Question

A family of pdf's that has been used to approximate the distribution of income, city population size, and size of firms is the Pareto family. One such member of the Pareto family is the following pdf.
f(x) = { c//x^3 x≥2
0 x<2
Find the mean of X.
(a) 2
(b) 3
(c) 5
(d) 6
(e) 4

Answer 1

Answer:

$Mean = 4$

Step-by-step explanation:

Given

$f(x) =\left \{ {{\frac{c}{x^3} \ x\ge 2} \atop {0\ x<2}} \right.$

Required

The mean of x

Given that:

$f(x) = \frac{c}{x^3}$     $x \ge 2$

First, solve for c using;

$\int\limits^a_b {f(x)} \, dx = 1$

Substitute $f(x) = \frac{c}{x^3}$ and $x \ge 2$

$\int\limits^{\infty}_2 {\frac{c}{x^3}} \, dx = 1$

Isolate c

$c\int\limits^{\infty}_2 {\frac{1}{x^3}} \, dx = 1$

Rewrite as:

$c\int\limits^{\infty}_2 {x^{-3}} \, dx = 1$

$c[\frac{x^{-3+1}}{-3 +1}]|\limits^{\infty}_2 = 1$

$c[\frac{x^{-2}}{-2}]|\limits^{\infty}_2 = 1$

$-\frac{c}{2} [x^{-2}]|\limits^{\infty}_2 = 1$

Expand

$-\frac{c}{2} [{\infty}^{-2} - {2}^{-2}]= 1$

$-\frac{c}{2} [0 - \frac{1}{4}]= 1$

$-\frac{c}{2} *- \frac{1}{4}= 1$

$\frac{c}{8}= 1$

Solve for c

$x = 8 * 1$

$x = 8$

So, we have:

$f(x) = \frac{c}{x^3}$     $x \ge 2$

$f(x) = \frac{8}{x^3}$     $x \ge 2$

So, the mean is calculated as:

$Mean = \int\limits^a_b {x * f(x)} \, dx$

This gives;

$Mean = \int\limits^{\infty}_2 {x * \frac{8}{x^3}} \, dx$

$Mean = \int\limits^{\infty}_2 {\frac{8}{x^2}} \, dx$

$Mean = 8\int\limits^{\infty}_2 {\frac{1}{x^2}} \, dx$

Rewrite as:

$Mean = 8\int\limits^{\infty}_2 {x^{-2} \, dx$

Integrate

$Mean = 8 {\frac{x^{-2+1}}{-2+1}|\limits^{\infty}_2$

$Mean = 8 {\frac{x^{-1}}{-1}}|\limits^{\infty}_2$

$Mean = -8x^{-1}|\limits^{\infty}_2$

Expand

$Mean = -8[(\infty)^{-1} - 2^{-1}]$

$Mean = -8[0 - \frac{1}{2}]$

$Mean = -8* - \frac{1}{2}$

$Mean = 8* \frac{1}{2}$

$Mean = 4$