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Uniform distributions are probability distributions in which all events are equally likely. A discrete uniform distribution has discrete outcomes with the same probability. The outcomes of a continuous uniform distribution are both continuous and infinite.

\({{\rm{f}}_{\rm{u}}}{\rm{(u) = }}\left\{ {\begin{aligned}{{}{}}{{\rm{1 0}} \le {\rm{u}} \le {\rm{1}}}\\{{\rm{0 otherwise\;}}}\end{aligned}} \right.\)

It is also assumed that X is a linear function of U in the following way:

\({\rm{X = }}\frac{{{\rm{ - 1}}}}{{\rm{\lambda }}}{\rm{ \times ln(1 - U)}}\)

When we write the cdf of U and X as \({{\rm{F}}_{\rm{u}}}{\rm{(u)}}\) and \({{\rm{F}}_{\rm{x}}}{\rm{(x)}}\), we may write the cdf of X as:

\({{\rm{F}}_{\rm{x}}}{\rm{(x) = P(X}} \le {\rm{x)}}\)

\({\rm{ = P}}\left( {\frac{{{\rm{ - 1}}}}{{\rm{\lambda }}}{\rm{ \times ln(1 - u)}} \le {\rm{x}}} \right)\)

\({\rm{ = P}}\left( {\frac{{\rm{1}}}{{\rm{\lambda }}}{\rm{ \times ln(1 - u)}} \ge {\rm{ - x}}} \right)\)

\({\rm{ = P(ln(1 - u)}} \ge {\rm{ - \lambda x)}}\)

\({\rm{ = P}}\left( {{\rm{1 - u}} \ge {{\rm{e}}^{{\rm{ - \lambda x}}}}} \right)\)

\({\rm{ = P}}\left( {{\rm{u}} \le {\rm{1 - }}{{\rm{e}}^{{\rm{ - \lambda x}}}}} \right)\)

\({{\rm{F}}_{\rm{x}}}{\rm{(x) = }}{{\rm{F}}_{\rm{u}}}\left( {{\rm{1 - }}{{\rm{e}}^{{\rm{ - \lambda x}}}}} \right)\)

The pdf of X is denoted by \({{\rm{F}}_{\rm{x}}}{\rm{(x)}}\).

\({{\rm{f}}_{\rm{x}}}{\rm{(x) = }}\frac{{\rm{d}}}{{{\rm{dx}}}}{{\rm{F}}_{\rm{x}}}{\rm{(x) = }}\frac{{\rm{d}}}{{{\rm{dx}}}}{{\rm{F}}_{\rm{u}}}\left( {{\rm{1 - }}{{\rm{e}}^{{\rm{ - \lambda x}}}}} \right)\)

\({{\rm{f}}_{\rm{x}}}{\rm{(x) = }}\left( {{\rm{\lambda \times }}{{\rm{e}}^{{\rm{ - \lambda x}}}}} \right){\rm{ \times }}{{\rm{f}}_{\rm{u}}}\left( {{\rm{1 - }}{{\rm{e}}^{{\rm{ - \lambda x}}}}} \right)\)

As can be seen:

\({\rm{0}} \le {\rm{1 - }}{{\rm{e}}^{{\rm{ - \lambda x}}}} \le {\rm{1}}\;\;\;{\rm{\;for\;x}} \ge {\rm{0}}\)

\({\rm{1 - }}{{\rm{e}}^{{\rm{ - \lambda x}}}}{\rm{ < 0}}\;\;\;{\rm{\;for\;x < 0}}\)

Hence

\({f_u}\left( {1 - {e^{ - \lambda x}}} \right) = \left\{ {\begin{aligned}{{}{}}{{\rm{1 x}} \ge {\rm{0}}}\\{{\rm{0 x < \;0}}}\end{aligned}} \right.\)

Finally, considering the expressions \({{\rm{f}}_{\rm{u}}}\left( {{\rm{1 - }}{{\rm{e}}^{{\rm{ - \lambda x}}}}} \right)\) and eqn. (2), we can write:

\({{\rm{f}}_{\rm{x}}}{\rm{(x) = }}\left\{ {\begin{aligned}{{}{}}{{\rm{\lambda \times }}{{\rm{e}}^{{\rm{ - \lambda x}}}}{\rm{ x}} \ge {\rm{0}}}\\{{\rm{0 x < 0}}}\end{aligned}} \right.\)

We can deduce from the above equation that X is exponentially distributed with the parameter \({\rm{\lambda }}\).

(b) Because each random number has an equal chance of being chosen, the random numbers generated by the generator will have a uniform distribution.

Let's call the sequence of random numbers \({{\rm{u}}_{\rm{1}}}{\rm{,}}{{\rm{u}}_{\rm{2}}}{\rm{,}}{{\rm{u}}_{\rm{3}}}{\rm{,}}{{\rm{u}}_{\rm{4}}}{\rm{ \ldots \ldots \ldots }}{\rm{.}}\)

\({\rm{\lambda = 10}}\)is given.

Then do the following: \({{\rm{x}}_{\rm{i}}}{\rm{ = }}\frac{{{\rm{ - 1}}}}{{{\rm{10}}}}{\rm{ \times ln}}\left( {{\rm{1 - }}{{\rm{u}}_{\rm{i}}}} \right)\) for \({\rm{i = 1,2,3,4 \ldots }}{\rm{.}}\)

Then comes the \({{\rm{x}}_{\rm{1}}}{\rm{,}}{{\rm{x}}_{\rm{2}}}{\rm{,}}{{\rm{x}}_{\rm{3}}}{\rm{,}}{{\rm{x}}_{\rm{4}}}{\rm{ \ldots \ldots \ldots }}{\rm{.}}\) sequence of numbers with the parameter \({\rm{\lambda = 10}}\), will have an exponential distribution.