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 to us that \({\rm{t}}\) is the amount of sales tax a retailer owes the government for a certain period. It is also given that \({\rm{t}}\) is a normally distributed random variable with mean value \({\rm{\mu }}\)and standard deviation\({\rm{\sigma }}\).

\({\rm{T\gg N(\mu ,\sigma )}}\)

The proposed loss function is given as:

\(L(a,t) = \left\{ {\begin{array}{*{20}{l}}{k(a - t)}&{t£a}\\{t - a}&{t > a}\end{array}} \right.\)

Then the expected value of loss function can be written as:

\(\begin{aligned}E(T) &= \int_{ - ¥}^a k (a - t) \times f(t) \times dt + \int_a^¥ {(t - a)} \times f(t) \times dt \hfill \\&= ka\int_{ - ¥}^a f (t) \times dt - k\int_{ - ¥}^a t \times f(t) \times dt + \int_a^¥t \times f(t) \times dt - a\int_a^¥ f (t) \times dt \hfill \\\end{aligned} \)

Now to get the maximum value of the expected value as \({\rm{a}}\) is varied, we differentiate the expression of \({\rm{E(T)}}\) with respect to\({\rm{a}}\). We make use of following propositions:

Proposition-I: If \({\rm{c(x)}}\) and \({\rm{b(x)}}\)are functions of\({\rm{x}}\), then

\(\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {\int_{{\rm{c(x)}}}^{{\rm{b(x)}}} {\rm{f}} {\rm{(t) \times dt}}} \right){\rm{ = f(c(x)) \times }}\frac{{\rm{d}}}{{{\rm{dx}}}}{\rm{b(x) - f(b(x)) \times }}\frac{{\rm{d}}}{{{\rm{dx}}}}{\rm{c(x)}}\)

Multiplication rule: If \({\rm{g(x)}}\) and \({\rm{f(x)}}\) are functions of \({\rm{x}}\), then

\(\frac{{\rm{d}}}{{{\rm{dx}}}}{\rm{(g(x) \times f(x)) = g(x) \times }}\frac{{\rm{d}}}{{{\rm{dx}}}}{\rm{f(x) + f(x) \times }}\frac{{\rm{d}}}{{{\rm{dx}}}}{\rm{g(x)}}\)

Using these we differentiate both side in the expression of\({\rm{E(T)}}\):

\(\begin{aligned}E(T)&= ka\int_{ - ¥}^a f (t) \times dt - k\int_{ - ¥}^a t \times f(t) \times dt + \int_a^¥ t \times f(t) \times dt - a\int_a^¥ f (t) \times dt \\ E(T) &= ka \times F(a) - k\int_{ - ¥}^a t \times f(t) \times dt + \int_a^¥ t \times f(t) \times dt - a(1 - F(a))\frac{d}{{da}} \\E(T) &= \left[ {k \times F(a) + ka \times {F^¢}(a)} \right] - k\left[ {a \times f(a) \times \frac{d}{{da}}(a)} \right] + \left[ {a \times f(a) \times \frac{d}{{da}}(a)} \right] - \left[ {1 - F(a) - a \times {F^¢}(a)} \right] \\&= k \times F(a) + ka \times f(a) - ka \times f(a) + a \times f(a) - 1 + F(a) + a \times f(a) \\&= (k + 1) \times F(a) - 1 \\\end{aligned} \)

If \({\rm{E(T)}}\) is maximum at this critical value\({{\rm{a}}^{\rm{*}}}\), then: \({\left( {\frac{{\rm{d}}}{{{\rm{da}}}}{\rm{E(T)}}} \right)_{{\rm{a = }}{{\rm{a}}^{\rm{*}}}}}{\rm{ = 0}}\)

\(\begin{aligned}(k + 1) \times F\left( {{a^*}} \right) - 1 &= 0 \hfill \\F\left( {{a^*}} \right) &= \frac{1}{{k + 1}} \hfill \\\end{aligned} \)

Now we use the fact that\({\rm{T}}\)is normally distributed, hence it's cdf can be written in terms of cdf of standard normal distribution variable. i.e., \({\rm{f(z)}}\)

\({\rm{F}}\left( {{{\rm{a}}^{\rm{*}}}} \right){\rm{ = f}}\left( {\frac{{{{\rm{a}}^{\rm{*}}}{\rm{ - \mu }}}}{{\rm{\sigma }}}} \right)\)

Now equating both values of \({\rm{F(a*)}}\) :

\(\begin{aligned}f\left( {\frac{{{a^*} - \mu }}{\sigma }} \right) &= \frac{1}{{k + 1}}\frac{{{a^*} - \mu }}{\sigma } \\&= {f^{ - 1}}\left( {\frac{1}{{k + 1}}} \right) \\{a^*} &= \mu + \sigma \times {f^{ - 1}}\left( {\frac{1}{{k + 1}}} \right) \\\end{aligned} \)

(b) Then optimal value of \({\rm{a}}\)is:

\(\begin{aligned}{a^*} &= 100000 + (10000) \times {f^{ - 1}}\left( {\frac{1}{{2 + 1}}} \right) \\&= 100000 + (10000) \times {f^{ - 1}}(0.3333) \\\end{aligned} \)

Now using Appendix, A-\({\rm{3}}\), we observe that the z-value corresponding to probability of \({\rm{0}}{\rm{.3333}}\) is approximately\({\rm{ - 0}}{\rm{.43}}\)

\(\begin{array}{l}{{\rm{a}}^{\rm{*}}}{\rm{ = 100000 + (10000) \times ( - 0}}{\rm{.43)}}\\{{\rm{a}}^{\rm{*}}}{\rm{ = 95700}}\end{array}\)

Now over-assessment occurs if\({\rm{t < 95700}}\), hence the probability of over-assessment is represented as\({\rm{P(t < 95700)}}\).

\({\rm{P(t < 95700) = F(95700)}}\)

We already have derived following relation: \({\rm{F(a) = }}\frac{{\rm{1}}}{{{\rm{k + 1}}}}\), Hence

\({\rm{P(t < 95700) = }}\frac{{\rm{1}}}{{\rm{3}}}\)

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

\({\rm{P(Z < a) = F(a)}}\)