A Random Variable X Has The Pdf Shown Below: F X(x)={ Cx(1x 2)00x1elsewhere A. Find C And Plot The Pdf.

A Random Variable X Has The Pdf Shown Below: F X(x)={ Cx(1x 2)00x1elsewhere A. Find C And Plot The Pdf.

Understanding probability density functions (pdfs) is fundamental in the study of probability and statistics. When dealing with continuous random variables, the pdf provides the likelihood of the variable taking on a specific value within a range. In this article, we will analyze a given pdf, determine the constant \( C \), and illustrate the shape of the function through plotting. We will also explore the concepts involved in deriving the constant, verifying the validity of the pdf, and visualizing the distribution for better comprehension.

Introduction to Probability Density Functions (pdfs)

Before diving into the problem, it's essential to review what a probability density function entails:

• Definition: A pdf \( f_X(x) \) describes the likelihood of a continuous random variable \( X \) taking on a value near \( x \). Unlike probability mass functions for discrete variables, pdfs are used for continuous variables.


• Properties of a pdf:

1. Total probability: The integral over the entire support equals 1:

• Cumulative Distribution Function (CDF): The CDF \( F_X(x) \) is the integral of the pdf from \(-\infty\) up to \( x \):

In our problem, the given function appears to be a piecewise function, which is common when defining pdfs over specific intervals.

Understanding the Given Pdf: \( F_X(x) \)

The problem states:

> A Random Variable \( X \) has the pdf shown below:
>
> \[
> F_X(x) =
> \begin{cases}
> Cx(1 - x^2), & x \text{ in a certain interval} \\
> 0, & \text{elsewhere}
> \end{cases}
> \]

(Note: The original prompt seems to have some typographical issues, but based on context, it's likely that \( FX(x) \) was intended to be \( fX(x) \), the pdf, rather than the CDF. Alternatively, it could be referring to the CDF, but typically, the notation \( FX(x) \) is used for the CDF. For clarity, we'll assume it's the pdf, \( fX(x) \).)

Assumption: The function is a piecewise function defined over a specific interval, say \( x \in [a, b] \), with \( f_X(x) = C \cdot x(1 - x^2) \) within that interval, and zero elsewhere.

The key points are:

• \( f_X(x) \) is non-zero only over a specific range.


• The function involves a constant \( C \) which we need to determine.


• The function involves polynomial terms, which are common in pdfs.

Step 1: Clarify the Support and Function Form

Given the typical structure, the pdf might be:

\[
f_X(x) =
\begin{cases}
C \cdot x(1 - x^2), & 0 \leq x \leq 1 \\
0, & \text{elsewhere}
\end{cases}
\]

This assumption aligns with standard practice, but if the original problem specifies different support, adjust accordingly.

Step 2: Find the Constant \( C \)

To find \( C \), we utilize the fundamental property of pdfs:

\[
\int{-\infty}^\infty fX(x) \, dx = 1
\]

Given the support, this reduces to:

\[
\int_{0}^{1} C \cdot x(1 - x^2) \, dx = 1
\]

\[
C \int_0^1 x(1 - x^2) \, dx = 1
\]

Let's compute this integral step-by-step.

Calculating the Integral

\[
\int_0^1 x(1 - x^2) \, dx
\]

Expand the integrand:

\[
x(1 - x^2) = x - x^3
\]

Thus,

\[
\int0^1 (x - x^3) \, dx = \int0^1 x \, dx - \int_0^1 x^3 \, dx
\]

Compute each integral:

• \(\int0^1 x \, dx = \left[\frac{x^2}{2}\right]0^1 = \frac{1}{2}\)


• \(\int0^1 x^3 \, dx = \left[\frac{x^4}{4}\right]0^1 = \frac{1}{4}\)

\[
\int_0^1 (x - x^3) \, dx = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}
\]

Recall the original integral with \( C \):

\[
C \times \frac{1}{4} = 1
\]

Solve for \( C \):

\[
C = 4
\]

Result: The constant \( C \) is 4.

Step 3: Write the Final Pdf Expression

The pdf of the random variable \( X \):

\[
f_X(x) =
\begin{cases}
4 x (1 - x^2), & 0 \leq x \leq 1 \\
0, & \text{elsewhere}
\end{cases}
\]

This function is valid as it satisfies the properties of a pdf.

Step 4: Verify the Pdf

To ensure correctness, verify:

\[
\int{-\infty}^\infty fX(x) \, dx = 1
\]

which we've already confirmed through the calculation of \( C \). Additionally, check that \( f_X(x) \geq 0 \) over its support:

• For \( x \in [0, 1] \), \( x \geq 0 \) and \( 1 - x^2 \geq 0 \) for \( x \leq 1 \), so \( f_X(x) \geq 0 \).

Step 5: Plotting the Pdf

Visualizing the pdf helps in understanding the distribution's shape.

Method:

• Use graphing tools like Desmos, GeoGebra, or programming languages such as Python with matplotlib.


• Plot \( f_X(x) \) over the interval \( [0, 1] \).

• At \( x = 0 \): \( f_X(0) = 0 \).


• At \( x = 1 \): \( f_X(1) = 4 \times 1 \times (1 - 1) = 0 \).


• The function reaches a maximum somewhere in \( (0, 1) \).

Find the maximum by taking the derivative:

\[
f_X(x) = 4 x (1 - x^2) = 4x - 4x^3
\]

Derivative:

\[
f_X'(x) = 4 - 12 x^2
\]

Set \( f_X'(x) = 0 \):

\[
4 - 12 x^2 = 0 \Rightarrow 12 x^2 = 4 \Rightarrow x^2 = \frac{1}{3} \Rightarrow x = \pm \frac{1}{\sqrt{3}}
\]

Since the support is \( [0, 1] \), consider \( x = \frac{1}{\sqrt{3}} \approx 0.577 \).

Calculate \( f_X \) at this point:

\[
f_X\left(\frac{1}{\sqrt{3}}\right) = 4 \times \frac{1}{\sqrt{3}} \times \left(1 - \frac{1}{3}\right) = 4 \times \frac{1}{\sqrt{3}} \times \frac{2}{3} = \frac{8}{3} \times \frac{1}{\sqrt{3}} = \frac{8}{3 \sqrt{3}}
\]

Numerically:

\[
f_X\left(\frac{1}{\sqrt{3}}\right) \approx \frac{8}{3 \times 1.732} \approx \frac{8}{5.196} \approx 1.539
\]

Thus, the maximum pdf value is approximately 1.539 at