Assignment 8

Week 8, Sums of RV

Fundamentals of Computational Mathematics

CUNY MSDS DATA 605, Fall 2018

Rose Koh

10/21/2018

#11 and #14 on page 303 of probability text

The exercise 10 illustrates that th expected value for the density of the minimum value is \(\frac{\mu}{n}\) where \(n\) is independent random variable with an exponential density of mean \(\mu\).

The expected life is 1000 hours, thus \(\lambda = \frac{1}{1000}\)

A Company buys 100 lightbulbs, thus \(100* \lambda = \frac{1}{10}\)

This distribution is exponential with mean \(\frac{1000}{100} = 10\) The expected minimum is 10 hours.

\[f_Z(z) = \frac{1}{2}\lambda e^{-\lambda |z|}\]

The exponential probability density function (pdf) is:

\[f_{X1}(x) = f_{X2}(x) = \begin{cases} \lambda e^{-\lambda x} & \text{where } x \geq 0, \\ 0, & \text{otherwise. } \\ \end{cases}\]

Since \(x_1\) and \(x_2\) are two independent variables, the joint density function can be calculated as:

\[f(x_1, x_2) = f(x_1)f(x_2) = \lambda e^{-\lambda x_1}\lambda e^{-\lambda x_2} = \lambda^2 e^{-\lambda(x_1 + x_2)}\]

Since \(Z = X_1 − X_2\), we know that \(x_1 = z + x_2\)

Now, we can use substition to get the joint density of \(Z\) and \(X_2\), which is \(\lambda^2 e^{-\lambda((z + x_2) + x_2)}\) or \(\lambda^2 e^{-\lambda(z + 2x_2)}\)

We know that \(x_2 = x_1 - z\).

If \(z < 0\), \(x_2\) is greater than \(-z\)

If \(z \ge 0\), then \(x_2\) is positive.

Using WolframAlpha,

When z is negative, \(\int_{-z}^{\infty} \lambda^2 e^{-\lambda(z + 2x_2)} dx = \frac{\lambda}{2} e^{\lambda z}\).

When z is positive, \(\int_{0}^{\infty} \lambda^2 e^{-\lambda(z + 2x_2)} dx = \frac{\lambda}{2} e^{-\lambda z}\).

Combining two sides we get \(f_Z(z) = \begin{cases} \frac{1}{2}e^{-\lambda z}, & \mbox{if } z \ge 0, \\ \frac{1}{2}e^{\lambda z}, & \mbox{if }z <0. \end{cases}\)

#1 on page 320-321

Let X be a continuous random variable with mean µ = 10 and variance \(σ^2 = \frac{100}{3}\). Using Chebyshev’s Inequality, find an upper bound for the following probabilities.

  1. P(|X − 10| ≥ 2).

  2. P(|X − 10| ≥ 5).

  3. P(|X − 10| ≥ 9).

  4. P(|X − 10| ≥ 20).

Chebyshev Inequality

For any positive number \(\epsilon > 0\),

\[P(|X - \mu| \geq \epsilon) \leq \frac{\sigma^2}{\epsilon^2}\]

  1. P(|X − 10| ≥ 2) ≤ \(\frac{\frac{100}{3}}{4}\)

Since probability cannot be greater than 1, the upper bound is 1.

  1. P(|X − 10| ≥ 5) ≤ \(\frac{\frac{100}{3}}{25}\)

Since probability cannot be greater than 1, the upper bound is 1.

  1. P(|X − 10| ≥ 9) ≤ \(\frac{\frac{100}{3}}{81} \approx\) 0.4115226

  2. P(|X − 10| ≥ 20) ≤ \(\frac{\frac{100}{3}}{400} \approx\) 0.0833333