.
This is a unique property of the Cauchy distribution that distinguishes it from many other probability distributions.
Probability Density Function
A probability density function (PDF) describes the likelihood of a random variable to take on a certain value. For the Cauchy distribution, the PDF is:\[ f(x) = \frac{1}{\pi(1+(x-\alpha)^2)} \]This specific function leads to some interesting and peculiar characteristics:
  • It's centered around \( \alpha \), which is often confused as a "mean", but isn't since the mean is undefined.
  • The shape of the Cauchy PDF is characterized by a "peak" at \( x = \alpha \) and "heavy tails", which implies that even values far from the center can have substantial probabilities.
  • The non-converging nature of the integrals that represent statistical moments such as mean or variance results from the tail behavior of the PDF.
Understanding the Cauchy PDF gives deeper insight into why standard measures, like expected value and variance, do not conventionally apply to this distribution.

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Most popular questions from this chapter

Let \(X\) be an exponential r.v. . Show that \(P\\{X>s+t \mid X>s\\}=\) \(P\\{X>t\\}\) for \(s>0, t>0\). This is known as the "memoryless property" of the exponential.

Let \(X\) be a r.v, with the property that \(P\\{X>s+t \mid X>s\\}=P\\{X>\) \(t\\}\). Show that if \(h(t)=P\\{X>t\\}\), then \(h\) satisfies Cauchy's equation: $$ h(s+t)=h(s) h(t) \quad(s>0, t>0) $$ and show that \(X\) is exponentially distributed (Hint: use the fact that \(h\) is continuous from the right, so Cauchy's equation can be solved).

Let \(X\) be normal (or Gaussian) with parameters \(\mu=0\) and \(\sigma^{2}=1\). Show that \(P(X>x) \leq \frac{1}{x \sqrt{2 \pi}} e^{-\frac{1}{2} x^{2}}\), for \(x>0\).

Let \(\alpha\) be an integer and suppose \(X\) has distribution Gamma \((\alpha, \beta)\). Show that \(P(X \leq x)=P(Y \geq \alpha)\), where \(Y\) is Poisson with parameter \(\lambda=x \beta .\) (Hint: Recall \(\Gamma(\alpha)=(\alpha-1) !\) and write down \(P(X \leq x)\), and then use integration by parts with \(u=t^{\alpha-1}\) and \(d v=e^{-t / \beta} d t\).)

Let \(X:(\Omega, \mathcal{A}) \rightarrow(\mathbf{R}, \mathcal{B})\) be a r.v. Let $$ \mathcal{F}=\left\\{A: A=X^{-1}(B), \text { some } B \in \mathcal{B}\right\\}=X^{-1}(\mathcal{B}) $$ Show that \(X\) is measurable as a function from \((\Omega, \mathcal{F})\) to \((\mathbf{R}, \mathcal{B})\).
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