Office_Shredder said:I can't say much about the first part of your post, but for there to be a Taylor series expansion the function has to be differentiable at the origin. It's not, just draw a graph and take a look at it.
To formally prove it, just calculate what the derivate is to the left of 0, and the right of 0, and observe they are not equal.
I have looked at the plot of the function. What would that do to help? As there is a sharp turn at ##t=0##, surely that further enhances my conclusion?Office_Shredder said:Have you tried drawing a plot of the function yet? There's a sharp corner at t=0. On the left of 0 it has one slope, on the right of 0 it has another slope. Similar to how the graph of |t| has a slope of -1 at 0 on the left, and 1 at 0 on the right.
A characteristic function is a mathematical function that describes the probability distribution of a random variable. It is the Fourier transform of the probability density function (PDF) and provides information about the moments of the distribution.
To find the moments using the characteristic function, you can use the inverse Fourier transform. This will give you the PDF, from which you can calculate the moments using standard formulas.
The characteristic function allows for a more efficient and accurate way to find moments compared to traditional methods. It also works for both discrete and continuous distributions, making it a versatile tool for statistical analysis.
One limitation is that the characteristic function may not exist for all distributions. In addition, the inverse Fourier transform can be complex and require advanced mathematical skills. It is also important to have a good understanding of the properties of characteristic functions to accurately interpret the results.
Yes, the characteristic function can be used to find any order of moments. By taking the derivative of the characteristic function, you can find the moments of any order. This makes it a powerful tool for analyzing the properties of a distribution.