统计代写|统计推断代写Statistical inference代考|MAST20005

2022年9月29日

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统计代写|统计推断代写Statistical inference代考|Generating functions

For many distributions, all the moments $\mathbb{E}(X), \mathbb{E}\left(X^2\right), \ldots$ can be encapsulated in a single function. This function is referred to as the moment-generating function, and it exists for many commonly used distributions. It often provides the most efficient method for calculating moments. Moment-generating functions are also useful in establishing distributional results, such as the properties of sums of random variables, and in proving asymptotic results.

Definition 3.5.1 (Moment-generating function)
The moment-generating function of a random variable $X$ is a function $M_X: \mathbb{R} \longrightarrow$ $[0, \infty)$ given by
$$M_X(t)=\mathbb{E}\left(e^{t X}\right)= \begin{cases}\sum_x e^{t x} f_X(x) & \text { if } X \text { discrete } \ \int_{-\infty}^{\infty} e^{t x} f_X(x) d x & \text { if } X \text { continuous. }\end{cases}$$
where, for the function to be well defined, we require that $M_X(t)<\infty$ for all $t \in[-h, h]$ for some $h>0$.
A few things to note about moment-generating functions.

1. Problems involving moment-generating functions almost always use the definition in terms of expectation as a starting point.
2. The moment-generating function $M_X(t)=\mathbb{E}\left(e^{t X}\right)$ is a function of $t$. The $t$ is just a label, so $M_X(s)=\mathbb{E}\left(e^{s X}\right), M_X(\theta)=\mathbb{E}\left(e^{\theta X}\right), M_Y(p)=\mathbb{E}\left(e^{p Y}\right)$, and so on.
3. We need the moment-generating function to be defined in an interval around the origin. Later on we will be taking derivatives of the moment-generating function at zero, $M_X^{\prime}(0), M_X^{\prime \prime}(0)$, and so on.

The moment-generating function of $X$ is the expected value of an exponential function of $X$. Useful properties of moment-generating functions are inherited from the exponential function, $e^x$. The Taylor series expansion around zero, provides an expression for $e^x$ as a polynomial in $x$,
$$e^x=1+x+\frac{1}{2 !} x^2+\ldots+\frac{1}{r !} x^r+\ldots=\sum_{j=0}^{\infty} \frac{1}{j !} x^j$$
This expansion (and any Taylor series expansion around zero) is often referred to as the Maclaurin series expansion. All the derivatives of $e^x$ are equal to $e^x$,
$$\frac{d^r}{d x^r} e^x=e^x \text { for } r=1,2, \ldots$$

统计代写|统计推断代写Statistical inference代考|Cumulant-generating functions and cumulants

It is often convenient to work with the log of the moment-generating function. It turns out that the coefficients of the polynomial expansion of the log of the momentgenerating function have convenient interpretations in terms of moments and central moments.
Definition 3.5.7 (Cumulant-generating function and sumulants)
The cumulant-generating function of a random variable $X$ with moment-generating function $M_X(t)$, is defined as
$$K_X(t)=\log M_X(t) .$$
The $r^{\text {th }}$ cumulant, $\kappa_r$, is the coefficient of $t^r / r !$ in the expansion of the cumulantgenerating function $K_X(t)$ so
$$K_X(t)=\kappa_1 t+\kappa_2 \frac{t^2}{2 !}+\ldots+\kappa_r \frac{t^r}{r !}+\ldots=\sum_{j=1}^{\infty} \kappa_j \frac{t^j}{j !} .$$
It is clear from this definition that the relationship between cumulant-generating function and cumulants is the same as the relationship between moment-generating function and moments. Thus, to calculate cumulants we can either compare coefficients or differentiate.

1. Calculating the $r^{\text {th }}$ cumulant, $\kappa_r$, by comparing coefficients:
$$\text { if } K_X(t)=\sum_{j=0}^{\infty} b_j t^j \text { then } \kappa_r=r ! b_r \text {. }$$
2. Calculating the $r^{\text {th }}$ cumulant, $\kappa_r$, by differentiation:
$$\kappa_r=K_X^{(r)}(0)=\left.\frac{d^r}{d t^r} K_X(t)\right|_{t=0} .$$
Cumulants can be expressed in terms of moments and central moments. Particularly useful are the facts that the first cumulant is the mean and the second cumulant is the variance. In order to prove these results we will use the expansion, for $|x|<1$,
$$\log (1+x)=x-\frac{1}{2} x^2+\frac{1}{3} x^3-\ldots+\frac{(-1)^{j+1}}{j} x^j+\ldots$$

统计推断代考

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