Problem 24 - Calculus Concepts

Question

Indicate whether the function could be a probability density function. Explain. \(g(x)=\left\\{\begin{array}{ll}3 x\left(1-x^{2}\right) & \text { when } 0 \leq x \leq 1 \\ 0 & \text { elsewhere }\end{array}\right.\)

Step-by-Step Solution

Answer

Yes, \( g(x) \) can be a probability density function as it satisfies all three conditions.

Step 1: Check Support

The function \( g(x) \) is defined for \( 0 \leq x \leq 1 \) and is zero elsewhere. This indicates that the support of the function is \([0, 1]\). For a function to be a probability density function, its support must cover all the probability mass, and since it is zero elsewhere, the support is correct.

Step 2: Check Non-negativity

For a function to be a probability density function, it must be non-negative on its support. Examine \( g(x) = 3x(1-x^2) \) within the interval \([0, 1]\). Since \( 1-x^2 \) is non-negative when \( x \) is in \([0, 1]\) and \( 3x \) is non-negative within this range, \( g(x) \) is non-negative on its support.

Step 3: Check Integration to 1

The final criterion is that the integral of \( g(x) \) over its entire support should equal 1. Calculate the integral: \[\int_{0}^{1} 3x(1-x^2) \, dx\]Find an antiderivative of \( 3x(1-x^2) \), which is \(-x^4 + x^2\). Evaluate it from 0 to 1:\[\left[-x^4 + x^2\right]_{0}^{1} = (-(1)^4 + (1)^2) - (-(0)^4 + (0)^2) = 1 - 0 = 1 \]Thus, the integral equals 1, confirming it integrates correctly.

Key Concepts

Support of a function

In the context of a probability density function (PDF), the support of a function is crucial. The support of a function refers to the range of values for which the function is defined and not zero.For our example function,\[g(x) = \begin{cases} 3x(1-x^2) & \text{if } 0 \leq x \leq 1, \ 0 & \text{elsewhere} \end{cases}\]the support is the interval \([0, 1]\). This means that the function has probability mass only between these limits. Outside this interval, the function value is zero. When checking whether a function can be a PDF, we confirm that the entire probability density is contained within its support.

• A function can have a finite or infinite interval as its support.
• However, all PDF probability must exist within this support.

Non-negative function

For a function to be a valid probability density function, it must be non-negative over its support.This requirement ensures there are no 'negative probabilities,' which naturally don't make sense in a probabilistic context. For our function \( g(x) = 3x(1-x^2) \) defined on \([0, 1]\), let's delve a bit deeper:

• The term \(3x\) is non-negative when \( x \) is in \([0, 1]\) because it is a linear function that starts at zero and increases.
• The expression \(1-x^2\) is non-negative in the same interval, as \(x^2\) will always be less than or equal to 1 here.
• As both terms in the product are non-negative, \( g(x) \) is indeed non-negative on its support.

Thus, this must hold true for any proposed PDF.

Integration to one

A defining property of any probability density function is that it must integrate to one over its entire support. This is because the total probability represented by the PDF must sum to one, similar to the fact that all possible outcomes' probabilities in a probability distribution must sum to one.To check if \( g(x) = 3x(1-x^2) \) yields the correct total probability,we integrate over its support \([0, 1]\):\[\int_{0}^{1} 3x(1-x^2) \, dx\]On performing this integral, we find:\[\begin{align*}\int_{0}^{1} 3x(1-x^2) \, dx &= \int_{0}^{1} (3x - 3x^3) \, dx \&= \left[ \frac{3}{2}x^2 - \frac{3}{4}x^4 \right]_{0}^{1} \&= (\frac{3}{2}(1)^2 - \frac{3}{4}(1)^4) - (\frac{3}{2}(0)^2 - \frac{3}{4}(0)^4) \&= 1\end{align*}\]The whole computation shows the integral indeed equals 1, confirming our function satisfies the PDF property of total integration to one. This is fundamental in assuring that the function accounts for all probability it describes.