91Ó°ÊÓ

Probability simply refers to the likelihood of something occurring. We may talk about the probabilities of particular outcomes—how likely they are—when we're unclear about the result of an event. Statistics is the study of occurrences guided by probability.

a) Given: X has pdf

\()f\left( {x;{\lambda _1},{\lambda _2},p} \right) = \left\{ {\begin{array}{*{20}{c}}{p{\lambda _1}{e^{ {\lambda _1}x}} + (1 - p){\lambda _2}{e^{ - {\lambda _2}x}}}&{x³0} \\0&{{\text{ }}otherwise{\text{ }}}\end{array}} \right.\)

Since \({\lambda _1}{e^{ - {\lambda _1}x}}\) is the pdf of an exponential distribution with parameter \({\lambda _1}\) and \({\lambda _2}{e^{ - {\lambda _2}x}}\) is the pdf of an exponential distribution with parameter \({\lambda _2}\), we know:

\(\begin{gathered}X = p{X_1} + (1 - p){X_2} \hfill \\{X_1}\~Exponential\left( {{\lambda _1}} \right) \hfill \\

{X_2}\~Exponential\left( {{\lambda _2}} \right) \hfill \\\end{gathered} \)

The expected value of an exponential distribution is the reciprocal of its parameter \(\lambda \):

\(\begin{aligned} E\left( {{X_1}} \right) &= \frac{1}{{{\lambda _1}}} \hfill \\E\left( {{X_2}} \right) &= \frac{1}{{{\lambda _2}}} \hfill \\\end{aligned} \)

The variance of an exponential distribution is the reciprocal of the square of its parameter \(\lambda \):

\(\begin{aligned} V\left( {{X_1}} \right) &= \frac{1}{{\lambda _1^2}} \hfill \\ V\left( {{X_2}} \right) &= \frac{1}{{\lambda _2^2}} \hfill \\\end{aligned} \)

For the linear combination \(W = a{X_1} + b{X_2}\), the mean and variance have the following properties:

\(\begin{aligned}E(W) &= aE\left( {{X_1}} \right) + bE\left( {{X_2}} \right) \hfill \\V(W) &= {a^2}V\left( {{X_1}} \right) + {b^2}V\left( {{X_2}} \right) \hfill \\\end{aligned} \)

Use these properties with a=p and b=1-p :

\(\begin{aligned}E(X) = E\left( {p{X_1} + (1 - p){X_2}} \right) \\ &= pE\left( {{X_1}} \right) + (1 - p)E\left( {{X_2}} \right) \\ &= \frac{p}{{{\lambda _1}}} + \frac{{1 - p}}{{{\lambda _2}}} \\ V(X) = V\left( {p{X_1} + (1 - p){X_2}} \right) \\&= {p^2}V\left( {{X_1}} \right) + {(1 - p)^2}V\left( {{X_2}} \right) \\ &= \frac{{{p^2}}}{{\lambda _1^2}} + \frac{{{{(1 - p)}^2}}}{{\lambda _2^2}} \\\end{aligned} \)

b) Given: has pdf

Sinceis the pdf of an exponential distribution with parameterandis the pdf of an exponential distribution with parameter, we know:

The cumulative distribution function of an exponential distribution with parameteris:

The cumulative distribution function ofis then: