I am trying to solve a question which gives me a random variable with the distribution function below

$$ F(x) = 1 - \left(\frac{\mu}{x}\right)^{2n} $$

where $0 < \mu \le x < \infty$

I differentiate this to obtain the PDF

$$ f(x) = 2n \mu^{2n} x^{-2n-1} $$

At this point, I notice that the integral over the PDF does not sum to 1

$$ \int_{-\infty}^\infty f(x) dx = \int_{-\infty}^\infty 2n \mu^{2n} x^{-2n-1} dx = 2n\mu^{2n} \left[ \frac{x^{-2n}}{-2n} \right]_{-\infty}^\infty = \mu^{2n} \left[ x^{-2n} \right]_{-\infty}^\infty = 0 $$

Have I gone wrong somewhere above? This is part of a bigger question but the CDF is stated in the question as above.

  • 18
    $\begingroup$ 1. You are integrating from $-\infty$ to $\infty$, but the range of the variable is $[\mu,\infty)$. 2. You have dropped the $-$ sign in the last step of your integration. 3. Why are you taking the derivative and then integrating the result? You already have the functional form of $F(x)$! You can see by inspection that $F(\mu) = 0$ and $\lim_{x \to \infty}F(x) = 1$. $\endgroup$
    – jbowman
    Commented Jan 8, 2023 at 19:04
  • 2
    $\begingroup$ However, any estimator with zero variance is almost surely a constant, which means its distribution function cannot be this particular $F$! $\endgroup$
    – whuber
    Commented Jan 8, 2023 at 19:36
  • 1
    $\begingroup$ As far as the variance is concerned, the integral of $2x(1-F(x))\mathrm dx$ will compute it, assuming you correctly use the implied fact that $F(x)=0$ for all $x\le \mu.$ $\endgroup$
    – whuber
    Commented Jan 8, 2023 at 21:21
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    $\begingroup$ @Solomon Sometimes there are so many issues raised in a question that comments like jbowman's are effective ways to get to the point and understand how best to respond. $\endgroup$
    – whuber
    Commented Jan 9, 2023 at 14:31
  • 2
    $\begingroup$ @SolomonUcko - I thought about writing a full answer, but my third sub-comment was questioning whether the OP had gone down a pointless path in the approach to the entire problem rather than trying to point out mistakes, which I felt was better placed in a comment. Maybe I should have done both! $\endgroup$
    – jbowman
    Commented Jan 9, 2023 at 16:07

1 Answer 1


As pointed out in comments, the range of integration in your integral does not match the listed support of the random variable (which is $\mu \leqslant x < \infty$). Start by correcting the expression for your density, with explicit statement of the support:

$$f(x) = \begin{cases} \frac{2n}{\mu} \Big( \frac{\mu}{x} \Big)^{2n+1} & & & \text{for } x \geqslant \mu, \\[6pt] 0 & & & \text{otherwise}. \\[6pt] \end{cases}$$

Now correct your integral:

$$\begin{align} \int \limits_{-\infty}^\infty f(x) \ dx &= \int \limits_{-\infty}^\mu 0 \ dx + \int \limits_{\mu}^\infty \frac{2n}{\mu} \Big( \frac{\mu}{x} \Big)^{2n+1} \ dx \\[6pt] &= 2n \mu^{2n} \int \limits_{\mu}^\infty x^{-2n-1} \ dx \\[6pt] &= 2n \mu^{2n} \bigg[ -\frac{x^{-2n}}{2n} \bigg]_{x = \mu}^{x \rightarrow \infty} \\[6pt] &= 2n \mu^{2n} \bigg[ 0 - \Big( -\frac{\mu^{-2n}}{2n} \Big) \bigg] \\[6pt] &= 2n \mu^{2n} \cdot \frac{\mu^{-2n}}{2n} \\[12pt] &= 1. \\[6pt] \end{align}$$

  • $\begingroup$ I miss calculus. $\endgroup$
    – J. Doe.
    Commented Jan 10, 2023 at 14:31

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