Problem 1

Q) A company buys 100 light bulbs, each of which has an exponential lifetime of 1000 hours. What is the expected time for the first of these bulbs to burn out?

Ans)

# Manual calculation:

# The expected life is 1000 hours so lambda = 1/1000.
# If 100 are bought then it becomes 100* lambda = 1/10.
# The expected time  is 1/1/10 which is 10 hours.

# R code :

mu <- 1000
n <- 100
(Exp_time_hrs <- mu/n)
## [1] 10

Problem 2

Q) Assume that X1 and X2 are independent random variables, each having an exponential density with paramater λ. Show that Z=X1−X2 has density fZ(z)=(1/2).λ.e^−λ|z|

Ans) The densities of the three random variables X1, X2, and Z can be represented by fX1, fX2, and fZ, respectively.

fX1=λ.e−λx1 for X1≥0, and 0 elsewhere.

fX2=λ.e−λx2 for X2≥0, and 0 elsewhere.

Then, the convolution of X1 and X2 is

fZ(z)=∫+∞−∞ fX1(z−x1) fX2(x1).dx2

=∫∞0 λe^−λ(z+2x2).λ e^−λ(z+2x2).dx1

=∫∞0 λ^2 e^−λ(z+2x2) e^λz .dx1 =λ/2 e^λz When z is negative

=∫∞−z λ^2 e^−λ(z+2x2) e^λz .dx1 =λ/2 e^−λz When z is positive

=1/2.λ.e^ λ|z|

Problem 3

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

Ans) Chebyshev’s Inequality states the following. Suppose that μ = E(X) and σ^2 = V(X), then for any positive number ϵ > 0 we have

P(|X−μ|≥kσ) ≤ σ2/k2σ2=1/k2

(a) P(|X − 10|>= 2)

Ans)

#sigma
(s = 10/sqrt(3))
## [1] 5.773503
#value of k
(k = 2/s)
## [1] 0.3464102
#Upper boundary
(ksq = 1/(k^2))
## [1] 8.333333

Since k is less than 1, Chebyshev’s Inequality is not accurate

(b) P(|X − 10|>= 5)

Ans)

#sigma
(s = 10/sqrt(3))
## [1] 5.773503
#value of k
(k = 5/s)
## [1] 0.8660254
#Upper boundary
(ksq = 1/(k^2))
## [1] 1.333333

Since k is less than 1, Chebyshev’s Inequality is not accurate

(c) P(|X − 10|>= 9)

Ans)

#sigma
(s = 10/sqrt(3))
## [1] 5.773503
#value of k
(k = 9/s)
## [1] 1.558846
#Upper boundary
(ksq = 1/(k^2))
## [1] 0.4115226

For any random variable X, the probability of a deviation from the mean of more than 1.56 standard deviations is 0.41

(d) P(|X − 10|>= 20)

Ans)

#sigma
(s = 10/sqrt(3))
## [1] 5.773503
#value of k
(k = 20/s)
## [1] 3.464102
#Upper boundary
(ksq = 1/(k^2))
## [1] 0.08333333

For any random variable X, the probability of a deviation from the mean of more than 3.46 standard deviations is 0.08