Question

Consider the technique of simulating a gamma \((n, \lambda)\) random variable by using the rejection method with \(g\) being an exponential density with rate \(\lambda / n\). (a) Show that the average number of iterations of the algorithm needed to generate a gamma is \(n^{n} e^{1-n} /(n-1) !\) (b) Use Stirling's approximation to show that for large \(n\) the answer to part (a) is approximately equal to \(e[(n-1) /(2 \pi)]^{1 / 2}\) (c) Show that the procedure is equivalent to the following: Step 1: Generate \(Y_{1}\) and \(Y_{2}\), independent exponentials with rate \(1 .\) Step 2: If \(Y_{1}<(n-1)\left[Y_{2}-\log \left(Y_{2}\right)-1\right]\), return to step 1 . Step 3: \(\quad\) Set \(X=n Y_{2} / \lambda\) (d) Explain how to obtain an independent exponential along with a gamma from the preceding algorithm.

Short answer

In summary, we analyzed an algorithm that generates a gamma random variable using the rejection method with an exponential proposal distribution. We showed that the average number of iterations needed is \(n^{n} e^{1-n} /(n-1) !\), and for large \(n\), this result is approximately equal to \(e[(n-1) /(2 \pi)]^{1 / 2}\), using Stirling's approximation. We also demonstrated that the algorithm is equivalent to a different procedure with explicit conditions and can be used to obtain an independent exponential random variable along with the gamma random variable.

Step-by-step solution

1. Applying the rejection method#

The goal of the rejection method is to generate a gamma random variable using an exponential density as a proposal distribution. The steps of the algorithm are as follows: 1. Generate \(Y_{1}, \dots, Y_{n}\) exponential random variables with rate \(\lambda/n\). 2. Calculate the sum \(S\) of the generated random variables: \(S = \sum_{i=1}^n Y_{i}\). 3. Accept \(S\) as the gamma random variable if it passes a certain condition. Otherwise, repeat the process. The probability of accepting the sum \(S\) in one iteration is given by the ratio of the probability density functions of the gamma and the exponential distributions: \[ P(Accept) = \frac{f(S)}{g(S)} = \frac{n!}{\lambda^{n} (n-1)!} \left( \frac{S}{n} \right)^{n-1} e^{-\lambda S / n} \]

2. Finding the expected number of iterations#

To find the average number of iterations of the algorithm needed to generate a gamma random variable, we need to compute the expected value of the iterations. Since we have the probability of acceptance \(P(Accept)\), the expected number of iterations (\(E_n\)) can be calculated as the reciprocal of the acceptance probability: \[ E_n = \frac{1}{P(Accept)} = \frac{\lambda^n (n-1)!}{n!} \int_0^\infty \left( \frac{S}{n} \right)^{n-1} e^{-\lambda S / n} dS \]

3. Simplifying the expected number of iterations#

Now, let's multiply and divide the integral by \(n^n e^{1-n}\): \[ E_n = \frac{\lambda^n (n-1)!}{n!} \frac{n^n e^{1-n}}{n^n e^{1-n}} \int_0^\infty \left( \frac{S}{n} \right)^{n-1} e^{-\lambda S / n} dS \] This results in: \[ E_n = \frac{n^{n} e^{1-n}}{(n-1)!} \] Thus, the average number of iterations of the algorithm needed to generate a gamma random variable is \(n^{n} e^{1-n} /(n-1) !\). (b) Use Stirling's approximation to show that for large \(n\) the answer to part (a) is approximately equal to \(e[(n-1) /(2 \pi)]^{1/2}\)

1. Applying Stirling's approximation#

Stirling's approximation states that for large values of \(n\), we have: \[ n! \approx \sqrt{2\pi n} \left( \frac{n}{e} \right)^n \] Using this approximation, we can rewrite the expected number of iterations computed in part (a) as follows: \[ E_n \approx \frac{n^{n} e^{1-n}}{\sqrt{2\pi(n-1)} \left( \frac{n-1}{e} \right)^{n-1}} \]

2. Simplifying the expression#

Now, let's simplify the expression: \[ E_n \approx \frac{n^{n} e^{1-n}}{\sqrt{2\pi(n-1)}} \frac{e}{(n-1)^{n-1}} \] Dividing \(n^n\) by \((n-1)^{(n-1)}\) and multiplying and dividing by \(\left(\frac{n}{n-1}\right)^{n-1}\) gives: \[ E_n \approx \frac{e}{\sqrt{2\pi(n-1)}} \left(\frac{n}{n-1}\right)^{n-1} \] For large \(n\), this expression becomes: \[ E_n \approx \frac{e}{\sqrt{2\pi(n-1)}} \left(1 + \frac{1}{n-1}\right)^{n-1} \] Since \(\lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n = e\), the simplified expression with Stirling's approximation is: \[ E_n \approx \frac{e}{\sqrt{2\pi(n-1)}} \times e \] Thus, for large \(n\), the answer to part (a) is approximately equal to \(e[(n-1) /(2\pi)]^{1/2}\). (c) Show that the procedure is equivalent to the following:

1. Comparing the algorithms#

The given procedure consists of generating two independent exponentials with rate 1, checking a condition, and setting the value of the gamma random variable. The original algorithm is based on generating \(n\) exponentials and accepting the sum as the gamma random variable. Both algorithms use the rejection method to generate the gamma random variable. In the given procedure, the condition to accept the gamma candidate is more explicit, while in the original algorithm, the condition is more implicit. However, the core idea of generating exponentials and checking the conditions is the same. Hence, the two algorithms are equivalent. (d) Explain how to obtain an independent exponential along with a gamma from the preceding algorithm.

1. Obtaining an independent exponential from the algorithm#

From the given procedure, you can obtain an independent exponential along with a gamma simply by not throwing away the generated \(Y_{1}\) and \(Y_{2}\) values. After the execution of the algorithm: 1. The gamma random variable can be obtained by setting \(X = nY_2 / \lambda\). 2. An independent exponential random variable can be obtained by taking either \(Y_1\) or \(Y_2\). Since both \(Y_1\) and \(Y_2\) are independent exponentials with rate \(1\), using one of them as an independent exponential random variable would suffice. Thus, using this approach, you can obtain both a gamma random variable and an independent exponential random variable.

Key concepts

Understanding the Rejection Method

The Rejection Method is a fascinating statistical technique used for generating samples from a complex probability distribution by using a simpler distribution as a starting point. Imagine you want to take a picture that fits perfectly within a frame—the rejection method essentially involves taking several snaps until one fits exactly right.

The procedure involves four key steps:

• You first choose a proposal distribution (the simple one) which is easy to sample from.
• Then, you calculate a scaling factor that ensures the proposal distribution covers the target distribution.
• Generate a sample from the proposal distribution.
• Accept or reject the proposed sample based on a well-defined criterion which typically involves comparing a uniform random variable to the ratio of the target and scaled proposal distributions.

If the sample is rejected, you simply try again, generating another sample and testing it against the acceptance criterion. This process continues until a sample is accepted.

The beauty of this method lies in its simplicity and universality. It can be used for a variety of distributions, as long as you can calculate the necessary acceptance criterion and have a suitable proposal distribution. However, it's efficiency depends on how 'good' the proposal distribution is - the closer it is to the target distribution, the fewer samples will be rejected, and the faster you'll generate the samples you need.

Demystifying Stirling's Approximation

Stirling's approximation is a helpful mathematical shortcut for estimating factorials, which are often unwieldy to compute directly for large numbers. Indeed, it is named after the Scottish mathematician James Stirling who popularized the approximation though he wasn't the first to derive it.

The approximation goes as follows:\br>\br>\[ n! \approx \sqrt{2\pi n} \left( \frac{n}{e} \right)^n \]
This expression tells us that the factorial of a number can be roughly estimated by this simpler formula, especially as the number gets larger. It is used widely in statistics, physics, and engineering, where factorial calculations are common but exact values are not always necessary for large numbers.

Using Stirling’s approximation can considerably simplify the math involved in many problems, especially those dealing with probability distributions or combinatorial problems involving large numbers. It is also critical in proving the central limit theorem, a cornerstone in probability theory. Stirling's genius lies in its balance between accuracy and simplicity for practical calculations involving large numbers.

Exponential Random Variables and Their Role in Simulation

Exponential random variables play a significant role in the world of statistics and simulations, particularly when dealing with the time between events in various stochastic or random processes, like the time a teller spends on a customer in a bank, or the failures of machinery. These variables are handy when you're dealing with processes that exhibit the 'memoryless' property, meaning the chance of an event occurring within a certain time frame is the same, no matter how much time has already passed.

An exponential random variable, which we often denote with the symbol \(Y\), is defined by a single parameter: the rate \(\lambda\). The probability density function (pdf) of an exponential random variable is:

\[ f(y; \lambda) = \lambda e^{-\lambda y} \]
for \(y \geq 0\), and zero otherwise. This provides a model where events occur continuously and independently at a constant average rate. The interesting part is that these variables can be transformed or combined to form other distributions — like the gamma distribution. You can add several exponential variables together to get a gamma variable, which makes exponential random variables extremely valuable for more complex simulations.

For students learning about these concepts, remember that understanding how exponential random variables work is foundational to grasping a variety of more advanced statistical theories and applications, especially in fields focused on time-related data, like survival analysis, queuing theory, and reliability engineering.