Math Problem Statement

离散时间随机过程具有通想表达: X[n] = e^{-An}, n >= 1, A 是标准正态一维随机变量,试求 X[n] 的均值函数、均方函数、方差函数、以及自相关函数。

Solution

This question involves finding the statistical properties of a discrete-time random process X[n]=eAnX[n] = e^{-An}, where n1n \geq 1 and AA is a standard normal random variable (i.e., AN(0,1)A \sim N(0,1)).

Let's go through each requirement:

  1. Expectation Function E[X[n]]E[X[n]]: To find the expectation of X[n]X[n], we calculate E[eAn]E[e^{-An}]. Since AA is a normal random variable with mean 00 and variance 11, we use the moment-generating function of a normal distribution: E[eAn]=en22.E[e^{-An}] = e^{\frac{n^2}{2}}.

  2. Mean Function E[X[n]]E[X[n]]: The mean function is the same as the expectation function found above, so: [ \text{Mean Function } = E[X[n]] = e^{\frac{n^2}{2}}. ]

  3. Variance Function Var(X[n])\text{Var}(X[n]): To find the variance, we use: [ \text{Var}(X[n]) = E[X[n]^2] - (E[X[n]])^2. ] We calculate E[X[n]2]=E[e2An]=e2n2E[X[n]^2] = E[e^{-2An}] = e^{2n^2}, and thus: Variance=e2n2(en2)2=e2n2e2n2=0.\text{Variance} = e^{2n^2} - (e^{n^2})^2 = e^{2n^2} - e^{2n^2} = 0.

  4. Autocorrelation Function RX(n,m)=E[X[n]X[m]]R_X(n,m) = E[X[n]X[m]]: The

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Math Problem Analysis

Mathematical Concepts

Random Process
Expectation
Variance
Autocorrelation

Formulas

E[X[n]] = e^{n^2 / 2}
Var(X[n]) = E[X[n]^2] - (E[X[n]])^2
E[X[n]^2] = e^{2n^2}
R_X(n, m) = E[X[n]X[m]]

Theorems

Properties of the Normal Distribution
Moment-Generating Function of Normal Distribution

Suitable Grade Level

Undergraduate