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Chapter 10: Problem 11

Use the rejection method with \(g(x)=1,0

Short Answer

Expert verified
To simulate a random variable with density function \(f(x) = 60x^3(1-x)^2\) for \(0 < x < 1\) using the rejection method, follow these steps: 1. Find the constant c: \(c \geq \frac{80}{243}\); 2. Generate two random variables U1 and U2, uniformly distributed between 0 and 1; 3. Set \(x = U1\) and \(y = \frac{60x^3(1-x)^2}{c}\); 4. Accept x as a realization of the random variable with density function f(x) if \(U2 \leq \frac{y}{c}\). Otherwise, repeat steps 2-4.

Step by step solution

01

Find the constant c

We need to find a constant c such that: \(c \cdot g(x) \geq f(x)\) Using the given g(x) and f(x): \(c \geq \frac{60x^3(1-x)^2}{1}\), for all \(0 < x < 1\) Optimize the function in the domain (0, 1) to find the maximum value: \(h(x) = 60x^3(1-x)^2\) We can find the maximum by taking the first derivative of h(x) with respect to x and setting it to zero: \(h'(x) = 0\) Then, solving for x will give us the point where h(x) reaches its maximum value. \(h'(x) = 180x^2(1-x)(1-3x)\) Solve the equation for x: \(180x^2(1-x)(1-3x) = 0\) This gives us x = 0, x = 1/3, and x = 1 as potential critical points. Evaluate h(x) at these points: - \(h(0) = 0\) - \(h\left(\frac{1}{3}\right) = \frac{80}{243}\) - \(h(1) = 0\) The maximum value of h(x), within the given domain, is \(\frac{80}{243}\). Thus, the minimum value of c should be: \(c \geq \frac{80}{243}\) Now we have the value of c, we can proceed to use the rejection method to create an algorithm for simulating the random variable with the given density function.
02

Rejection Method Algorithm

Using the rejection method, we have the following algorithm: 1. Generate a random variable U1 uniformly distributed between 0 and 1; 2. Generate a random variable U2 uniformly distributed between 0 and 1; 3. Set \(x = U1\) and \(y = \frac{60x^3(1-x)^2}{c}\); 4. If \(U2 \leq \frac{y}{c}\), then accept x as a realization of the random variable with density function f(x). Otherwise, go back to step 1. This algorithm will generate random variables following the given density function, f(x).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability Density Function
The Probability Density Function (PDF) is a vital concept in statistics and probability theory. It describes the likelihood of a random variable to take on a particular value. More specifically, for a continuous random variable, the PDF is a function that shows the probability distribution across different outcomes in a continuous range.

The area under the PDF curve, over a specific interval, gives the probability that a random variable falls within that interval. The function that we want to simulate in this exercise is:
  • For \(0 < x < 1\), the PDF is \(f(x) = 60x^3(1-x)^2\).
  • Outside this interval, the PDF is zero.
In this problem, we want an algorithm that generates numbers according to this density function using rejection sampling.
Optimization
Optimization plays a crucial role in rejection sampling, as it helps determine the scale factor 'c'. This factor ensures that the function \(c \cdot g(x)\) always exceeds or equals the target PDF \(f(x)\) for the domain \(0 < x < 1\). This way, the rejection algorithm ensures that the generated numbers truly follow the desired distribution.

To find this constant, we optimize the function \(h(x) = 60x^3(1-x)^2\), which represents the PDF within our domain, to determine its maximum value. By taking the derivative \(h'(x) = 180x^2(1-x)(1-3x)\) and solving for x, we identify critical points where the function may achieve a peak.
  • x=0 and x=1 lead to h(x)=0, which indicates outer boundaries where the function hits zero.
  • x=1/3 results in the highest function value \left(\frac{80}{243}\right).\
Thus, the lower bound for 'c' is determined as \(c \geq \frac{80}{243}\). This ensures the proposed scaling function captures the true behavior of the PDF accurately.
Algorithm Design
Algorithm design is about creating efficient ways to solve problems, and here, the challenge is to simulate a random variable following a specific density function using rejection sampling. Rejection sampling is an intuitive way to generate random numbers from a complex distribution by using a simpler proposal distribution, denoted as \(g(x)\).

Here's how we proceed:
  • Choose \(g(x) = 1\) as our proposal distribution since it is uniform over our interval of interest.
  • Determine constant 'c' so that \(c \cdot g(x) \geq f(x)\) for \(0 < x < 1\).
  • The algorithm iterates through random sampling from a uniform distribution until a point meets acceptance criteria determined by our PDF.
  • If \(U2 \leq \frac{f(x)}{c}\), accept the sample. Otherwise, restart.
Discarding certain samples ensures that the retained samples follow the intended probability distribution of \(f(x)\). Overall, this method is a flexible tool for simulating phenomena in scientific computations.
Uniform Distribution
The uniform distribution is fundamental for rejection sampling algorithms as it serves as the base distribution from which we initially draw samples. In the context of this exercise, \(g(x) = 1\) represents a uniform probability density function over the range \(0 < x < 1\).

This simplicity allows its cumulative distribution to be conveniently handled.
Some essential properties include:
  • Every interval of the same length within the distribution's range is equally probable.
  • It ensures that our sampling framework starts with no favoritism toward any sub-interval.
In the bigger picture of algorithmic design, drawing initial samples from a uniform distribution ensures that each potential outcome is sampled fairly. This randomness spreads the starting sample points evenly over the interval, providing a reliable foundation for the subsequent rejection step.

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