Question 1

μ = 1000 (mean of exponential density of individual variable )

n = 100 (number of light bulbs)

The density of minimum value among n independent random variables with an exponential density has mean \(\frac{μ}{n}\),

So the mean of the minimum value (first bulb to burn out) is:

μ = 1000
n = 100

print(μ/n)
## [1] 10

Question 2

\(fz(z) = (\frac{1}{2})e^{-λ|z|}\) can be rewritten as: \[ fz(z) = \begin{cases} (\frac{1}{2})e^{-λ|z|}, z \geq 0\\ (\frac{1}{2})e^{λ|z|}, z < 0 \end{cases} \]

Since \(X_1\) and \(X_2\) have exponential density, their PDF is:

\[ fx(x_1) = fx(x_2) = \begin{cases} λe^{-λx}, x \geq 0\\ 0, otherwise \end{cases} \]

So the combined density is \(λ^2e^{-λ(χ_1 + χ_2)}\)

If \(z = x_1 - x_2\) then \(x_1 = z + x_2\) \(=>\)

If \(z \geq 0\), then \(x_1 = (z + x_2) \geq 0\) and

\(fz(z) = \int_{z}^{∞} λ^2e^{-λ(z + 2x_2)}, dx_2 = \frac{1}{2}λe^{-λz}\)

If \(z < 0\), then \(x_1 = (z + x_2) > 0\) and

\(fz(z) = \int_{0}^{∞} λ^2e^{λ(z + 2x_2)}, dx_2 = \frac{1}{2}λe^{λz}\)

then \[ fz(z) = \begin{cases} (\frac{1}{2})e^{-λz}, z \geq 0\\ (\frac{1}{2})e^{λz}, z < 0 \end{cases} \]

Question 3

Chebyschev inequality states:

\(P(| X - μ | \geq kσ ) \leq \frac{1}{k^2}\)

\(μ\) is given as 10, with \(σ^2\) as \(\frac{100}{3}\), so \(σ = \frac{10}{\sqrt(3)}\)

Thus,

  1. \(k(\frac{10}{\sqrt(3)}) = 2 \implies k = 0.3464\)

so, \(\frac{1}{k^2} = 8.338\)

Since the upper bound cannot exceed 1, the upper bound is 1.

  1. \(k(\frac{10}{\sqrt(3)}) = 5 \implies k = 0.86602\)

so, \(\frac{1}{k^2} = 1.33334\)

Since the upper bound cannot exceed 1, the upper bound is 1.

  1. \(k(\frac{10}{\sqrt(3)}) = 9 \implies k = 1.55884\)

so, \(\frac{1}{k^2} = 0.41152\)

The upper bound is 0.41152.

  1. \(k(\frac{10}{\sqrt(3)}) = 20 \implies k = 3.4641\)

so, \(\frac{1}{k^2} = 0.08333\)

The upper bound is 0.08333.