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Chapter 1: Problem 1

Find the mean and variance, if they exist, of each of the following distributions. (a) \(p(x)=\frac{3 !}{x !(3-x) !}\left(\frac{1}{2}\right)^{3}, x=0,1,2,3\), zero elsewhere. (b) \(f(x)=6 x(1-x), 0

Short Answer

Expert verified
The mean for distribution (a) is \(\frac{3}{2}\) and variance is \(\frac{3}{4}\); the mean for distribution (b) is \(\frac{1}{2}\) and variance is \(\frac{1}{20}\); the mean and variance for distribution (c) do not exist.

Step by step solution

01

Find the mean of the first distribution

Given that the distribution \(p(x)=\frac{3 !}{x !(3-x) !}\left(\frac{1}{2}\right)^3\), for x=0,1,2,3, we can find the mean by calculating \(E[X] = \sum x*p(x)\) over all x. This gives: \(E[X] = \sum_{x=0}^{3} x*\frac{3 !}{x !(3-x) !}\left(\frac{1}{2}\right)^3 = \frac{3}{2}\)
02

Find the variance of the first distribution

Having calculated E[X], we can now find the variance using the formula: \(Var(X) = E[X^2] - (E[X])^2\). We first find \(E[X^2] = \sum_{x=0}^{3} x^2*p(x) = \frac{3}{2}\), and then substitute in the values to calculate the variance as \(Var(X) = \frac{3}{2} - \left(\frac{3}{2}\right)^2 = \frac{3}{4}\)
03

Find the mean of the second distribution

The distribution \(f(x)=6x(1-x), 0<x<1\) is continuous, so we use the formula \(E[X] = \int xf(x)dx\) to find the mean. The integral evaluates as \(E[X] = \int_{0}^{1} x*6x(1-x) dx = \frac{1}{2}\)
04

Find the variance of the second distribution

Similarly, we will use the variance formula. The integral for \(E[X^2] = \int_{0}^{1} x^2*6x(1-x) dx = \frac{3}{10}\), which leads to \(Var(X) = \frac{3}{10} - \left(\frac{1}{2}\right)^2 = \frac{1}{20}\)
05

Find the mean of the third distribution

The distribution \(f(x)=2 / x^3, 1<x<\infty\) is also continuous. Following the same method: \(E[X] = \int_{1}^{\infty} x*2 / x^3 dx\). Unfortunately, the integral does not converge, which means the mean does not exist for this distribution
06

Find the variance of the third distribution

Since we are not able to find the mean, the variance calculation does not apply either. So, the variance also does not exist for this distribution

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