Problem 3 - A First Course in Probability

Question

Consider the function $$ f(x)=\left\\{\begin{array}{ll} C\left(2 x-x^{3}\right) & 0

Step-by-Step Solution

Answer

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 1: 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

Step 2: 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: $$ \int_{0}^{\sqrt{2}} C(2x - x^3) dx = 1 $$

Step 3: 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.

Step 4: 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

Step 5: 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: $$ \int_{0}^{2} C(2x - x^2) dx = 1 $$

Step 6: 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 \(\frac{3}{4}\).

Key Concepts

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.

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.

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 \( C \).

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 \( x = 2 \).

• 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 \( 0 < x < 2 \).

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Problem 2

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Problem 4

Other exercises in this chapter

Problem 1

Let \(X\) be a random variable with probability density function $$ f(x)=\left\\{\begin{array}{ll} c\left(1-x^{2}\right) & -1

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Problem 2

A system consisting of one original unit plus a spare can function for a random amount of time \(X .\) If the density of \(X\) is given (in units of months) by

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Problem 4

The probability density function of \(X\), the lifetime of a certain type of electronic device (measured in hours), is given by $$ f(x)=\left\\{\begin{array}{ll

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Problem 5

A filling station is supplied with gasoline once a week. If its weekly volume of sales in thousands of gallons is a random variable with probability density fun

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Key Concepts

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.

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.

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 \( C \).

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 \( x = 2 \).

• 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 \( 0 < x < 2 \).