# Checking regular variance around 0, hypergeometric fucntion

• jjhyun90
In summary, the conversation discusses the definition of a function f that is 2\alpha-regularly varying around zero and a method for showing this property. The conversation also mentions using the properties of hypergeometric series and considering the limit of the ratio of two approximations to prove that f is 2\alpha-regularly varying around zero.
jjhyun90

## Homework Statement

A function g is $\alpha$-regularly varying around zero if for all $\lambda > 0$, $\lim_{x\to 0} \frac{g(\lambda x)}{g(x)}=\lambda^{\alpha}$
For real s and $\alpha \in (0,1)$, define f:
$f(s)=1-\alpha \int_{0}^{\infty} e^{\alpha t} \frac{\frac{1}{1+s^2}}{e^t(1-\frac{1}{1+s^2})+\frac{1}{1+s^2}} dt = 1 + \alpha \sum_{n=1}^{\infty} \frac{(-s^{-2})^{n}}{n+\alpha}$.
Show f is $2\alpha$-regularly varying around zero.

## Homework Equations

Note $\frac{1}{1+s^2}$ is a characteristic function of Laplace distribution.

## The Attempt at a Solution

I am not familiar with hypergeometric series. For example, I do not know how to show that the series converges to $-\frac{1}{\alpha}$ when s goes to 0. Are there any properties of hypergeometric series that might be useful for proving this? Attempt to directly compute the limit failed.

Last edited:
so, i may be reading it wrong but can you make the following simplifications
$$f(s) = 1-\alpha \int_0^{\infty} e^{\alpha t} \frac{\frac{1}{1+s^2}} {e^t(1- \frac{1}{1+s^2}) + \frac{1}{1+s^2}}dt$$

$$= 1-\alpha \int_0^{\infty} e^{(\alpha -1)t} \frac{1} {(1+s^2- 1) + 1}dt$$

$$= 1-\alpha \int_0^{\infty} \frac{e^{(\alpha -1)t}} {1+s^2}dt$$

Last edited:
then i would be thinking consider expanding in the function in either integral or summation for around s=0 in terms of s

substitute $s = s\lambda$ and look at taking the limit or something along those lines...

Thank you for the reply, but the $e^t$ on the denominator is multiplied to the first term only, so the simplification doesn't quite work.

ok so that points you in the direction of using the sum, which is looking easier anyway
$$f(s) = 1 + \alpha \Sum_{n=1}^{infty} \frac{(-s^{-2})^n}{n+\alpha} = 1 + \alpha \Sum_{n=1}^{infty} \frac{(-1)^n s^{-2n}}{n+\alpha}$$

now when s is small you can approximate f(s) by
$$f(s) \approx f(0) + \frac{df(0)}{ds}s$$

try and come up with a similar approximation for $f(s\lambda)$ then consider the limit of the ratio

Last edited:

## 1. What is regular variance around 0?

Regular variance around 0 refers to the measure of how much a set of data points deviate from the mean value of 0. It is a statistical concept used to understand the spread or variability of a dataset.

## 2. How is regular variance around 0 calculated?

To calculate regular variance around 0, the squared difference between each data point and the mean of 0 is summed and divided by the total number of data points. This value is known as the variance and is commonly denoted by σ².

## 3. What is the significance of regular variance around 0 in data analysis?

Regular variance around 0 is important in data analysis as it helps to understand the distribution of data points and identify any patterns or outliers. It is also used in various statistical tests to determine the significance of results.

## 4. What is the hypergeometric function?

The hypergeometric function is a special mathematical function that is used to describe the relationship between two variables. It is commonly used in probability theory and statistics to calculate the probability of obtaining a specific number of successes in a sample of a specific size from a population with a known number of successes and failures.

## 5. How is the hypergeometric function related to regular variance around 0?

The hypergeometric function is related to regular variance around 0 as it is used in statistical tests, such as the hypergeometric test, to calculate the probability of obtaining a certain variance around 0 or a more extreme value. This helps to determine the statistical significance of the results and make conclusions about the data.

• Calculus and Beyond Homework Help
Replies
10
Views
940
• Calculus and Beyond Homework Help
Replies
1
Views
556
• Calculus and Beyond Homework Help
Replies
14
Views
690
• Calculus and Beyond Homework Help
Replies
9
Views
871
• Calculus and Beyond Homework Help
Replies
2
Views
985
• Calculus and Beyond Homework Help
Replies
4
Views
1K
• Calculus and Beyond Homework Help
Replies
1
Views
893
• Calculus and Beyond Homework Help
Replies
3
Views
675
• Calculus and Beyond Homework Help
Replies
18
Views
2K
• Calculus and Beyond Homework Help
Replies
1
Views
428