Solution:

Under Exponential Distribution, expected value \(E[X] = \frac{1}{\lambda}\).

Proof:

If \(X\) is exponential random variable, then density \(f_{x}(x) = \lambda e^{-\lambda x}\), where \(x\ge0\).

\(E[X] = \int_0^\infty xf_x(x)dx\)

= \(\int_0^\infty x\lambda e^{-\lambda x}dx\)

= \(\bigg[-xe^{-\lambda x} \bigg]_0^\infty + \int_0^\infty e^{-\lambda x}dx\)

= \((-\infty *e^{-\lambda*\infty} - 0*e^{-\lambda*0}) + \int_0^\infty e^{-\lambda x}dx\)

Since \(e^{-\infty} = \frac{1}{e^{\infty}} = \frac{1}{\infty} = 0\)

= \((0 -0) + \bigg[-\frac{1}{\lambda}e^{-\lambda x} \bigg]_0^\infty\)

= \(0 + (-\frac{1}{\lambda}e^{\lambda * \infty} + \frac{1}{\lambda}e^{\lambda * 0})\)

= \(0 + (0 + \frac{1}{\lambda}) = \frac{1}{\lambda}\)

Hence expected value \(E[X] = \frac{1}{\lambda}\), also \(E[X] = \mu\)

Subituting values \(\mu = 1000, n = 100\) then \(\lambda = \frac{1}{1000}\)

\(min\{X_1, X_2, ....X_i\}\) is exponentially distributed by \(\sum \lambda_i = n * \lambda\)

Minimum expected = \(\frac{1}{n * \lambda} = \frac{1}{n*\frac{100}{1000}} = 10\)

Expected lifetime of the first burnt bulb is \(10\) hours.

Solution:

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

Difference of two variables \(Z = X_1 - X_2\), and we will be deriving \(f_z(z)\). Let \(X_1 = X\) and \(X_2 = Y\), then \(Z = X-Y\)

Using convolution formula of sums \(Z = X + Y\), when \(z>0\), function of \(z\) is denoted by

\(f_Z(z) = \int^\infty_{-\infty} f_X(x)f_{Y}(z-x)dx\)

With given data we are intrested in difference, \(Z = X-Y)\), this can be rewritten as \(Z = X+(-Y)\), then

\(f_Z(z) = \int^\infty_{-\infty} f_X(x)f_{-Y}(z-x)dx\)

As \(P(-Y=-1) = P(Y=1)\), \(f_{-Y}(z-x) = f_Y(x-z)\), substituting the values

\(f_Z(z) = \int^\infty_{-\infty} f_X(x)f_{Y}(x-z)dx\)

applying probability density function(pdf)

\[ f_X(x)=\begin{cases} \lambda e^{-\lambda x}, & \text{if }x\geq 0\,, \\ 0, & \text{if }x< 0\,. \end{cases} \]

When \(z<0\)

\(f_Z(z) = \int^\infty_{-\infty} f_X(x)f_{Y}(x-z)dx\)

\(f_Z(z<0) = \int^\infty_{0} \lambda e^{-\lambda x}\lambda e^{-\lambda ({x-z})}dx\)

\(f_Z(z<0) = \int^\infty_{0} \lambda e^{-\lambda x}\lambda (e^{-\lambda {x}}*e^{\lambda {z}})dx\)

\(f_Z(z<0) = \lambda e^{\lambda {z}}\int^\infty_{0} \lambda e^{-2\lambda x}dx\)

\(f_Z(z<0) = \lambda e^{\lambda {z}}\int^\infty_{0} \lambda e^{-2\lambda x}dx\)

Applying simple calculas \(\int e^{-2x}dx = -\frac{1}{2}e^{-2x}\)

\(f_Z(z<0) = \bigg(\lambda e^{\lambda {z}} *-\frac{\lambda}{2\lambda} e^{-2\lambda x}\bigg)\bigg|^\infty_0\)

\(f_Z(z<0) = \bigg(-\frac{1}{2}\lambda e^{\lambda {z}} e^{-2\lambda x}\bigg)\bigg|^\infty_0\)

solving for the values \([0,\infty], \frac{1}{e^0} = 1, \frac{1}{e^\infty} = 0\)

\(f_Z(z<0) = \bigg(-\frac{1}{2}\lambda e^{\lambda {z}} e^{-2\lambda \infty}\bigg ) - \bigg(-\frac{1}{2}\lambda e^{\lambda {z}} e^{-2\lambda * 0}\bigg )\)

\(f_Z(z<0) = \bigg(0\bigg) - \bigg(-\frac{1}{2}\lambda e^{\lambda {z}} \bigg )\)

\(f_Z(z<0) = \frac{1}{2}\lambda e^{\lambda {z}}\)

\(f_Z(z) = \frac{1}{2}\lambda e^{\lambda {z}},~ z<0\)

Since \(z = x-y, -z = y-z\),

Area under the curve will be same for \(f_Z(z) = f_Z(-z)\). We can generalize,

\(f_Z(-z) = \frac{1}{2}\lambda e^{-\lambda {z}},~ z\ge0\)

\[ f_Z(z)=\begin{cases} \frac{1}{2}\lambda e^{-\lambda {z}}, & \text{if }z\geq 0\,, \\ \frac{1}{2}\lambda e^{\lambda {z}}, & \text{if }x< 0\,. \end{cases} \]

This proves, \(Z = X_1 - X_2\) has density \(f_Z(z) = \frac{1}{2}\lambda e^{-\lambda {|z|}}\).

Solution:

According to Chebyshev’s Inequality for any random variable \(X\) with \(E(X) = \mu\) and \(V(X) = \sigma^2\), \(\epsilon = k\sigma\).

Also \(k > 1\)

Probability \(Pr(|X-\mu|\ge k\sigma) \le \frac{1}{k^2}\).

Using R code,

if (!require('knitr')) install.packages('knitr')

ci <- "Pr(|X-m|>= kv)"

output <- matrix(NA, nrow=4, ncol=4)

kv <- c(2,5,9,20)
m <- 10
sd <- sqrt(100/3)

for (i in 1:length(kv)){
  k <- kv[i]/sd
  ks <- 1/(k^2)
  
  fun <- gsub("m", m, ci)
  fun <- gsub("kv", kv[i], fun)
  con <- paste("for any random variable X, the probability of a deviation from the mean of more than", round(k,2), sep=" ") 
  con <- paste(con, "standard deviations is", sep=" ")
  con <- paste(con, round(ks,2), sep=" ")
  if (k < 1){
      con <- "Since k is less than 1, Chebyshev's Inequality is not accurate"
  }

  output[i,1] <- fun
  output[i,2] <- round(k,2)
  output[i,3] <- round(ks,2)
  output[i,4] <- con

}

kable (output,col.names = c("Function", "k", "1/k^2","Conclusion"), caption = "Chebyshev's Inequality", escape = T)
Chebyshev’s Inequality
Function k 1/k^2 Conclusion
Pr(|X-10|>= 2) 0.35 8.33 Since k is less than 1, Chebyshev’s Inequality is not accurate
Pr(|X-10|>= 5) 0.87 1.33 Since k is less than 1, Chebyshev’s Inequality is not accurate
Pr(|X-10|>= 9) 1.56 0.41 for any random variable X, the probability of a deviation from the mean of more than 1.56 standard deviations is 0.41
Pr(|X-10|>= 20) 3.46 0.08 for any random variable X, the probability of a deviation from the mean of more than 3.46 standard deviations is 0.08

References: