Probability & Statistics Ch.2 Section 4 Question 5

· Mohammad-Ali Bandzar

Solutions for “Probability and Statistics: The Science of Uncertainty” (Second Edition). These are solutions I have come up with; I offer no guarantee of accuracy.

Question

Is the function defined by f(x)=x3f(x) = \frac{x}{3} for −1<x<2-1 < x < 2 and 0 otherwise, a density? Why or why not?

Solution

For a function to be a valid probability density, two conditions must be met:

  1. f(x)≥0f(x) \geq 0 for all x
  2. ∫−∞∞f(x) dx=1\int_{-\infty}^{\infty} f(x) \, dx = 1

Let’s check the integral:

∫−12x3 dx=x26∣−12=46−16=36=12\int_{-1}^{2} \frac{x}{3} \, dx = \frac{x^2}{6}\Big|_{-1}^{2} = \frac{4}{6} - \frac{1}{6} = \frac{3}{6} = \frac{1}{2}

Also, note that f(x)<0f(x) < 0 for x∈(−1,0)x \in (-1, 0).

No, this is not a valid density for two reasons:

  1. The function takes negative values on the interval (−1,0)(-1, 0)
  2. The integral equals 12\frac{1}{2}, not 1

This violates Definition 2.4.2 which states that for any density function ∫−∞∞f(x) dx=1\int_{-\infty}^{\infty} f(x) \, dx = 1.