Chapter 12: Problem 1
Show that \(f(x)=\left\\{\begin{array}{cl}3 e^{-3 x} & \text { for } x>0 \\ 0 & \text { for } x \leq 0\end{array}\right.\) is a density function. Find the corresponding distribution function.
Short Answer
Expert verified
The function is a valid PDF with CDF: \( F(x) = 0 \) for \( x \leq 0 \) and \( F(x) = 1 - e^{-3x} \) for \( x > 0 \).
Step by step solution
01
Understand the Density Function
A density function, also known as a probability density function (PDF), describes the likelihood of a continuous random variable taking on a particular value. The main condition for a function to be a density function is that the integral over its entire space should equal 1.
02
Define the Integral
The function is defined as \( f(x) = 3e^{-3x} \) for \( x > 0 \) and \( f(x) = 0 \) for \( x \leq 0 \). Since the function is zero when \( x \leq 0 \), we integrate from 0 to infinity for its non-zero part.
03
Integrate the Density Function
Compute the integral of \( f(x) = 3e^{-3x} \) from 0 to infinity:\[\int_0^{\infty} 3e^{-3x} \, dx\]This integral should equal 1 if \( f(x) \) is a valid density function.
04
Solve the Integral
Evaluate the integral \( \int 3e^{-3x} \, dx \). To do this, use substitution. Let \( u = -3x \), then \( du = -3 \, dx \), or \( dx = -\frac{du}{3} \). This gives:\[\int 3e^{-3x} \, dx = \int -e^u \, du\]Integrating gives \( -e^u + C \), or substituting back, \(-e^{-3x} + C\). Evaluate this from 0 to \( \infty \).
05
Evaluate the Definite Integral
Substitute the bounds into \(-e^{-3x}\):\[\begin{aligned}\lim_{{x \to \infty}} -e^{-3x} &= 0, \-e^{-3 \times 0} &= -1.\end{aligned}\]Thus the integral yields \( 0 - (-1) = 1 \). This confirms \( f(x) \) is a valid probability density function.
06
Define the Cumulative Distribution Function
The cumulative distribution function (CDF) \( F(x) \) is determined by integrating the PDF up to \( x \) for all valid values. Since \( f(x) = 0 \) for \( x \leq 0 \), \( F(x) = 0 \) for \( x \leq 0 \). For \( x > 0 \), \[F(x) = \int_0^x 3e^{-3t} \, dt = [-e^{-3t}]_0^x = 1 - e^{-3x}.\]
07
Distribution Function Conclusion
Combining the findings:- For \( x \leq 0 \), \( F(x) = 0 \).- For \( x > 0 \), \( F(x) = 1 - e^{-3x} \).This defines the cumulative distribution function for the given density.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Cumulative Distribution Function
The cumulative distribution function (CDF) is a fundamental concept in probability and statistics. It represents the probability that a continuous random variable takes a value less than or equal to a given point. The CDF is a non-decreasing function and ranges between 0 and 1. For a given probability density function (PDF) like \( f(x) \), the CDF \( F(x) \) is calculated by integrating the PDF from the lower bound of the variable up to \( x \).
For the exercise above, the PDF is \( f(x) = 3e^{-3x} \) for \( x > 0 \) and 0 for \( x \leq 0 \). Therefore, the CDF can be expressed as:
\[ F(x) = \int_{0}^{x} 3e^{-3t} \, dt = 1 - e^{-3x} \text{ for } x > 0, \] and \( F(x) = 0 \) for \( x \leq 0 \).
For the exercise above, the PDF is \( f(x) = 3e^{-3x} \) for \( x > 0 \) and 0 for \( x \leq 0 \). Therefore, the CDF can be expressed as:
\[ F(x) = \int_{0}^{x} 3e^{-3t} \, dt = 1 - e^{-3x} \text{ for } x > 0, \] and \( F(x) = 0 \) for \( x \leq 0 \).
- The function \( F(x) = 1 - e^{-3x} \) shows that as \( x \) increases, \( e^{-3x} \) decreases, making \( F(x) \) approach 1.
- This signifies that as we move further along the \( x \)-axis, the probability of observing a value less than or equal to \( x \) increases.
Integration
Integration is a key mathematical process used to calculate areas under curves, which in the context of probability, relate to finding probabilities.
For the probability density function \( f(x) = 3e^{-3x} \), we need to integrate from 0 to infinity to ensure the total probability equals 1. This integration process confirms whether a function can serve as a valid probability density function.
In the step-by-step solution, integration was performed as follows:
\[ \int_{0}^{\infty} 3e^{-3x} \, dx = 1 \]This outcome verifies that the function can represent a continuous random variable.
For the probability density function \( f(x) = 3e^{-3x} \), we need to integrate from 0 to infinity to ensure the total probability equals 1. This integration process confirms whether a function can serve as a valid probability density function.
In the step-by-step solution, integration was performed as follows:
- We defined \( u = -3x \) and thus \( du = -3 \, dx \), making \( dx = -\frac{du}{3} \).
- Changing variables allowed us to transform the integral \( \int 3e^{-3x} \, dx \) into \( \int -e^u \, du \).
\[ \int_{0}^{\infty} 3e^{-3x} \, dx = 1 \]This outcome verifies that the function can represent a continuous random variable.
Continuous Random Variable
A continuous random variable is one that can take on an infinite number of possible values within a given range.
Unlike discrete random variables, which are countable, continuous variables are represented by probability density functions. These functions do not give probabilities for specific outcomes; instead, they provide a probability density, which must be integrated over a range to find actual probabilities.
In this exercise, the function \( f(x) = 3e^{-3x} \) describes a continuous random variable for \( x > 0 \).
Key points about continuous random variables include:
Unlike discrete random variables, which are countable, continuous variables are represented by probability density functions. These functions do not give probabilities for specific outcomes; instead, they provide a probability density, which must be integrated over a range to find actual probabilities.
In this exercise, the function \( f(x) = 3e^{-3x} \) describes a continuous random variable for \( x > 0 \).
Key points about continuous random variables include:
- The value of the probability density function at a single point is zero because there are infinitely many potential individual outcomes for a continuous random variable.
- To find the probability of the variable falling within a specified interval, we calculate the area under the curve of that interval using integration.