Question

Consider the power calculation done in Example 9.5.5.

a. Simulate \({v_0} = 1000\) i.i.d. noncentral t pseudo-random variables with 14 degrees of freedom and noncentrality parameter \(1.936.\)

b. Estimate the probability that a noncentral t random variable with 14 degrees of freedom and noncentrality parameter \(1.936\) is at least \(1.761.\) Also, compute the standard simulation error.

c. Suppose that we want our estimator of the noncentral t probability in part (b) to be closer than \(0.01\) the true value with probability \(0.99.\) How many noncentral t random variables do we need to simulate?

Short answer

(a) Use\(\,T = X/(\sqrt Y /df).\)

(b) Estimate of the probability 0.58; Simulation standard error: 0.0049 for\(n = 10000\)

(c) n=10094.

Step-by-step solution

(a) To find the value of random numbers.

One way to generate a sample size \({\nu _0} = 1000\) from a student t distribution with a non-centrality parameter is to use a build-in function \(r t(n, d f,\) non-centrality). However, this exercise aims to simulate it using only the noncentral student $t$ distribution and another distribution. The second one that shall be used is a chi-square distribution. Let

\(T = \frac{X}{{\sqrt {Y/df} }}\)

where X is a standard random variable with\(\mu = 1.936\)(the noncentrality parameter) and standard deviation 1, and Y is a chi-square random variable with 14 degrees of freedom. Such random variable T students t distribution with 14 degrees of freedom and noncentrality parameter\(\mu = 1.936.\)

Then, the simulation generates random samples from the distributions Z and V. The code below generates such random numbers.

(b) To find the simulation standard error gets higher

To estimate the probability, after generating the random samples\({{\rm T}^{*\iota }},\iota = 1,2, \ldots ,\nu \) , see how many of those are more significant, and it will be the estimate of the probability. The code below gives the estimates of the probability close to 0.58. The simulation standard error is around \(0.0049\) for n=10000. For lower n, the simulation standard error gets higher.

(c) To find the simulation sample size.

Lastly, the necessary simulation sample size may be found from

As \(\sigma \) is unknown in this case, use the estimate which is \(1.19.\) Here, \(\gamma = 0.99\) and Remember that \(\Phi \) is the cdf of the standard normal distribution. Then, the necessary sample size which satisfies the requirements is

\(\begin{aligned}{c}n = {\Phi ^{ - 1}}\frac{{1 + \gamma }}{2}\frac{{{s^2}}}{ \in }\\ = {\Phi ^{ - 1}}\frac{{1 + 0.99}}{2}\frac{{1.1{9^2}}}{{0.01}}\\ = 10094\end{aligned}\)

However, the simulation sample size will differ every time you run the code (unless you place "seed" before).