Question

Let \(X_{1}, X_{2}, \ldots, X_{10}\) be independent Poisson random variables with mean \(1 .\) (a) Use the Markov inequality to get a bound on \(P\left\\{X_{1}+\cdots+X_{10} \geq 15\right\\}\). (b) Use the central limit theorem to approximate \(P\left(X_{1}+\cdots+X_{10} \geq 15\right\\}\).

Short answer

In summary, we used the Markov inequality to get a bound on the probability \(P\left(X_{1}+\cdots+X_{10}\geq 15\right)\) and found that it is \(\leq \frac{2}{3}\). Then, we used the central limit theorem to approximate the same probability and found that it is approximately \(0.0139\).

Step-by-step solution

Understanding the properties of Poisson random variables

Poisson random variables have the property that their mean and variance are equal to their parameter, which is 1 for all of the given random variables in this case. Therefore, the mean and variance of each random variable are both 1. ─ Mean: \(E\left[X_i\right] = 1\) ─ Variance: \(Var\left[X_i\right] = 1\)

Calculating the mean and variance of the sum

We have 10 independent Poisson random variables, say \(S_{10} = X_{1} + X_{2} + \cdots + X_{10}\), and we need to find the mean and variance of \(S_{10}\). For independent random variables, Mean: \(E[S_{10}] = E[X_1] + E[X_2] + \cdots + E[X_{10}] = 1 + 1 + \cdots + 1 = 10\) Variance: \(Var\left[S_{10}\right] = Var[X_1] + Var[X_2] + \cdots + Var[X_{10}] = 1 + 1 + \cdots + 1 = 10\) Therefore, the mean of \(S_{10}\) is 10, and the variance of \(S_{10}\) is 10.

Step (a) 3: Applying the Markov inequality

Now that we have the mean of the sum \(S_{10}\), we can apply the Markov inequality to get a bound on the probability. The Markov inequality states that for any non-negative random variable \(X\) and any \(a > 0\), \(P\left[X \ge a\right] \le \frac{E[X]}{a}\) Here, \(a=15\), so we apply the Markov inequality to obtain: \(P\left[S_{10} \ge 15\right] \le \frac{E[S_{10}]}{15} = \frac{10}{15} = \frac{2}{3}\) Thus, using the Markov inequality, we have a bound on the probability: \(P\left(S_{10}\geq15\right) \leq \frac{2}{3}\).

Step (b) 4: Applying the Central Limit Theorem

The central limit theorem states that, for a large number of independent and identically distributed random variables, the sum or average of the variables is approximately normally distributed. In this case, \(S_{10}\) represents the sum of 10 independent Poisson random variables, so the central limit theorem applies. To use the central limit theorem, we need to standardize the sum \(S_{10}\) to a standard normal random variable \(Z\). \(Z = \frac{S_{10} - E[S_{10}]}{\sqrt{Var\left[S_{10}\right]}} = \frac{S_{10} - 10}{\sqrt{10}}\) We want the probability that \(S_{10} \ge 15\). Transforming 15 into the standard normal scale using the same formula, we get: \(Z_{15} = \frac{15 - 10}{\sqrt{10}} = \frac{5}{\sqrt{10}}\) Now, we need to find the probability that \(Z \ge Z_{15}\), i.e., \(P\left(Z \ge \frac{5}{\sqrt{10}}\right)\). Let's denote the cumulative distribution function of the standard normal distribution as Φ. The probability we are looking for is: \(P\left(Z \ge \frac{5}{\sqrt{10}}\right) = 1 - P\left(Z < \frac{5}{\sqrt{10}}\right) = 1 - Φ\left(\frac{5}{\sqrt{10}}\right)\) Using a standard normal distribution table or calculator, we can find: Φ\(\left(\frac{5}{\sqrt{10}}\right) \approx 0.9861\) So, the approximate probability is: \(P\left(Z \ge \frac{5}{\sqrt{10}}\right) \approx 1 - 0.9861 = 0.0139\) Therefore, using the central limit theorem, we find that the approximate probability of the sum \(S_{10}\) being greater than or equal to 15 is about 0.0139.