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Chapter 1: Problem 13

If \(E|X|^{k}<\infty\) then for \(0

Short Answer

Expert verified
Firstly, the finiteness condition \(E|X|^{k}<\infty\) allows us to conclude that \(E|X|^{j}<\infty\) using the inequality \(|x|^j \leq |x|^k\). Then Jensen's Inequality, applied to the function \(-t^{j/k}\), gives us the required inequality \(E|X|^{j} \leq\left(E|X|^{k}\right)^{j / k}\).

Step by step solution

01

- Proving finiteness

Firstly, from the inequality \(|x|^j \leq |x|^k\) for all \(x\), it follows directly that if \(E|X|^{k}<\infty\), then \(E|X|^{j}<\infty\).
02

- Establishing inequality

The next part involves establishing the inequality. We will use Jensen's Inequality here since it deals with the expectation of a function of a random variable. Jensen's Inequality states that for a convex function \( \phi \), we have \(E[\phi(X)] \geq \phi(E[X])\). Now consider the function \( \phi(t) = t^{j/k}\), which is concave if \(0<j<k\). This means the -phi(t) function is convex, so we can use Jensen's Inequality here and find that \(-E[\phi(X)] \geq -\phi(E[X])\). This can be rewritten as \(E[X^{j/k}] \leq (E[X])^{j/k}\). Taking both sides to the power of \(k\), we get \(E|X|^{j} \leq\left(E|X|^{k}\right)^{j / k}\).
03

- Final Conclusion

We have thus shown both that if \(E|X|^{k}<\infty\), then \(E|X|^{j}<\infty\), and also that \(E|X|^{j} \leq\left(E|X|^{k}\right)^{j / k}\). This concludes the proof.

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Key Concepts

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

Jensen's Inequality
Jensen's Inequality is an important principle in mathematical expectation. It serves as a foundation for evaluating the expected value of a convex function when applied to a random variable.
  • For any convex function, the function's output of the expected input is less than or equal to the expected output of the function applied directly to the input.
  • Mathematically, if \( \phi \) is a convex function and \( X \) is a random variable, Jensen's Inequality states: \( E[\phi(X)] \geq \phi(E[X]) \).
In the context of our exercise, considering \( \phi(t) = t^{j/k} \), this particular function is concave for \(0
Convex and Concave Functions
Understanding the nature of convex and concave functions is key to applying principles like Jensen's Inequality. Convex Functions
- A function \( \phi \) is considered convex on an interval if the line segment between any two points on the graph of the function lies above or on the graph.
  • Mathematically, \( \phi((1 - \lambda)x + \lambda y) \leq (1 - \lambda)\phi(x) + \lambda \phi(y)\) for all \(x, y\) in the interval and \(0 \leq \lambda \leq 1\).
Concave Functions
- A function is concave if the line segment between any two points on the graph of the function lies below or on the graph. Concave functions are the opposite of convex functions.- Mathematically, for a concave function \(f\), the inequality becomes \( f((1-\lambda)x + \lambda y) \geq (1-\lambda)f(x) + \lambda f(y) \).In our exercise, the function \(t^{j/k}\) is concave for \(0
Random Variables
Random variables are a fundamental concept in probability and statistics. They are used to quantify outcomes of random phenomena, turning the abstract into something mathematical that can be manipulated. - A random variable \( X \) is a function that assigns a numerical value to each outcome in a sample space.- We deal with two main types: discrete and continuous.
  • Discrete random variables have countable outcomes, like the number of heads in tosses of a coin.
  • Continuous random variables have uncountable outcomes, often measured, such as heights or weights.
In our exercise, \(X\) represents a random variable with a certain property such as being finite under expectations of different powers. Studying such scenarios aids in understanding the behavior of random variables and functions applied to them under varying operations and powers.

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Most popular questions from this chapter

Inclusion-exclusion formula. Let \(A_{1}, A_{2}, \ldots A_{n}\) be events and \(A=\) \(U_{i=1}^{n} A_{i} .\) Prove that \(1_{A}=1-\prod_{i=1}^{n}\left(1-1_{A_{i}}\right) .\) Expand out the right hand side, then take expected value to conchude $$ \begin{aligned} P\left(\cup_{i=1}^{n} A_{i}\right)=& \sum_{i=1}^{n} P\left(A_{i}\right)-\sum_{i

Suppose \(X\) has continuous density \(f, P(\alpha \leq X \leq \beta)=1\) and \(g\) is a function that is strictly increasing and differentiable on \((\alpha, \beta)\). Then \(g(X)\) has density \(f\left(g^{-1}(x)\right) / g^{\prime}\left(g^{-1}(x)\right)\) for \(x \in(g(\alpha), g(\beta)\) and 0 otherwise. When \(g(x)=a x+b\) with \(a>0\), the answer is \(f((y-b) / a)\).

Let \(\psi(x)=x^{2}\) when \(|x| \leq 1\) and \(=|x|\) when \(|x| \geq 1\). Show that if \(X_{1}, X_{2}, \ldots\) are independent with \(E X_{n}=0\) and \(\sum_{n=1}^{\infty} E \psi\left(X_{n}\right)<\infty\) then \(\sum_{n=1}^{\infty} X_{n}\) converges a.s.

Let \(\Omega=\mathbf{R}, \mathcal{F}=\) all subsets so that \(A\) or \(A^{c}\) is countable, \(P(A)=0\) in the first case and \(=1\) in the second. Show that \((\Omega, \mathcal{F}, P)\) is a probability space.

Suppose \(E X_{i}=0\) and \(E \exp \left(\theta X_{i}\right)=\infty\) for all \(\theta>0\). Then $$ \frac{1}{n} \log P\left(S_{n} \geq n a\right) \rightarrow 0 \text { for all } a>0 $$
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