A company buys 100 lightbulbs, 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: let x1,x2…xn be independent exponential random variable.
\[ X _i = \mu. e^{-\mu x} where x \geq 0 \]
\[P(X>x) = e^{-\mu x} \]
\[ so, E(X) = 1/\ { \sum \limits_{1}^{n} \mu i } \]
X = first of these bulbs to burn out
n=100
\[\mu i=1/1000\]
i =1,2,…100
So, E(X) = 10 HRS.
Assume that X1 and X2 are independent random variables, each having an exponential density with parameter \(\lambda\). Show that Z = X1 - X2 has density
\[ f_z(Z) = (1/2) \lambda e- ^ {\lambda{|z|} } \]
ANS: The density function of \(X_1\) is \(\lambda e^{-\lambda x}\) \((for\ x\geq 0)\) , and 0 elsewhere. There is a similar expression for the density function of \(X_2\). By independence, the joint density function of \(X_1\ and\ X_2\) is \(\lambda^2 e^{-\lambda x}e^{-\lambda y}\) in the first quadrant, and 0 elsewhere.
Let \(Z=X_1-X_2\). We want to find the density function of Z. First we will find the cumulative distribution function \(F_Z(z)\) of Z, that is, the probability that \(Z\leq z\).
So we want the probability that \(X_1-X_2\leq z\).
Consider z fixed and positive, and draw the line \(x_1-x_2=z\). We want to find the probability that the ordered pair (X_1,X_2) ends up below that line or on it. The only relevant region is in the first quadrant. So let D be the part of the first quadrant that lies below or on the line \(x_1=x_2+z\). Then
We will evaluate this integral, by using an iterated integral. First we will integrate with respect to y, and then with respect to x. Note that yy travels from \(0 to x_2+z\), and then x travels from 0 to infinity. Thus
P(Z<=X) = \[\int_{0}^{\infty} - \lambda x (\int_{y=0}^{x+2} \lambda e^{-\lambda y} ) dx\]
P(Z<=X) = \[\int_{0}^{\infty} ( \lambda e^{-\lambda x} - \lambda e^{-\lambda z} e^{-2\lambda x} ) dx\]
P(Z<=X) = \[1 - \lambda/2 e^{-\lambda z} \]
For the density function f_Z(z), we get
\[F_Z(z) = \lambda/2 e^{-\lambda z} for z>=0 \]
Let X be a continuous random variable with mean \(\mu = 10\) and variance \(\sigma ^2 = 100/3\). Using Chebyshev’s Inequality, find an upper bound for the following probabilities.
Chebychev inequality:
Let X (integrable) be a random variable with finite expected value ?? and finite non-zero variance ??2. Then for any real number k > 0,
\[P(|X-\mu|) \geq k\sigma) \leq 1/k^2 \]
\(\mu =2\) ,\(\sigma ^2 = 100/3\)
Let us find k first
s = 10/sqrt(3)
#sigma
s## [1] 5.773503#value of k
k = 2/s
k## [1] 0.3464102#value of k square -- upper bound
ksq = 1/(k*k)
ksq## [1] 8.333333S0, P(|X - 10|>=2) <= 8.33 which is not true.
s = 10/sqrt(3)
#sigma
s## [1] 5.773503#value of k
k = 5/s
k## [1] 0.8660254#value of k square -- upper bound
ksq = 1/(k*k)
ksq## [1] 1.333333So, P(|X - 10|>=5) <= 1.33 which is not true
s = 10/sqrt(3)
#sigma
s## [1] 5.773503#value of k
k = 9/s
k## [1] 1.558846#value of k square -- upper bound
ksq = 1/(k*k)
ksq## [1] 0.4115226So, P(|X - 10|>=9) <= 0.4115 which is not true
s = 10/sqrt(3)
#sigma
s## [1] 5.773503#value of k
k = 20/s
k## [1] 3.464102#value of k square -- upper bound
ksq = 1/(k*k)
ksq## [1] 0.08333333So, P(|X - 10|>=20) <= 0.083 which is not true.