\[\begin{array}{*{20}{c}} {{F_X}(x;b,c,k) = \frac{1}{{{{\left[ {1 + {{\left( {\frac{b}{x}} \right)}^c}} \right]}^k}}}{\rm{ }}}&{b,c,k,x > 0} \end{array}\]

Let $X \sim Burr(b,c,k)$ have the *pdf* given in the box above. It is well known that the distribution of $Y = \frac{1}{X}$ is the *inverse Burr* distribution (also known as the *SinghMaddala* distribution) for which:\[\begin{array}{*{20}{c}} {{f_Y}(y;b,c,k) = \frac{{c{\kern 1pt} {\kern 1pt} k{{\left( {\frac{y}{b}} \right)}^c}}}{{y{\kern 1pt} {{\left[ {1 + {{\left( {\frac{y}{b}} \right)}^c}} \right]}^{k + 1}}}}}&{b,c,k,y > 0} \end{array}\] \[\begin{array}{*{20}{c}} {{F_Y}(y;b,c,k) = 1 - \frac{1}{{{{\left[ {1 + {{\left( {\frac{y}{b}} \right)}^c}} \right]}^k}}}}&{b,c,k,y > 0} \end{array}\] We now consider the limiting distribution when $c \to \infty$ and $k \to 0$ in such a way that the product $ck$ remains constant, i.e. $ck = \lambda$. Now, \[\begin{array}{l} \mathop {\mathop {\lim }\limits_{(c,k) \to (\infty ,0)} }\limits_{ck = \lambda } \left\{ {{F_Y}(y;b,c,k)} \right\} = 1 - \mathop {\mathop {\lim }\limits_{(c,k) \to (\infty ,0)} }\limits_{ck = \lambda } \frac{1}{{{{\left[ {1 + {{\left( {\frac{y}{b}} \right)}^c}} \right]}^k}}}\\ \\ and\\ \\ \mathop {\mathop {\lim }\limits_{(c,k) \to (\infty ,0)} }\limits_{ck = \lambda } {\left[ {1 + {{\left( {\frac{y}{b}} \right)}^c}} \right]^k} = \mathop {\mathop {\lim }\limits_{(c,k) \to (\infty ,0)} }\limits_{ck = \lambda } \left\{ {{{\left( {\frac{y}{b}} \right)}^{ck}}{{\left[ {1 + {{\left( {\frac{b}{y}} \right)}^c}} \right]}^k}} \right\}\\ \\ and\\ \\ \mathop {\mathop {\lim }\limits_{(c,k) \to (\infty ,0)} }\limits_{ck = \lambda } \left\{ {{{\left( {\frac{y}{b}} \right)}^{ck}}{{\left[ {1 + {{\left( {\frac{b}{y}} \right)}^c}} \right]}^k}} \right\} = \mathop {\mathop {\lim }\limits_{(c,k) \to (\infty ,0)} }\limits_{ck = \lambda } \left\{ {{{\left( {\frac{y}{b}} \right)}^{ck}}} \right\}\mathop {\mathop {\lim }\limits_{(c,k) \to (\infty ,0)} }\limits_{ck = \lambda } \left\{ {{{\left[ {1 + {{\left( {\frac{b}{y}} \right)}^c}} \right]}^k}} \right\}\\ = \mathop {\mathop {\lim }\limits_{(c,k) \to (\infty ,0)} }\limits_{ck = \lambda } \left\{ {{{\left( {\frac{y}{b}} \right)}^{ck}}} \right\}\; \cdot \,1\\ = {\left( {\frac{y}{b}} \right)^\lambda } \end{array}\] Therefore, \[\begin{array}{*{20}{c}} {\mathop {\mathop {\lim }\limits_{(c,k) \to (\infty ,0)} }\limits_{ck = \lambda } \left\{ {{F_Y}(y;b,c,k)} \right\} = 1 - {{\left( {\frac{b}{y}} \right)}^\lambda }}&{y \ge b} \end{array}\] which we recognise as the (American) Pareto distribution. So, if the limiting distribution of $Y = \frac{1}{X}$ is a Pareto distribution, then the limiting distribution of $X = \frac{1}{Y}$ is the (American) *inverse Pareto* distribution: \[\begin{array}{l} {f_X}\left( {x;\alpha ,\beta } \right) = \lambda {b^\lambda }{x^{\lambda - 1}};{\rm{ }}0 \le x \le {\textstyle{1 \over b}};{\rm{ }}\lambda {\rm{,}}b > 0\\ {F_X}\left( {x;\alpha ,\beta } \right) = {\left( {xb} \right)^\lambda };{\rm{ }}0 \le x \le {\textstyle{1 \over b}};{\rm{ }}\lambda {\rm{,}}b > 0 \end{array}\] For completeness, the MLEs of this distribution have closed-form expressions and are given by: \[\begin{array}{l} \hat \lambda = {\left[ {\ln \left( {\frac{{{g_X}}}{{\hat b}}} \right)} \right]^{ - 1}}\\ \hat b = \frac{1}{{\max \left\{ {{X_i}} \right\}}}{\rm{ }} \end{array}\] and ${\rm{ }}{g_X}$is the *geometric mean* of the data. ### Inverse Weibull distribution

Let $X \sim Burr(b,c,k)$ have the *pdf* given in the box above. We make the transformation \[Y = \frac{{b{\kern 1pt} {k^{\tfrac{1}{c}}}{\kern 1pt} \theta }}{X}\] where $\theta$ is a parameter (constant). The distribution of $Y$ is also a Burr distribution and has *cdf* \[{G_Y}\left( y \right) = 1 - \frac{1}{{{{\left[ {1 + {{\left( {\frac{y}{{{k^{\tfrac{1}{c}}}{\kern 1pt} \theta }}} \right)}^c}} \right]}^k}}}\].

We are interested in the limiting behaviour of this Burr distribution as $k \to \infty$.

Now,\[\mathop {\lim }\limits_{k \to \infty } {G_Y}\left( y \right) = 1 - \mathop {\lim }\limits_{k \to \infty } {\left[ {1 + {{\left( {\frac{y}{{{k^{\tfrac{1}{c}}}{\kern 1pt} \theta }}} \right)}^c}} \right]^{ - k}}\] \[{ = 1 - \mathop {\lim }\limits_{k \to \infty } {{\left[ {1 + \frac{{{{\left( {\frac{y}{\theta }} \right)}^c}}}{{k{\kern 1pt} }}} \right]}^{ - k}}}\] \[\begin{matrix} =1-\exp \left[ -{{\left( \frac{y}{\theta } \right)}^{c}} \right] \\ \left\{ \text{using the fact that }\underset{n\to \infty }{\mathop{\lim }}\,{{\left( 1+{}^{z}\!\!\diagup\!\!{}_{n}\; \right)}^{-n}}={{e}^{-z}} \right\} \\ \end{matrix}\] We recognise the last expression as the *cdf* of a Weibull distribution with parameters $c$ and $\theta$. ## References ```{r, results = "asis", echo = FALSE} cat(ssdtools::ssd_licensing_md()) ```