91Ó°ÊÓ

The exponential distribution in probability theory and statistics is the probability distribution of the time between events in a Poisson point process, that is, a process in which events occur continuously and independently at a constant average rate. It's a special case of gamma distribution.

Random variable \({\rm{X}}\)with pdf

\({{\rm{f}}_{\rm{X}}}{\rm{(x) = }}\left\{ {\begin{array}{*{20}{l}}{{\rm{\lambda }}{{\rm{e}}^{{\rm{ - \lambda x}}}}}&{{\rm{,x}} \ge {\rm{0}}}\\{\rm{0}}&{{\rm{,x < 0}}}\end{array}} \right.\)

With parameter, is said to have an exponential distribution \({\rm{\lambda }}\).

(a):

The exponentially distributed random variable's cumulative density function is

\(\begin{array}{c}{{\rm{F}}_{\rm{X}}}{\rm{(x) = P(X}} \le {\rm{x)}}\\{\rm{ = 1 - }}{{\rm{e}}^{{\rm{ - \lambda x}}}}{\rm{,}}\;\;\;{\rm{x}} \ge {\rm{0}}\end{array}\)

The cdf of \({\rm{Y}}\)is

\(\begin{array}{l}{\rm{F(y) = P(Y}} \le {\rm{y)}}\\{\rm{ = P}}\left( {\left\{ {{X_1} \le y} \right\} \cup \left( {\left\{ {{X_2} \le y} \right\} \cap \left\{ {{X_3} \le y} \right\}} \right)} \right)\end{array}\)

where union represents world "or" when reading the event, and intersection represents "and". Let's find the probability now. The following holds

\(\begin{array}{*{20}{c}}{{\rm{F(y) = P}}\left( {\left\{ {{X_1} \le y} \right\} \cup \left( {\left\{ {{X_2} \le y} \right\} \cap \left\{ {{X_3} \le y} \right\}} \right)} \right)}&{}&{}\\{\mathop = \limits^{(1)} P\left( {{X_1} \le y} \right) + P\left( {\left\{ {{X_2} \le y} \right\} \cap \left\{ {{X_3} \le y} \right\}} \right)}&{}&{}\\{ - P\left( {\left( {\left\{ {{X_1} \le y} \right\} \cap \left\{ {{X_2} \le y} \right\} \cap \left\{ {{X_3} \le y} \right\}} \right)} \right.}&{}&{}\\{\mathop = \limits^{(2)} \left( {1 - {e^{ - \lambda y}}} \right) + {{\left( {1 - {e^{ - \lambda y}}} \right)}^2} - {{\left( {1 - {e^{ - \lambda y}}} \right)}^3},}&{}&{y > 0}\\{{\rm{F(y) = 0,}}}&{}&{y \le 0}\end{array}\)

(1): Proposition: For every two events A and B

\({\rm{P(A}} \cup {\rm{B) = P(A) + P(B) - P(A}} \cap {\rm{B)}}\)

(2) : Property of Multiplication: Two events If and only if, \({\rm{A}}\)and \({\rm{B}}\)are independent.

\(P(A \cap B){\rm{ = P(A) \times P(B)}}\)

Therefore, the cdf of \({\rm{Y}}\)

is

\({\rm{F(y) = }}\left\{ {\begin{array}{*{20}{l}}{\left( {{\rm{1 - }}{{\rm{e}}^{{\rm{ - \lambda y}}}}} \right){\rm{ + }}{{\left( {{\rm{1 - }}{{\rm{e}}^{{\rm{ - \lambda y}}}}} \right)}^{\rm{2}}}{\rm{ - }}{{\left( {{\rm{1 - }}{{\rm{e}}^{{\rm{ - \lambda y}}}}} \right)}^{\rm{3}}}}&{{\rm{,y > 0}}}\\{\rm{0}}&{{\rm{, otherwise }}}\end{array}} \right.\)

The pdf of \({\rm{Y}}\)is obtained by differentiating the cdf of \({\rm{Y}}\)as follows:

\(\begin{array}{c}{\rm{f(y) = }}{F^\prime }{\rm{(y)}}\\\mathop {\rm{ = }}\limits^{{\rm{(3)}}} {\rm{\lambda }}{{\rm{e}}^{{\rm{ - \lambda y}}}}{\rm{ + 2}}\left( {{\rm{1 - }}{{\rm{e}}^{{\rm{ - \lambda y}}}}} \right)\left( {{\rm{\lambda }}{{\rm{e}}^{{\rm{ - \lambda y}}}}} \right){\rm{ - 3}}{\left( {{\rm{1 - }}{{\rm{e}}^{{\rm{ - \lambda y}}}}} \right)^{\rm{2}}}\left( {{\rm{\lambda }}{{\rm{e}}^{{\rm{ - \lambda y}}}}} \right)\end{array}\)

\(\mathop {\rm{ = }}\limits^{{\rm{(4)}}} {\rm{4\lambda }}{{\rm{e}}^{{\rm{ - 2\lambda y}}}}{\rm{ - 3\lambda }}{{\rm{e}}^{{\rm{ - 3\lambda y}}}}\),

\({\rm{f(y) = 0}}\),

\({\rm{y > 0}}\)

\(y \le 0\),

(3) : rules for differentiating,

(4) : simplified.

the pdf of \(Y\)

is

Therefore , the function distribution is \({\rm{f(y) = }}\left\{ {\begin{array}{*{20}{l}}{{\rm{4\lambda }}{{\rm{e}}^{{\rm{ - 2\lambda y}}}}{\rm{ - 3\lambda }}{{\rm{e}}^{{\rm{ - 3\lambda y}}}}}&{{\rm{,y > 0}}}\\{\rm{0}}&{{\rm{, otherwise }}}\end{array}} \right.\)

(b):

With pdf \({\rm{f(x)}}\), the Expected Value (mean value) of a continuous random variable \({\rm{X}}\)is

\({\rm{E(X) = }}{{\rm{\mu }}_{\rm{X}}}{\rm{ = }}\int_{ - \infty }^\infty x {\rm{ \times f(x)dx}}\)

As a result, the system's estimated lifetime is

\(\begin{array}{l}{\rm{E(Y) = }}\int_{\rm{0}}^\infty {\left( {{\rm{y \times 4 \times \lambda }}{{\rm{e}}^{{\rm{ - 2\lambda y}}}}{\rm{ - 3\lambda }}{{\rm{e}}^{{\rm{ - 3\lambda y}}}}} \right)} {\rm{dy}}\\{\rm{ = }}\int_{\rm{0}}^\infty {\rm{4}} {\rm{y \times \lambda }}{{\rm{e}}^{{\rm{ - 2\lambda y}}}}{\rm{dy - }}\int_{\rm{0}}^\infty {\rm{3}} {\rm{y \times \lambda }}{{\rm{e}}^{{\rm{ - 3\lambda y}}}}{\rm{dy}}\\{\rm{ = 4\lambda }}\underbrace {\int_{\rm{0}}^\infty {\rm{y}} {{\rm{e}}^{{\rm{ - 2\lambda y}}}}{\rm{dy}}}_{{{\rm{I}}_{\rm{1}}}}{\rm{ - 3\lambda }}\underbrace {\int_{\rm{0}}^\infty {\rm{y}} {{\rm{e}}^{{\rm{ - 3\lambda y}}}}{\rm{dy}}}_{{{\rm{I}}_{\rm{2}}}}{\rm{.}}\end{array}\)

The first integral is

\({{\rm{I}}_{\rm{1}}}{\rm{ = }}\int_{\rm{0}}^ \to {\rm{y}} {{\rm{e}}^{{\rm{ - 2\lambda y}}}}{\rm{dy = }}\left| {\begin{array}{*{20}{c}}{{\rm{u = y}}}& \to &{{\rm{du = dy}}}\\{{\rm{dv = }}{{\rm{e}}^{{\rm{ - 2\lambda y}}}}{\rm{dy}}}& \to &{{\rm{v = - }}\frac{{\rm{1}}}{{{\rm{2\lambda }}}}{{\rm{e}}^{{\rm{ - 2\lambda y}}}}}\\{{\rm{ integration by parts: }}}&{{\rm{uv}}}&{{\rm{ = }}\int {\rm{v}} {\rm{du}}}\end{array}} \right|\)

\( = \left. {{\rm{y \times }}\left( {{\rm{ - }}\frac{{\rm{1}}}{{{\rm{2\lambda }}}}{{\rm{e}}^{{\rm{ - 2\lambda y}}}}} \right)} \right|_0^\infty - \int_0^\infty {\left( { - \frac{{\rm{1}}}{{{\rm{2\lambda }}}}{{\rm{e}}^{{\rm{ - 2\lambda y}}}}} \right)} {\rm{dy}}\)

\(\mathop {\rm{ = }}\limits^{{\rm{(1)}}} {\rm{0 + }}\frac{{\rm{1}}}{{{\rm{2\lambda }}}}{\rm{ \times }}\int_0^\infty {{e^{ - {\rm{2\lambda }}y}}} {\rm{dy}}\)

\({\rm{ = }}\left. {\frac{{\rm{1}}}{{{\rm{2\lambda }}}}{\rm{ \times }}\frac{{\rm{1}}}{{{\rm{2\lambda }}}}{{\rm{e}}^{{\rm{ - 2\lambda y}}}}} \right|_0^\infty \)

\({\rm{ = }}\frac{{\rm{1}}}{{{\rm{4}}{{\rm{\lambda }}^{\rm{2}}}}}\),(1)

here we used L'Hopital's rule to obtain the limit:

\(\mathop {\lim }\limits_{y \to \infty } \frac{{\rm{y}}}{{{{\rm{e}}^{{\rm{2\lambda y}}}}}}{\rm{ = }}\mathop {\lim }\limits_{y \to \infty } \frac{{\rm{1}}}{{{\rm{c \times }}{{\rm{e}}^{{\rm{2\lambda y}}}}}}{\rm{ = 0}}\)

When we derivative the denominator, we get \({\rm{c}}\), which is constant.

Similarly, we have the second interval.

\({{\rm{I}}_{\rm{1}}}{\rm{ = }}\int_{\rm{0}}^ \to {\rm{y}} {{\rm{e}}^{{\rm{ - 3\lambda y}}}}{\rm{dy = }}\left| {\begin{array}{*{20}{c}}{{\rm{u = y}}}& \to &{{\rm{du = dy}}}\\{{\rm{dv = }}{{\rm{e}}^{{\rm{ - 3\lambda y}}}}{\rm{dy}}}& \to &{{\rm{v = - }}\frac{{\rm{1}}}{{{\rm{3\lambda }}}}{{\rm{e}}^{{\rm{ - 3\lambda y}}}}}\\{{\rm{ integration by parts: }}}&{{\rm{uv}}}&{{\rm{ = }}\int {\rm{v}} {\rm{du}}}\end{array}} \right|\)

\(\begin{array}{l}{\rm{ = }}\left. {{\rm{y \times }}\left( {{\rm{ - }}\frac{{\rm{1}}}{{{\rm{2\lambda }}}}{{\rm{e}}^{{\rm{ - 3\lambda y}}}}} \right)} \right|_0^\infty - \int_0^\infty {\left( { - \frac{{\rm{1}}}{{{\rm{3\lambda }}}}{{\rm{e}}^{{\rm{ - 3\lambda y}}}}} \right)} {\rm{dy}}\\\mathop {\rm{ = }}\limits^{{\rm{(1)}}} {\rm{0 + }}\frac{{\rm{1}}}{{{\rm{3\lambda }}}}{\rm{ \times }}\int_0^\infty {{e^{{\rm{ - 3\lambda y}}}}} dy = \left. {\frac{{\rm{1}}}{{{\rm{3\lambda }}}}{\rm{ \times }}\frac{{\rm{1}}}{{{\rm{3\lambda }}}}{{\rm{e}}^{{\rm{ - 3\lambda y}}}}} \right|_0^\infty \\{\rm{ = }}\frac{{\rm{1}}}{{{\rm{9}}{{\rm{\lambda }}^{\rm{2}}}}}\end{array}\)

(1) : here we used L'Hopital's rule to obtain the limit:

\(\mathop {\lim }\limits_{y \to \infty } \frac{{\rm{y}}}{{{{\rm{e}}^{{\rm{3\lambda y}}}}}} = \mathop {\lim }\limits_{y \to \infty } \frac{{\rm{1}}}{{{\rm{c \times }}{{\rm{e}}^{{\rm{3\lambda y}}}}}}{\rm{ = 0}}\)

Finally we have

\(\begin{array}{l}{\rm{E(Y) = 4\lambda }}{{\rm{I}}_{\rm{1}}}{\rm{ - 3\lambda }}{{\rm{I}}_{\rm{2}}}\\{\rm{ = 4\lambda }}\frac{{\rm{1}}}{{{\rm{4}}{{\rm{\lambda }}^{\rm{2}}}}}{\rm{ - 3\lambda }}\frac{{\rm{1}}}{{{\rm{9}}{{\rm{\lambda }}^{\rm{2}}}}}\\{\rm{ = }}\frac{{\rm{1}}}{{\rm{\lambda }}}{\rm{ - }}\frac{{\rm{1}}}{{{\rm{3\lambda }}}}\\{\rm{ = }}\frac{{\rm{2}}}{{{\rm{3\lambda }}}}\end{array}\)

Therefore , the expected system is\({\rm{E(Y) = }}\frac{{\rm{2}}}{{{\rm{3\lambda }}}}\)