91影视

The number of values below 'x' divided by the total number of values yields the percentile.

(a)Assume that\({\rm{X}}\)has a uniform distribution on the interval\({\rm{(A,B)}}\). As a result,\({\rm{f(x)}}\)is a uniform depth distribution on the interval\({\rm{(A,B)}}\). In this interval, the value of\({\rm{f(x)}}\)is equal to:

\({\rm{f(x) = }}\frac{{\rm{1}}}{{{\rm{B - A}}}}\)

Otherwise, a zero. Then write\({\rm{f(x)}}\)as:

\({\rm{f(x) = }}\left\{ {\begin{array}{*{20}{l}}{\frac{{\rm{1}}}{{{\rm{B - A}}}}}&{{\rm{A}} \le {\rm{x}} \le {\rm{B}}}\\{\rm{0}}&{{\rm{ otherwise }}}\end{array}} \right.\)

For each x between A and B,

\(\begin{aligned}F(X) &= \int_{\rm{A}}^{\rm{x}} {\frac{{\rm{1}}}{{{\rm{B - A}}}}} {\rm{ \times dy}}\\&= \left( {\frac{{\rm{y}}}{{{\rm{B - A}}}}} \right)_{\rm{A}}^{\rm{x}}\\F(X) &=\frac{{{\rm{x - A}}}}{{{\rm{B - A}}}}\end{aligned}\)

As a result,\({\rm{F(X)}}\)can be written as:

\({\rm{F(x) = }}\left\{ {\begin{array}{*{20}{l}}{\frac{{{\rm{x - A}}}}{{{\rm{B - A}}}}}&{{\rm{A}} \le {\rm{x}} \le {\rm{B}}}\\{\rm{0}}&{{\rm{ otherwise }}}\end{array}} \right.\)

Recall that the percentile of any given distribution is defined as follows: Let p be a positive integer between \({\rm{0}}\) and \({\rm{1}}\). The \({{\rm{(100p)}}^{{\rm{th}}}}\) percentile of a continuous \({\rm{rv}}\) X's distribution (denoted by \({{\rm{\eta }}_{\rm{p}}}\)) is defined by

\(\begin{array}{c}{\rm{p = F}}\left( {{{\rm{\eta }}_{\rm{p}}}} \right)\\{\rm{ = }}\int_{{\rm{ - }}\infty }^{{{\rm{\eta }}_{\rm{p}}}} {\rm{f}} {\rm{(y)dy}}\end{array}\)

We can state that for any\({{\rm{(100p)}}^{{\rm{th}}}}\)percentile (let us call it\({{\rm{\eta }}_{\rm{p}}}\)):

\(\begin{array}{c}{\rm{F}}\left( {{{\rm{\eta }}_{\rm{p}}}} \right){\rm{ = p}}\\\frac{{{{\rm{\eta }}_{\rm{p}}}{\rm{ - A}}}}{{{\rm{B - A}}}}{\rm{ = p}}\\{{\rm{\eta }}_{\rm{p}}}{\rm{ - A = p(B - A)}}\\{{\rm{\eta }}_{\rm{p}}}{\rm{ = A + p(B - A)}}\end{array}\)

Therefore, \({{\rm{\eta }}_{\rm{p}}}{\rm{ = A + p(B - A)}}\).

(b)The following is a formula for calculating the mean depth value:

\(\begin{aligned}E(X) &= \int_{{\rm{ - }}\infty }^\infty {\rm{x}} {\rm{ \times f(x) \times dx}}\\ &= \int_{\rm{A}}^{\rm{B}} {\rm{x}} {\rm{ \times }}\frac{{\rm{1}}}{{{\rm{B - A}}}}{\rm{ \times dx}}\\ &= \frac{{\rm{1}}}{{{\rm{B - A}}}}\int_{\rm{A}}^{\rm{B}} {\rm{x}} {\rm{ \times dx}}\\&= \frac{{\rm{1}}}{{{\rm{B - A}}}}\left( {\frac{{{{\rm{x}}^{\rm{2}}}}}{{\rm{2}}}} \right)_{\rm{A}}^{\rm{B}}\\&= \frac{{\rm{1}}}{{{\rm{B - A}}}}\left( {\frac{{{{\rm{B}}^{\rm{2}}}}}{{\rm{2}}}{\rm{ - }}\frac{{{{\rm{A}}^{\rm{2}}}}}{{\rm{2}}}} \right)\\&= \frac{{{{\rm{B}}^{\rm{2}}}{\rm{ - }}{{\rm{A}}^{\rm{2}}}}}{{{\rm{2(B - A)}}}}\\ &= \frac{{{\rm{(B - A)(B + A)}}}}{{{\rm{2(B - A)}}}}\\E(X) &= \frac{{{\rm{(A + B)}}}}{{\rm{2}}}\end{aligned}\)

The expected or mean value of a continuous\({\rm{rv}}\)X with pdf\({\rm{f(x)}}\)is defined as

\(\begin{aligned}\mu &= E(X) \\ &= \int_{{\rm{ - }}\infty }^\infty {\rm{x}} {\rm{ \times f(x) \times dx}}\end{aligned}\)

To calculate variance for the pdf\({\rm{f(x)}}\), we first calculate\({\rm{E}}\left( {{{\rm{X}}^{\rm{2}}}} \right)\):

\(\begin{aligned}{\rm{E}}\left( {{{\rm{X}}^{\rm{2}}}} \right) &= \int_{{\rm{ - }}\infty }^\infty {{{\rm{x}}^{\rm{2}}}} {\rm{ \times f(x) \times dx}}\\ &= \int_{\rm{A}}^{\rm{B}} {{{\rm{x}}^{\rm{2}}}} {\rm{ \times }}\frac{{\rm{1}}}{{{\rm{B - A}}}}{\rm{ \times dx}}\\&= \frac{{\rm{1}}}{{{\rm{B - A}}}}\int_{\rm{A}}^{\rm{B}} {{{\rm{x}}^{\rm{2}}}} {\rm{ \times dx}}\\&=\frac{{\rm{1}}}{{{\rm{B - A}}}}\left( {\frac{{{{\rm{x}}^{\rm{3}}}}}{{\rm{3}}}} \right)_{\rm{A}}^{\rm{B}}\\&= \frac{{\rm{1}}}{{{\rm{B - A}}}}\left( {\frac{{{{\rm{B}}^{\rm{3}}}}}{{\rm{3}}}{\rm{ - }}\frac{{{{\rm{A}}^{\rm{3}}}}}{{\rm{3}}}} \right)\\& = \frac{{{{\rm{B}}^{\rm{3}}}{\rm{ - }}{{\rm{A}}^{\rm{3}}}}}{{{\rm{3(B - A)}}}}\\&= \frac{{{\rm{(B - A)}}\left( {{{\rm{B}}^{\rm{2}}}{\rm{ + AB + }}{{\rm{A}}^{\rm{2}}}} \right)}}{{{\rm{3(B - A)}}}}\\{\rm{E}}\left( {{{\rm{X}}^{\rm{2}}}} \right) & = \frac{{{{\rm{B}}^{\rm{2}}}{\rm{ + AB + }}{{\rm{A}}^{\rm{2}}}}}{{\rm{3}}}\end{aligned}\)

We apply the following proposition because we had calculated\({\rm{E(X)}}\)in the previous section:\({\rm{E(X) = }}\frac{{{\rm{A + B}}}}{{\rm{2}}}\)

\({\rm{V(X) = E}}\left( {{{\rm{X}}^{\rm{2}}}} \right){\rm{ - E(X}}{{\rm{)}}^{\rm{2}}}\)

This allows us to write:

\(\begin{aligned}V(X) &= \frac{{{{\rm{B}}^{\rm{2}}}{\rm{ + AB + }}{{\rm{A}}^{\rm{2}}}}}{{\rm{3}}}{\rm{ - }}{\left( {\frac{{{\rm{A + B}}}}{{\rm{2}}}} \right)^{\rm{2}}}\\ &= \frac{{{{\rm{A}}^{\rm{2}}}{\rm{ + AB + }}{{\rm{B}}^{\rm{2}}}}}{{\rm{3}}}{\rm{ - }}\left( {\frac{{{{\rm{A}}^{\rm{2}}}{\rm{ + 2AB + }}{{\rm{B}}^{\rm{2}}}}}{{\rm{4}}}} \right)\\&= \frac{{{\rm{4}}\left( {{{\rm{A}}^{\rm{2}}}{\rm{ + AB + }}{{\rm{B}}^{\rm{2}}}} \right){\rm{ - 3}}\left( {{{\rm{A}}^{\rm{2}}}{\rm{ + 2AB + }}{{\rm{B}}^{\rm{2}}}} \right)}}{{{\rm{12}}}}\\&= \frac{{{{\rm{B}}^{\rm{2}}}{\rm{ - 2AB + }}{{\rm{A}}^{\rm{2}}}}}{{{\rm{12}}}}\\V(X) &= \frac{{{{{\rm{(B - A)}}}^{\rm{2}}}}}{{{\rm{12}}}}\end{aligned}\)

Standard deviation (\({{\rm{\sigma }}_{\rm{X}}}\)) can now be represented as follows:

\(\begin{aligned}{{\rm{\sigma }}_{\rm{X}}} &= \sqrt {{\rm{V(X)}}} \\ &= \frac{{{{{\rm{(A - B)}}}^{\rm{2}}}}}{{{\rm{12}}}}\\{{\rm{\sigma }}_{\rm{X}}} &= \frac{{{\rm{(B - A)}}}}{{\sqrt {{\rm{12}}} }}\end{aligned}\)

Therefore, the values are \(\frac{{{\rm{(A + B)}}}}{{\rm{2}}}\) and \(\frac{{{\rm{(B - A)}}}}{{\sqrt {{\rm{12}}} }}\).

(c) We'll use the following definition as a guide:

As,\({\rm{h(X)}}\)is any function of X if X is a continuous\({\rm{rv}}\)with pdf\({\rm{f(x)}}\). Then,

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

We use\({\rm{h(x) = }}{{\rm{x}}^{\rm{n}}}\)to get\({\rm{E}}\left( {{{\rm{x}}^{\rm{n}}}} \right)\).

\(\begin{aligned}{\rm{E}}\left( {{{\rm{X}}^{\rm{n}}}} \right) &= \int_{{\rm{ - }}\infty }^\infty {{{\rm{x}}^{\rm{n}}}} {\rm{ \times f(x) \times dx}}\\&= \int_{\rm{A}}^{\rm{B}} {{{\rm{x}}^{\rm{n}}}} {\rm{ \times }}\frac{{\rm{1}}}{{{\rm{B - A}}}}{\rm{ \times dx}}\\&=\frac{{\rm{1}}}{{{\rm{B - A}}}}\int_{\rm{A}}^{\rm{B}} {{{\rm{x}}^{\rm{n}}}} {\rm{ \times dx}}\\&= \frac{{\rm{1}}}{{{\rm{B - A}}}}\left( {\frac{{{{\rm{x}}^{{\rm{n + 1}}}}}}{{{\rm{(n + 1)}}}}} \right)_{\rm{A}}^{\rm{B}}\\&= \frac{{\rm{1}}}{{{\rm{B - A}}}}\left( {\frac{{{{\rm{B}}^{{\rm{n + 1}}}}}}{{{\rm{(n + 1)}}}}{\rm{ - }}\frac{{{{\rm{A}}^{{\rm{n + 1}}}}}}{{{\rm{(n + 1)}}}}} \right)\\{\rm{E}}\left( {{{\rm{X}}^{\rm{n}}}} \right)&= \frac{{{{\rm{B}}^{{\rm{n + 1}}}}{\rm{ - }}{{\rm{A}}^{{\rm{n + 1}}}}}}{{{\rm{(n + 1)(B - A)}}}}\end{aligned}\)

Therefore, the value is \({\rm{E}}\left( {{{\rm{X}}^{\rm{n}}}} \right){\rm{ = }}\frac{{{{\rm{B}}^{{\rm{n + 1}}}}{\rm{ - }}{{\rm{A}}^{{\rm{n + 1}}}}}}{{{\rm{(n + 1)(B - A)}}}}\).