Homework 08

Week08 Problem Set -1
Solution

Let \(X_1\), \(X_2\), …., \(X_n\) be \(n\) independent random variables each of which has an exponential density with mean \(\mu\).

\(P(X > x)\) = \(e^{-\frac{x}{\mu}}\)

\(P\)(\(min\) (\(X_1\), \(X_2\), … \(X_n\) > \(x\)) = \(P(X_i > x))^n\) \(for\) \(i\)=\(1\),..\(n\)

= \(e^{-\frac{x}{\mu_1}}\). \(e^{-\frac{x}{\mu_2}}\) … \(e^{-\frac{x}{\mu_n}}\) = \(e^{-(\frac{1}{\mu_1} + \frac{1}{\mu_2} + \frac{1}{\mu_3}).x}\)

Since this distribution is also a exponential distribution

Therefore, \(E(X)\) = \(\frac{1}{\displaystyle\sum_{i=1}^{n} \mu_i}\)

\(n\) = \(100\), \(\mu_i\) = \(\frac{1}{1000}\) \(for\) \(i\)=\(1\),\(2\), … \(100\)

Therefore, \(E(X)\) = \(\frac{1}{\mu_1+\mu_2 + ... + \mu_{100}}\) = \(\frac{1}{\frac{1}{1000}+\frac{1}{1000} + ... + \frac{1}{1000}}\) = \(\frac{1}{\frac{100}{1000}}\)
= \(\frac{1000}{100}\) = \(10\)
Therefore \(E(X)\) = \(10\) hours 

n   <- 100
mu  <- 1/1000
sum <- 0

for (i in 1:n) {
  sum <- sum + mu
}

expected_value = 1/sum
print (expected_value)
## [1] 10
Separator - 01
Week08 Problem Set -2
Solution

\[ f(x) = \left\{ \begin{array}{ll} x & \mbox{if $x \geq 0$};\\ -x & \mbox{if $x < 0$}.\end{array} \right. \]

Let \(X_1\) and \(X_2\) are two independent random variables with exponential density with parameter \(\lambda\).

The PDF of \(X_1\) can be defined as
\[ f(x_1) = \left\{ \begin{array}{ll} \lambda e^{-\lambda x_1} & \mbox{if $x_1 \geq 0$}\\ 0 & \mbox{otherwise}.\end{array} \right. \]

The PDF of \(X_2\) can be defined as
\[ f(x_2) = \left\{ \begin{array}{ll} \lambda e^{-\lambda x_2} & \mbox{if $x_2 \geq 0$}\\ 0 & \mbox{otherwise}.\end{array} \right. \]

Since \(X_1\) and \(X_2\) are independent, the joint density function of \(X_1\) and \(X_2\) is:
\(f(x_1, x_2)\) = \(f(x_1) f(x_2)\)
\(f(x_1, x_2)\) = \(\lambda e^{-\lambda x_1}\) \(\lambda e^{-\lambda x_2}\)
\(f(x_1, x_2)\) = \(\lambda^2 e^{-\lambda (x_1+x_2)}\)

Since \(Z\) = \(X_1\) - \(X_2\) \(\Rightarrow\) \(X_1\) = \(Z\) + \(X_2\); \(X_2\) = \(V\)
The PDF of \(Z\) and \(V\) will be as follows:
\(g(z, v)\) = \(\lambda^2 e^{-\lambda (z+2v)}\) for \(v > 0\) and \(-\infty < z < \infty\)

When \(z\) is negative (\(\infty < z < 0\))
\(f_Z(z)\) = \(\int_{-z}^{\infty} g(z, v) dv\)
\(f_Z(z)\) = \(\int_{-z}^{\infty} \lambda^2 e^{-\lambda (z+2v)} dv\) = \(\frac{\lambda}{2}e^{\lambda z}\)

When \(z\) is positive (\(z > 0\))
\(f_Z(z)\) = \(\int_{0}^{\infty} g(z, v) dv\)
\(f_Z(z)\) = \(\int_{0}^{\infty} \lambda^2 e^{-\lambda (z+2v)} dv\) = \(\frac{\lambda}{2}e^{-\lambda z}\)

Thus we can define \(f_Z(z)\) as:
\[ f_Z(z) = \left\{ \begin{array}{ll} \frac{\lambda}{2}e^{\lambda z} & \mbox{$-\infty < z < 0$}\\ \frac{\lambda}{2}e^{-\lambda z} & \mbox{$0 < z < \infty$}. \end{array} \right. \] Therefore \(f_Z(z)\) = \(\frac{\lambda}{2}e^{\lambda |z|}\)

Separator - 02
Week08 Problem Set -3
Solution

The following values are given:
\(\mu\) = \(10\)
\(\sigma^2\) = \(\frac{100}{3}\)

Using Chebyshev Inequality, for any positive number \(\epsilon\) > \(0\) we have \(P(\left| X - \mu \right| \geqslant \epsilon)\) \(\leqslant\) \(\frac{\sigma^2}{\epsilon^2}\)

(a)

\(P(\left| X - 10 \right| \geqslant 2)\) \(\leqslant\) \(\frac{\sigma^2}{\epsilon^2}\)
\(P(\left| X - 10 \right| \geqslant \epsilon)\) \(\leqslant\) \(\frac{\sigma^2}{\epsilon^2}\)
\(\epsilon\) = \(2\)
\(P(\left| X - 10 \right| \geqslant 2)\) \(\leqslant\) \(\frac{\frac{100}{3}}{(2)^2}\) = \(\frac{100}{12}\) = \(8.3333\)
Since the probability cannot be more than 1.
\(P(\left| X - 10 \right| \geqslant 2)\) = \(1\)

(b)

\(P(\left| X - 10 \right| \geqslant 5)\) \(\leqslant\) \(\frac{\sigma^2}{\epsilon^2}\)
\(P(\left| X - 10 \right| \geqslant \epsilon)\) \(\leqslant\) \(\frac{\sigma^2}{\epsilon^2}\)
\(\epsilon\) = \(5\)
\(P(\left| X - 10 \right| \geqslant 5)\) \(\leqslant\) \(\frac{\frac{100}{3}}{(5)^2}\) = \(\frac{100}{75}\) = \(1.3333\)
Since the probability cannot be more than 1.
\(P(\left| X - 10 \right| \geqslant 5)\) = \(1\)

(c)

\(P(\left| X - 10 \right| \geqslant 9)\) \(\leqslant\) \(\frac{\sigma^2}{\epsilon^2}\)
\(P(\left| X - 10 \right| \geqslant \epsilon)\) \(\leqslant\) \(\frac{\sigma^2}{\epsilon^2}\)
\(\epsilon\) = \(9\)
\(P(\left| X - 10 \right| \geqslant 9)\) \(\leqslant\) \(\frac{\frac{100}{3}}{(9)^2}\) = \(\frac{100}{243}\) = \(0.4115\)
\(P(\left| X - 10 \right| \geqslant 9)\) = \(0.4115\)

(d)

\(P(\left| X - 10 \right| \geqslant 20)\) \(\leqslant\) \(\frac{\sigma^2}{\epsilon^2}\)
\(P(\left| X - 10 \right| \geqslant \epsilon)\) \(\leqslant\) \(\frac{\sigma^2}{\epsilon^2}\)
\(\epsilon\) = \(20\)
\(P(\left| X - 10 \right| \geqslant 20)\) \(\leqslant\) \(\frac{\frac{100}{3}}{(20)^2}\) = \(\frac{100}{1200}\) = \(0.08333\)
\(P(\left| X - 10 \right| \geqslant 20)\) = \(0.08333\)