Chapter 5: Problem 3
Consider the function
$$
f(x)=\left\\{\begin{array}{ll}
C\left(2 x-x^{3}\right) & 0
Short Answer
Expert verified
In summary, both given functions can be probability density functions. For the first function, \(f(x) = C(2x - x^3)\), the constant C is \(\frac{2}{3}\). For the second function, \(f(x) = C(2x - x^2)\), the constant C is \(\frac{3}{4}\).
Step by step solution
01
Check Non-Negativity
First, we need to check the non-negativity condition. Observe that the function is zero outside the range \(0 < x < \frac{5}{2}\). Also, during the interval \(0 < x < \frac{5}{2}\), we have:
$$
f(x) = C(2x - x^3)
$$
We should find the intervals in which \(f(x)\ge 0\). For that, we check the critical points by setting f(x) to 0:
$$
C(2x - x^3) = 0 \Rightarrow 2x - x^3 = 0 \Rightarrow x(2-x^2) = 0
$$
The critical points are \(x=0\) and \(x=\pm\sqrt{2}\). As we are looking for non-negative x, our option is \(x=\sqrt{2}\). So, we should check the interval \(0 < x < \sqrt{2}\):
$$
0<x<\sqrt{2}\Rightarrow 0<2-x^2
$$
Therefore, \(f(x) \geq 0\) for \(0 < x < \sqrt{2}\)
02
Check Total Probability
Now, we need to calculate the integral of the function over the interval \(0 < x < \sqrt{2}\) to check for total probability:
$$
\int_{0}^{\sqrt{2}} C(2x - x^3) dx = 1
$$
03
Find the Constant C
To find C, we solve the integral above:
$$
C\int_{0}^{\sqrt{2}} (2x - x^3) dx = 1
$$
$$
C \left[\int_{0}^{\sqrt{2}} 2x dx - \int_{0}^{\sqrt{2}} x^3 dx \right] = 1
$$
$$
C \left[\left. x^2 \right|_{0}^{\sqrt{2}} - \left.\frac{1}{4} x^4 \right|_{0}^{\sqrt{2}} \right] = 1
$$
$$
C \left[2 - \frac{1}{2} \right] = 1
$$
$$
C \left[\frac{3}{2}\right] = 1
$$
$$
C = \frac{2}{3}
$$
The constant C for the first function is \(\frac{2}{3}\). Now, let's move on to the next function given by:
$$
f(x) = \left\{
\begin{array}{ll}
C(2x - x^2) & 0 < x < \frac{5}{2} \\
0 & \text{otherwise}
\end{array}
\right.
$$
We will repeat the steps for the second function by first checking the non-negativity condition and then finding the value of constant C.
04
Check Non-Negativity for the Second Function
For the second function, it's also zero outside the range \(0 < x < \frac{5}{2}\). Inside this range, we have:
$$
f(x) = C(2x - x^2)
$$
Again, we need to find the intervals in which \(f(x)\ge 0\). We set f(x) to 0:
$$
C(2x - x^2) = 0 \Rightarrow 2x - x^2 = 0 \Rightarrow x(2 - x) = 0
$$
The critical points are \(x=0\) and \(x=2\) which is non-negative x. So, we should check the interval \(0 < x < 2\):
$$
0<x<2\Rightarrow 0<2-x
$$
Thus, \(f(x)\ge 0\) for \(0 < x < 2\)
05
Check Total Probability for the Second Function
Next, we calculate the integral of the function over the interval \(0 < x < 2\) to check for total probability:
$$
\int_{0}^{2} C(2x - x^2) dx = 1
$$
06
Find the Constant C for the Second Function
To find C for the second function, we solve the integral above:
$$
C\int_{0}^{2} (2x - x^2) dx = 1
$$
$$
C \left[\int_{0}^{2} 2x dx - \int_{0}^{2} x^2 dx \right] = 1
$$
$$
C \left[\left. x^2 \right|_{0}^{2} - \left.\frac{1}{3} x^3 \right|_{0}^{2} \right] = 1
$$
$$
C \left[4 - \frac{8}{3} \right] = 1
$$
$$
C \left[\frac{4}{3}\right] = 1
$$
$$
C = \frac{3}{4}
$$
The constant C for the second function is \(\frac{3}{4}\).
So, both given functions can be probability density functions with the respective values of constant C as \(\frac{2}{3}\) and \(\frac{3}{4}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Non-Negativity Condition
In probability density functions (PDF), the non-negativity condition assures us that the function is always zero or positive. This reflects the idea that probabilities cannot be negative. For a function like \( f(x) \), defined for certain intervals, checking for non-negativity involves finding where \( C(2x - x^3) \) is greater than or equal to zero.
By solving the equation \( C(2x - x^3) = 0 \), we determine critical points. Critical points occur where the function changes from positive to negative or vice versa. For instance, solving \( x(2-x^2) = 0 \) gives potential points where the sign of \( f(x) \) could change.
Remember: Only non-negative intervals matter for PDFs. So if our critical points are \( x = 0 \) and \( x = \sqrt{2} \), we focus on \( 0 < x < \sqrt{2} \), ensuring \( f(x) \) stays non-negative.
By solving the equation \( C(2x - x^3) = 0 \), we determine critical points. Critical points occur where the function changes from positive to negative or vice versa. For instance, solving \( x(2-x^2) = 0 \) gives potential points where the sign of \( f(x) \) could change.
Remember: Only non-negative intervals matter for PDFs. So if our critical points are \( x = 0 \) and \( x = \sqrt{2} \), we focus on \( 0 < x < \sqrt{2} \), ensuring \( f(x) \) stays non-negative.
Total Probability
For a function to qualify as a probability density function, the integral of the function across its whole range must equal one. Think of it as summing all the possibilities - they should total one full probability.
For the example function \( f(x) = C(2x - x^3) \), its range for non-zero values is \( 0 < x < \sqrt{2} \). Thus, we calculate:
\[ \int_{0}^{\sqrt{2}} C(2x - x^3) dx = 1 \]
This equation ensures that the total area under the PDF curve sums up to one, providing a solid basis for describing probabilities continuously over an interval.
For the example function \( f(x) = C(2x - x^3) \), its range for non-zero values is \( 0 < x < \sqrt{2} \). Thus, we calculate:
\[ \int_{0}^{\sqrt{2}} C(2x - x^3) dx = 1 \]
This equation ensures that the total area under the PDF curve sums up to one, providing a solid basis for describing probabilities continuously over an interval.
Integral Calculation
Integral calculation is a fundamental process in confirming whether a function can serve as a valid probability density function. It involves finding the antiderivative and evaluating it over a specific interval.
Consider the function \( C(2x - x^3) \). We break down the integral:
\[ C \left[ \int_{0}^{\sqrt{2}} 2x \ dx - \int_{0}^{\sqrt{2}} x^3 \ dx \right] \]
Each integral is solved separately. For \( \int_{0}^{\sqrt{2}} 2x \ dx \), the antiderivative is \( x^2 \), evaluated from 0 to \(\sqrt{2}\). For \( \int_{0}^{\sqrt{2}} x^3 \ dx \), the antiderivative is \( \frac{1}{4}x^4 \), also evaluated from 0 to \( \sqrt{2} \). Combining these evaluative steps yields a full solution, leading to finding the constant \( C \).
Consider the function \( C(2x - x^3) \). We break down the integral:
\[ C \left[ \int_{0}^{\sqrt{2}} 2x \ dx - \int_{0}^{\sqrt{2}} x^3 \ dx \right] \]
Each integral is solved separately. For \( \int_{0}^{\sqrt{2}} 2x \ dx \), the antiderivative is \( x^2 \), evaluated from 0 to \(\sqrt{2}\). For \( \int_{0}^{\sqrt{2}} x^3 \ dx \), the antiderivative is \( \frac{1}{4}x^4 \), also evaluated from 0 to \( \sqrt{2} \). Combining these evaluative steps yields a full solution, leading to finding the constant \( C \).
Critical Points
Critical points are where the derivative of a function is zero or undefined; it helps determine changes in direction or slope of a function. These points can indicate maximums, minimums, or points of inflection.
In the context of probability, finding critical points is crucial when determining intervals of non-negativity. For example, when finding critical points for \( C(2x - x^2) \), solve \( 2x - x^2 = 0 \) yielding \( x = 0 \) or \( x = 2 \).
In the context of probability, finding critical points is crucial when determining intervals of non-negativity. For example, when finding critical points for \( C(2x - x^2) \), solve \( 2x - x^2 = 0 \) yielding \( x = 0 \) or \( x = 2 \).
- This tells us where the function either "flattens out" or could change from increasing to decreasing.
- Assistance in designating valid intervals for the PDF, ensuring \( f(x) \) remains non-negative in \( 0 < x < 2 \).