91影视

An unknown number, unknown value, or unknown quantity is represented by a variable, which is an alphabet or word. In the context of algebraic expressions or algebra, the variables are particularly useful.

(a) We are given a pdf of x as follows:

\({\rm{f(x;k,\theta ) = }}\left\{ {\begin{array}{*{20}{l}}{\frac{{{\rm{k \times }}{{\rm{\theta }}^{\rm{k}}}}}{{{{\rm{x}}^{{\rm{k + 1}}}}}}}&{{\rm{x}} \ge {\rm{\theta }}}\\{\rm{0}}&{{\rm{x < \theta }}}\end{array}} \right.\)

The anticipated value\({\rm{E(X)}}\)can be calculated as follows:

\(\begin{array}{c}{\rm{E(X) = }}\int_{{\rm{ - }}\infty }^\infty {\rm{x}} {\rm{ \times f(x;k,\theta ) \times dx}}\\{\rm{ = }}\int_{\rm{\theta }}^\infty {\rm{x}} {\rm{ \times }}\frac{{{\rm{k \times }}{{\rm{\theta }}^{\rm{k}}}}}{{{{\rm{x}}^{{\rm{k + 1}}}}}}{\rm{ \times dx}}\\{\rm{ = }}\int_{\rm{\theta }}^\infty {\frac{{{\rm{k \times }}{{\rm{\theta }}^{\rm{k}}}}}{{{{\rm{x}}^{\rm{k}}}{\rm{ \times dx}}}}} {\rm{ (1}}{\rm{.1)}}\\{\rm{ = k \times }}{{\rm{\theta }}^{\rm{k}}}\left( {\frac{{{\rm{ - 1}}}}{{{\rm{(k - 1)}}{{\rm{x}}^{{\rm{k - 1}}}}}}} \right)_{\rm{\theta }}^\infty {\rm{ (since k is greater than 1)}}\\{\rm{ = }}\frac{{{\rm{k \times }}{{\rm{\theta }}^{\rm{k}}}}}{{{\rm{k - 1}}}}\left( {\frac{{\rm{1}}}{{{{\rm{\theta }}^{{\rm{k - 1}}}}}}} \right)\\{\rm{E(X) = }}\frac{{{\rm{k \times \theta }}}}{{{\rm{k - 1}}}}\end{array}\)

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

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

Therefore, the value is \({\rm{E(X) = }}\frac{{{\rm{k \times \theta }}}}{{{\rm{k - 1}}}}\).

(b)If\({\rm{k = 1}}\), our analysis will be the same until we get to equation\({\rm{(1}}{\rm{.1)}}\), as shown in the preceding section. As a result, we pick up where we left off and substitute\({\rm{k = 1}}\), yielding:

\(\begin{array}{c}{\rm{E(X) = }}\int_{\rm{\theta }}^\infty {\frac{{{\rm{k \times }}{{\rm{\theta }}^{\rm{k}}}}}{{{{\rm{x}}^{\rm{k}}}}}} {\rm{ \times dx ( from 1}}{\rm{.1)}}\\{\rm{ = }}\int_{\rm{\theta }}^\infty {\frac{{\rm{\theta }}}{{\rm{x}}}} {\rm{ \times dx (putting k = 1 in above equation)}}\\{\rm{ = \theta (ln x)}}_{\rm{\theta }}^\infty \\{\rm{E(X) = \theta (ln(}}\infty {\rm{) - ln(\theta ))}}\end{array}\)

As a result, in this scenario, \({\rm{E(X)}}\) will be undefined.

(c)To begin, we must calculate\({\rm{E}}\left( {{{\rm{X}}^{\rm{2}}}} \right)\):

\(\begin{array}{c}{\rm{E}}\left( {{{\rm{X}}^{\rm{2}}}} \right){\rm{ = }}\int_{{\rm{ - }}\infty }^\infty {{{\rm{x}}^{\rm{2}}}} {\rm{ \times f(x;k,\theta ) \times dx}}\\{\rm{ = }}\int_{\rm{\theta }}^\infty {{{\rm{x}}^{\rm{2}}}} {\rm{ \times }}\frac{{{\rm{k \times }}{{\rm{\theta }}^{\rm{k}}}}}{{{{\rm{x}}^{{\rm{k + 1}}}}}}{\rm{ \times dx}}\\{\rm{ = }}\int_{\rm{\theta }}^\infty {\frac{{{\rm{k \times }}{{\rm{\theta }}^{\rm{k}}}}}{{{{\rm{x}}^{{\rm{k - 1}}}}}}} {\rm{ \times dx (1}}{\rm{.1)}}\\{\rm{ = k \times }}{{\rm{\theta }}^{\rm{k}}}\left( {\frac{{{\rm{ - 1}}}}{{{\rm{(k - 2)}}{{\rm{x}}^{{\rm{k - 2}}}}}}} \right)_{\rm{\theta }}^\infty {\rm{ (since k is greater than 2)}}\\{\rm{ = }}\frac{{{\rm{k \times }}{{\rm{\theta }}^{\rm{k}}}}}{{{\rm{k - 2}}}}\left( {\frac{{\rm{1}}}{{{{\rm{\theta }}^{{\rm{k - 2}}}}}}} \right)\\{\rm{E}}\left( {{{\rm{X}}^{\rm{2}}}} \right){\rm{ = }}\frac{{{\rm{k \times }}{{\rm{\theta }}^{\rm{2}}}}}{{{\rm{k - 2}}}}\end{array}\)

\({\rm{V(X)}}\)can then be written as:

\(\begin{array}{c}{\rm{V(X) = E}}\left( {{{\rm{X}}^{\rm{2}}}} \right){\rm{ - (E(X)}}{{\rm{)}}^{\rm{2}}}\\{\rm{ = }}\frac{{{\rm{k \times }}{{\rm{\theta }}^{\rm{2}}}}}{{{\rm{k - 2}}}}{\rm{ - }}{\left( {\frac{{{\rm{k \times \theta }}}}{{{\rm{k - 1}}}}} \right)^{\rm{2}}}\\{\rm{ = }}\frac{{\left( {{\rm{k \times }}{{\rm{\theta }}^{\rm{2}}}{{{\rm{(k - 1)}}}^{\rm{2}}}} \right){\rm{ - }}\left( {{{\rm{k}}^{\rm{2}}}{{\rm{\theta }}^{\rm{2}}}{\rm{(k - 2)}}} \right)}}{{{\rm{(k - 2)(k - 1}}{{\rm{)}}^{\rm{2}}}}}\\{\rm{V(X) = }}\frac{{{\rm{k \times }}{{\rm{\theta }}^{\rm{2}}}}}{{{\rm{(k - 2)(k - 1}}{{\rm{)}}^{\rm{2}}}}}\end{array}\)

Therefore, the value is \({\rm{V(X) = }}\frac{{{\rm{k \times }}{{\rm{\theta }}^{\rm{2}}}}}{{{\rm{(k - 2)(k - 1}}{{\rm{)}}^{\rm{2}}}}}\).

(d) Since the integral in the expression \({\rm{E}}\left( {{{\rm{X}}^{\rm{2}}}} \right)\) evaluates to infinity if \({\rm{k = 2}}\), \({\rm{V(X)}}\) will be infinite.

(e)We can deduce from sections (b) and (d) that to assure that\({\rm{E}}\left( {{{\rm{X}}^{\rm{n}}}} \right)\)is finite:

\({\rm{k > n}}\)