The characteristic function

In summary: Yes, I just want to find the CDF of the sum of the random variables, and yes it will be the product of the characteristic functions of the individual random variables because they are independent.
  • #1
EngWiPy
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Suppose I have a random variable whose moments are not defined, can I still use the characteristic function to find the CDF of that random variable?
 
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  • #2
EngWiPy said:
Suppose I have a random variable whose moments are not defined,

Distinguish between "not defined" and "not known".

Do you have a random variable known to be from a family of distributions (e.g. Cauchy) whose moments do not exist?

-or do you have a random variable whose moments exist but are not known numerical values?
 
  • #3
Stephen Tashi said:
Distinguish between "not defined" and "not known".

Do you have a random variable known to be from a family of distributions (e.g. Cauchy) whose moments do not exist?

-or do you have a random variable whose moments exist but are not known numerical values?

The random variable in question has a distribution is similar to Cauchy distribution, but not exactly the same. Its CDF and PDF are given by

[tex]F_X(x)=1-\frac{1}{1+x}\\f_X(x)=\frac{1}{(1+x)^2}[/tex]

respectively, for ##0\leq x<\infty## I searched the table of integral for the integration

[tex]\int_0^{\infty}\frac{x}{(1+x)^2}\,dx[/tex]

but the conditions to evaluate the integral are not met in my case.
 
  • #4
Since you know the CDF, I don't understand your question in post #1 about "finding" the CDF.
 
  • #5
Stephen Tashi said:
Since you know the CDF, I don't understand your question in post #1 about "finding" the CDF.

Right, I need to find the CDF of the summation of such random variables

[tex]Y=\sum_{k=1}^KX_k[/tex]

where ##\{X_k\}## are i.i.d. random variables with CDF and PDF as given previously.
 
  • #6
EngWiPy said:
can I still use the characteristic function to find the CDF of that random variable?

So what you mean to ask is "If I have a random variable X whose moments do not exist, can I use its characterisic function to find the CDF of Y = X1 + X2 + ...XN where each Xi is an independent realization of X?"

The characteristic function of X exists even if its moments do not. Y has a characteristic function that is the product of N copies of the characteristic function of X. Whether these facts suggest a practical way to find the CDF of Y in your specific problem, I don't know.
 
  • #7
Stephen Tashi said:
So what you mean to ask is "If I have a random variable X whose moments do not exist, can I use its characterisic function to find the CDF of Y = X1 + X2 + ...XN where each Xi is an independent realization of X?"

The characteristic function of X exists even if its moments do not. Y has a characteristic function that is the product of N copies of the characteristic function of X. Whether these facts suggest a practical way to find the CDF of Y in your specific problem, I don't know.

Yes, I just want to find the CDF of the sum of the random variables, and yes it will be the product of the characteristic functions of the individual random variables because they are independent. I wanted to know if this is a valid way to find the CDF, given that the individual random variables' moments don't exist.
 

1. What is the characteristic function?

The characteristic function is a mathematical function that is used to describe the probability distribution of a random variable. It is defined as the Fourier transform of the probability density function.

2. How is the characteristic function related to the moment generating function?

The characteristic function and the moment generating function are closely related, as both are used to describe the probability distribution of a random variable. However, the characteristic function is the Fourier transform of the probability density function, while the moment generating function is the Laplace transform. This means that while the moment generating function only exists for certain types of distributions, the characteristic function exists for all distributions.

3. What is the significance of the characteristic function in statistics?

The characteristic function is an important tool in statistics, as it allows for the calculation of various statistical properties of a random variable, such as the mean, variance, and higher moments. It also allows for the derivation of various probability distributions, making it a valuable tool for analyzing data and making statistical inferences.

4. Can the characteristic function be used for non-random variables?

No, the characteristic function is only applicable to random variables. It cannot be used for non-random variables, as it relies on the concept of probability and the probability distribution of a random variable.

5. How is the characteristic function used in hypothesis testing?

The characteristic function is used in hypothesis testing by allowing for the calculation of test statistics, such as the likelihood ratio test. These test statistics are then compared to critical values to determine the probability of obtaining the observed data under the null hypothesis, and thus make a decision on whether to reject or accept the null hypothesis.

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