具有相同矩的不同概率分布

hbghlyj

不同概率分布存在具有可能相同的矩。换言之，知道所有矩并不总是唯一确定概率分布。
Steven J. Miller - The Probability Lifesaver-Princeton University Press (2017) pp.518
Example 19.6.6: The standard examples given are the following two densities, defined for $x \geq 0$ by
\[\tag{19.2}
\begin{aligned}
f_1(x) &=\frac{1}{\sqrt{2 \pi x^2}} e^{-\left(\log ^2 x\right) / 2} \\
f_2(x) &=f_1(x)[1+\sin (2 \pi \log x)]
\end{aligned}
\]
It's a nice calculation to show that these two densities have the same moments; they're clearly different (see Figure 19.1).
What went wrong? It should seem absurd that two probability distributions could have the same moments without being the same. This example above isn’t a mere annoying curiousity to be forgotten, but rather a warning as to how difficult and technical the subject really is. This example isn’t an isolated problem, but rather indicative as to how strange and non-intuitive real valued functions can be.
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In the online supplemental chapter, “Complex Analysis and the Central Limit Theorem,” we explore what goes wrong with the functions defined in Equation (19.2). After seeing what the problem is, we discuss what additional properties we need to assume to prevent such an occurrence. The solution involves results from complex analysis, which will tell us when a moment generating function (of a continuous random variable) uniquely determines a probability distribution.

hbghlyj

Derive the PDF of the log-normal distribution?
Log-normal distribution \begin{multline*}\mu _k'=\int_0^\infty x^k\cdot\frac{1}{\sqrt{2 \pi x^2}} e^{-\left(\log ^2 x\right) / 2} dx\xlongequal{x=e^y}\int_0^\infty e^{ky}\cdot\frac{1}{\sqrt{2 \pi}} e^{-y^2/ 2} dy\\=\int_0^\infty\frac{1}{\sqrt{2 \pi}} e^{-y^2/ 2+ky} dx=e^{k^2/2}\int_0^\infty\frac{1}{\sqrt{2 \pi}} e^{-(y-k)^2/ 2} dx= e^{k^2/2}\end{multline*}这意味着矩生成函数是\[ M_X(t) \ = \ \sum _{k=0}^\infty \frac {\mu _k' t^k}{k!} \ = \ \sum _{k=0}^\infty \frac {e^{k^2/2} t^k}{k!}. \] 这个级数对什么 \(t\) 收敛？令人惊讶的是，这个级数仅在 \(t=0\) 时收敛！
要看到这一点，只需证明通项不趋于零。
因为 \(k!\le k^k\)，有 \(t^k/k!\ge (t/k)^k\)；而 \(e^{k^2/2} = (e^{k/2})^k\)，所以第 \(k\) 项 \(\ge(e^{k/2} t / k)^k\)。
对于任何 \(t \neq 0\)，这显然不会趋于零，因此 $f_1$ 的矩生成函数的收敛半径为零。