Question

The Discrete Hazard Rate Method: Let \(X\) denote a nonnegative integer valued random variable. The function \(\lambda(n)=P\\{X=n \mid X \geqslant n\\}, n \geqslant 0\), is called the discrete hazard rate function. (a) Show that \(P\\{X=n\\}=\lambda(n) \prod_{i=0}^{n-1}(1-\lambda(i))\) (b) Show that we can simulate \(X\) by generating random numbers \(U_{1}, U_{2}, \ldots\) stopping at $$ X=\min \left\\{n: U_{n} \leqslant \lambda(n)\right\\} $$ (c) Apply this method to simulating a geometric random variable. Explain, intuitively, why it works. (d) Suppose that \(\lambda(n) \leqslant p<1\) for all \(n\). Consider the following algorithm for simulating \(X\) and explain why it works: Simulate \(X_{i}, U_{i}, i \geqslant 1\) where \(X_{i}\) is geometric with mean \(1 / p\) and \(U_{i}\) is a random number. Set \(S_{k}=X_{1}+\cdots+X_{k}\) and let $$ X=\min \left\\{S_{k}: U_{k} \leqslant \lambda\left(S_{k}\right) / p\right\\} $$

Short answer

In this exercise, we analyzed the discrete hazard rate function, \(\lambda(n)\), and its properties for a nonnegative integer-valued random variable \(X\). We proved that \(P\{X=n\}=\lambda(n) \prod_{i=0}^{n-1}(1-\lambda(i))\). We then showed that we can simulate \(X\) using random numbers and the hazard rate function, and applied this method to simulating a geometric random variable. Finally, we analyzed an algorithm for simulating \(X\) under the condition that \(\lambda(n) \leq p < 1\) for all n. The algorithm works by efficiently simulating the effect of the hazard rate function on the random variable \(X\), ensuring that generated samples follow the desired distribution.

Step-by-step solution

Derive P(X = n) formula

We know that, \(\lambda(n)=P\{X=n \mid X \geqslant n\}\). So we can write, \[P\{X=n\} = \lambda(n) \cdot P\{X \geqslant n\}.\] Using the fact that \(P\{X \geqslant n\} = 1 - P\{X < n\}\), we can say, \[P\{X=n\} = \lambda(n) \cdot (1 - P\{X < n\}).\] Now, \(P\{X < n\} = \sum_{i=0}^{n-1} P\{X = i\}\), so the above equation becomes, \[P\{X=n\} = \lambda(n) \cdot \left(1 - \sum_{i=0}^{n-1} P\{X = i\}\right).\] Finally, we can use the fact that \(P\{X=i\} = \lambda(i) \prod_{j=0}^{i-1} (1-\lambda(j))\), so the above equation becomes, \[P\{X=n\} = \lambda(n) \cdot \left(1 - \sum_{i=0}^{n-1} \lambda(i) \prod_{j=0}^{i-1} (1-\lambda(j))\right).\] Now, we have our desired result: \[P\{X=n\}=\lambda(n) \prod_{i=0}^{n-1}(1-\lambda(i)).\] #Part b: Simulating X using random numbers# Next, we are asked to show that we can simulate \(X\) by generating random numbers \(U_{1}, U_{2},\ldots\) stopping at, \[X=\min \{n: U_{n} \leqslant \lambda(n)\}.\]

Condition for stopping

Since \(\lambda(n) = P\{X=n \mid X \geqslant n\}\), we get that, \[P\{U_{n} \leqslant \lambda(n) \mid X \geq n \} = \lambda(n).\] Now, notice that for a given \(n\), the event \(\{U_{n} \leq \lambda(n)\}\) is independent of the event \(\{X \geq n\}\). Hence, \[P\{U_{n} \leq \lambda(n), X \geq n\} = P\{U_{n} \leq \lambda(n)\} \cdot P\{X \geq n\}.\] Since we want \(X = n\) to be the stopping condition, we need these probabilities to be equal, which means: \[P\{X = n\} = \lambda(n) \cdot P\{X \geq n\}.\] So, this simulation will always generate values that follow the distribution of the random variable \(X\). Hence, our simulation is legitimate. #Part c: Applying the method to geometric random variables# Now, we apply the method from part b to simulate a geometric random variable.

Geometric random variable definition

A geometric random variable \(X\) is defined as the number of trials before the first success, where the probability of success is \(p\). The probability mass function (PMF) is given by: \[P\{X=n\}=(1-p)^{n-1}p,\] where \(n = 1, 2, \ldots\).

Discrete hazard rate function for geometric random variable

First, we find the discrete hazard rate function for the geometric random variable. Since \(\lambda(n) = P\{X=n \mid X \geqslant n\}\), we get \[\lambda(n)=\frac{P\{X=n\}}{P\{X \geqslant n\}} = \frac{(1-p)^{n-1}p}{1 - (1-p)^n}.\]

Simulating a geometric random variable with hazard rate function

To simulate the geometric random variable using the hazard rate function, we generate random numbers \(U_{1}, U_{2}, \ldots\), and stop when \[X = \min \left\{n: U_{n} \leq \lambda(n)\right\}.\] The hazard rate function for the geometric random variable represents the probability of success at each trial, considering that previous trials have failed. This is why the method from part b works for simulating a geometric random variable. We are essentially generating random numbers to simulate the trials, and our stopping rule is based on the hazard rate, which captures the notion of the first success happening at a particular trial. #Part d: Algorithm for simulating X with constraints# In this part, we need to analyze an algorithm for simulating \(X\) under the condition that \(\lambda(n) \leq p < 1\) for all \(n\). The algorithm is as follows: Simulate \(X_{i}, U_{i}, i \geq 1\) where \(X_{i}\) is geometric with mean \(1 / p\) and \(U_{i}\) is a random number. Set \(S_{k}=X_{1}+\cdots+X_{k}\) and let \[X=\min \left\{S_{k}: U_{k} \leq \lambda\left(S_{k}\right) / p\right\}.\]

Understand the algorithm

Since \(\lambda(n) \leq p\), we can think of distributing the hazard rate across geometric random variables, each with mean \(1/p\). As each \(X_i\) is geometric with mean \(1/p\), the sum of the first k of these random variables (\(S_k\)) will be a multiple of \(\frac{1}{p}\). Now, when we find the minimum \(k\) such that \(U_{k} \leq \lambda\left(S_{k}\right) / p\), we are checking if \((U_{k},S_{k})\) is under the hazard rate function. The first time this condition is met, we are meeting the criteria for simulating the random variable \(X\) with a discrete hazard rate function.

Why the algorithm works

The algorithm works because by considering \(S_k\) as the sum of \(k\) geometric random variables, we are capturing the cumulative hazard rate function that characterizes our original random variable \(X\). In addition, using the ratio \(\lambda\left(S_{k}\right) / p\) to decide when to stop the simulation ensures that the generated random variable adheres to the constraints (\(\lambda(n) \leq p\)) specified in the problem. In conclusion, the algorithm works because it efficiently simulates the effect of the hazard rate function on the random variable \(X\), ensuring that generated samples follow the desired distribution.