0

I'm reading Casella and Berger's book. In chapter 5, they show that we can recover the distribution of the sample mean by using the following relationship between the MGF of the underlying random variable and the MGF of the sample mean distribution:

$$M_{\overline{X}}(t)=[M_{X}(t/n)]^{n}$$

However, this will not work when the random variable $X$ don't have a MGF. Reading the book, and other contents in the internet, I found out that Cauchy distribution has no MGF and, therefore, this procedure will fail in this case.

Why Cauchy distribution has no MGF? Are there others distributions with this property?

P.S: I read this question in my previous research. It's a very good question and now I understand why Cauchy distribution has no mean. I believe my question is not duplicate since the MGF is a more general concept than the mean.

Lucas
  • 63
  • 6
  • 1
    Sure: but since the Cauchy has no mean, the Taylor series of its MGF cannot possibly represent the MGF. https://stats.stackexchange.com/questions/389846, https://stats.stackexchange.com/questions/302044, https://stats.stackexchange.com/questions/106286, and https://stats.stackexchange.com/questions/32706 cover your question well. – whuber Apr 25 '21 at 19:03

0 Answers0