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I have the following P.D.F function:

$$g(x)=4 \cdot 38^{4} x^{-5}, \quad x \geq 38$$

By taking the inverse of the CDF i get:

$$G^{-1}(u)=\left\{\begin{array}{ll} \sqrt[4]{-38/u}, & \text { if } u \text { > } 0 \\ -\sqrt[4]{-38/u}, & \text { if } u \text { < } 0 \end{array}\right.$$

So when I try to code this in R and use values from a uniform distribution to feed them into this inverse function

   u <- runif(1000, -1, 0)

   inverse_function_n2 <- function(u){(-38/u)^(1/4)}

   values_derived_from_inverse_CDF <- inverse_function_n2(u)
   hist(values_derived_from_inverse_CDF)

I get a nice histogram that resembles the function g(x)

curve(4*38^4*x^-5, 38, 100, add = FALSE, col = "orange", lwd=1)

But when I plug the values that derive from the inverse CDF back into g(x) I do not get a uniform distribution.

hist(4*38^4*(values_derived_from_inverse_CDF^-5))

Perhaps there is something wrong with my math here?

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    $\begingroup$ Why did you expect to get a uniform? $\endgroup$
    – innisfree
    May 29, 2020 at 11:37
  • $\begingroup$ I suggest you clean the comments that are not relevant for other readers. $\endgroup$
    – Xi'an
    Jun 5, 2020 at 8:59

1 Answer 1

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The Pareto distribution $\mathcal{Pa}(38,5)$ has density (pdf) $g$ and cumulative function (cdf) $$G(x)=[1-(x/38)^{-4}]\mathbb{I}_{(38,\infty)}(x)$$ Solving $G(x)=u$ thus leads to $1-u=(x/38)^{-4}$ $$G^{-1}(u)=\frac{38}{(1-u)^{1/4}}$$ Since the Uniform distribution is symmetric, simulating the Pareto distribution $\mathcal{Pa}(38,5)$ can thus be done by generating $U\sim\mathcal U(0,1)$ and taking$$X=38U^{-1/4}$$

hist(38/runif(1e4)^(.25),nclass=567)

shows a perfect fit to $g$.

However, as pointed out by @innisfree there is no reason for $g(X)$ to be Uniform. Instead, $G(X)\sim\mathcal U(0,1)$, which is the very argument for using the inverse cdf as a simulation method.

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