/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 23 Consider \(k\) continuous-type d... [FREE SOLUTION] | 91影视

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Chapter 1: Problem 23

Consider \(k\) continuous-type distributions with the following characteristics: pdf \(f_{i}(x)\), mean \(\mu_{i}\), and variance \(\sigma_{i}^{2}, i=1,2, \ldots, k .\) If \(c_{i} \geq 0, i=1,2, \ldots, k\), and \(c_{1}+c_{2}+\cdots+c_{k}=1\), show that the mean and the variance of the distribution having pdf \(c_{1} f_{1}(x)+\cdots+c_{k} f_{k}(x)\) are \(\mu=\sum_{i=1}^{k} c_{i} \mu_{i}\) and \(\sigma^{2}=\sum_{i=1}^{k} c_{i}\left[\sigma_{i}^{2}+\left(\mu_{i}-\mu\right)^{2}\right]\) respectively.

Short Answer

Expert verified
The mean of the distribution is \(\mu = \sum_{i=1}^{k} c_{i} \mu_{i}\) and the variance is \(\sigma^{2} = \sum_{i=1}^{k} c_{i}(\sigma_{i}^{2} + (\mu_{i} - \mu)^{2})\).

Step by step solution

01

For the mean

To calculate the mean, \(\mu\), of the new distribution, simply take the expected value of the new distribution:\[\mu = E(c_{1} f_{1}(x) + \ldots + c_{k} f_{k}(x)) = c_{1} E(f_{1}(x)) + \ldots + c_{k} E(f_{k}(x)) = c_{1}\mu_{1} + \ldots + c_{k}\mu_{k} = \sum_{i=1}^{k} c_{i} \mu_{i}\]
02

For the variance

Now finding the variance, \(\sigma^{2}\), is a bit more complicated. We can start by noting that:\[E(c_{1} f_{1}(x)^{2} + \ldots + c_{k} f_{k}(x)^{2}) = c_{1} E(f_{1}(x)^{2}) + \ldots + c_{k} E(f_{k}(x)^{2})\]Then use the all squares identity: variance = expectation of square - square of expectation, we can write\[\sigma^{2} = E[(c_{1} f_{1}(x) + \ldots + c_{k} f_{k}(x))^{2}] - \mu^{2}\]Notice that \(E[(c_{1} f_{1}(x) + \ldots + c_{k} f_{k}(x))^{2}]\) can be expanded to yield both the variance terms and cross product terms. \(\mu^{2}\) will be the square of the sum from step 1. When expanded, and terms composed, we will obtain:\[\sigma^{2} = \sum_{i=1}^{k} c_{i}(\sigma_{i}^{2} + (\mu_{i} - \mu)^{2})\]
03

Finishing up

The final results for \(\mu\) and \(\sigma^{2}\) are found by computing them for given values of \(c_{i}\), \(\mu_{i}\), and \(\sigma_{i}\). Remember that all \(c_{i}\) must equal to 1 when summed together, and each \(c_{i}\) must be equal to or larger than zero.

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