I try to calculate the expected value of a Pareto distribution.

Suppose that we have a Pareto distribution for $x \ge \theta$:

$$f(x;k;\theta)= \frac{k\theta^k}{x^{k+1}} $$

We can calculate $ E[X]$ with the following $$ E[X] = \int_{\theta}^{\infty} \frac{k\theta^k}{x^k} dx = \frac{k\theta}{k-1} $$ provided $k>1$.

Now, what if we want to calculate the average value between two values $a$ and $b$.

We can do

$$ \int_a^b xf(x)dx = \int_a^b x \frac{k\theta^k}{x^{k+1}} dx = \frac{k\theta ^k}{1-k} \left[x^{-k+1}\right]_a^b $$ $$ = \frac{k\theta ^k}{1-k} (b^{-k+1}-a^{-k+1}) $$

Now, if b is infinite, we can fin the expected value. And let choose some values for a, intuitively, if a increases, the average value between a and b (infinite) should increase. Mabybe I am wrong, but then I don't know.

Now let's plot some values for a with r:

k = 1.1

a= seq(3,10,0.1)

mm = - k * theta ^k/(1-k) * a ^(-k+1)


we can see that mm decrease with the value of a.

enter image description here

Could you please help me to understand why, or there may be an error in the formulas.

Thank you


1 Answer 1


Note that $\mathbb P(X \ge \theta)=\int\limits_{\theta}^{\infty} \frac{k\theta^k}{x^{k+1}} \,dx = 1.$

That will not be the case for $\mathbb P(\theta \le a \le X \le b) = \int\limits_{a}^{b} \frac{k\theta^k}{x^{k+1}} \,dx = \theta^k\left(\frac1{a^k}-\frac1{b^k}\right)\not=1$ when either $a > \theta$ or $b < \infty$ and you need to adjust for this in finding the expectation of the truncated distribution. So you want $$\mathbb E[X \mid \theta \le a \le X \le b]=\dfrac{\int\limits_{a}^{b} x\frac{k\theta^k}{x^{k+1}} \,dx}{\int\limits_{a}^{b} \frac{k\theta^k}{x^{k+1}} \,dx} = \frac{k}{k-1}\left(\dfrac{\frac1{a^{k-1}}-\frac1{b^{k-1}}}{\frac1{a^k}-\frac1{b^k}}\right)$$ which you can also write as $\dfrac{k(ab^k - ba^k)}{(k-1)(b^k - a^k)}$ so long as $b$ is finite.

When $b=\infty$ and $k>1$, this gives an expected value of $\dfrac{ka}{k-1}$, as all you are doing is shifting the lower limit of a Pareto distribution, i.e. rescaling. It increases with $a$, as you intuitively thought likely.

  • $\begingroup$ thank you for your answer, I see the fundamental flaw in my reasoning. but in you formula, it seems that you forgot the term with $\theta$ ? $\endgroup$
    – John Smith
    Commented Jun 6, 2023 at 15:51
  • 2
    $\begingroup$ @JohnSmith You get $\theta^k$ in the numerator and denominator so it cancels. This makes intuitive sense as $\theta$ is simply the bottom end of the original distribution while $k$ determines the shape, but in your truncated distribution the bottom end becomes $a$ and so long as $a \ge \theta$ you find the original $\theta$ becomes irrelevant $\endgroup$
    – Henry
    Commented Jun 6, 2023 at 15:59

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