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) It is given that rv \({\rm{X}}\) denotes current gain and given as:

\({\rm{X = ln}}\left( {\frac{{{{\rm{I}}_{\rm{o}}}}}{{{{\rm{I}}_{\rm{i}}}}}} \right)\)

It is also given that \({\rm{X}}\) is normally distributed with parameters:

\(\begin{array}{l}{{\rm{\mu }}_{\rm{x}}}{\rm{ = 1}}\\{{\rm{\sigma }}_{\rm{x}}}{\rm{ = 0}}{\rm{.05}}\end{array}\)

Then by definition, ratio \(\frac{{{{\rm{I}}_{\rm{o}}}}}{{{{\rm{I}}_{\rm{i}}}}}\) will have lognormal distribution.

Definition: A continuous rv \({\rm{X}}\) is said to have a lognormal distribution if the random variable \({\rm{Y}}\) have normal distribution such that:

\({\rm{Y = ln(X)}}\)

(b) We have to calculate the probability that the output current is more than twice the input current, which can be denoted as \({\rm{P}}\left( {\frac{{{{\rm{I}}_{\rm{o}}}}}{{{{\rm{I}}_{\rm{i}}}}}{\rm{ > 2}}} \right)\)

We know that \({\rm{X}}\) is given as:

\({\rm{X = ln}}\left( {\frac{{{{\rm{I}}_{\rm{o}}}}}{{{{\rm{I}}_{\rm{i}}}}}} \right)\)

Hence if: \(\left( {\frac{{{{\rm{I}}_{\rm{o}}}}}{{{{\rm{I}}_{\rm{i}}}}}{\rm{ > 2}}} \right)\)

Then \({\rm{:X > ln(2)}}\)

\({\rm{ Hence: P}}\left( {\frac{{{{\rm{I}}_{\rm{o}}}}}{{{{\rm{I}}_{\rm{i}}}}}{\rm{ > 2}}} \right){\rm{ = P(X > ln(2)) = P(X > 0}}{\rm{.6931)}}\)

Since \({\rm{X}}\) is distributed normally with \({\rm{\mu = 1}}\) and\({\rm{\sigma = 0}}{\rm{.05}}\), hence standardizing gives:

\({\rm{X > 0}}{\rm{.6931}}\)

if and only if

\(\frac{{{\rm{X - 1}}}}{{{\rm{0}}{\rm{.05}}}}{\rm{ > }}\frac{{{\rm{0}}{\rm{.6931 - 1}}}}{{{\rm{0}}{\rm{.05}}}}\frac{{{\rm{X - 1}}}}{{{\rm{0}}{\rm{.05}}}}{\rm{ > }}\frac{{{\rm{ - 0}}{\rm{.3069}}}}{{{\rm{0}}{\rm{.05}}}}{\rm{Z > - 6}}{\rm{.138}}\)

Thus

\({\rm{P(X > 0}}{\rm{.6931) = P(Z > - 6}}{\rm{.138)}}\)

Here \({\rm{Z}}\) is a standard normal distribution rv with cdf\({\rm{f(z)}}\). Hence

\({\rm{P(X > 0}}{\rm{.6931) = P(Z > - 6}}{\rm{.138) = 1 - f( - 6}}{\rm{.138)}}\)

We check Appendix Table \({\rm{A}}{\rm{.3}}\), from there

\({\rm{f( - 6}}{\rm{.138)\gg 0}}\)

Hence

\({\rm{P(X > 0}}{\rm{.6931) = P}}\left( {\frac{{{{\rm{I}}_{\rm{o}}}}}{{{{\rm{I}}_{\rm{i}}}}}{\rm{ > 2}}} \right){\rm{\gg 1}}\)

Proposition: Let \({\rm{Z}}\) be a continuous rv with cdf\({\rm{f(z)}}\). Then for any number a,

\({\rm{P(Z > a) = 1 - f(a)}}\)

(c) We will tackle this problem using two ways. First, we will utilize the usual lognormal distribution mean and variance statement. We shall generate them using the pdf of a lognormal distribution in the second technique.

Calculation of \({\rm{V(Y)}}\) will require calculation of\({\rm{E}}\left( {{{\rm{Y}}^{\rm{2}}}} \right)\). Which is much more tedious then\({\rm{E(Y)}}\). It is similar but we will use standard result for it given in the textbook. Accordingly mean and variance of lognormal distribution are given as:

\(\begin{aligned}E(Y) &= \exp \left( {\mu + \frac{{{\sigma ^2}}}{2}} \right) \\&= \exp \left( {1 + \frac{{{{(0.05)}^2}}}{2}} \right) \\&= 2.722V(Y) = {e^{2\mu + {\sigma ^2}}} \times \left({{e^{{\sigma ^2}}} - 1} \right) \\&= {e^{2(1) + {{(0.05)}^2}}} \times \left( {{e^{{{0.05}^2}}} - 1} \right) \\&= 0.0185 \\\end{aligned} \)

Hence the expected value and variance of the ratio of output to input current are \({\rm{2}}{\rm{.722}}\)and \({\rm{0}}{\rm{.0185}}\) respectively.