Expected value of a Pareto distribution between two values

Question

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:

theta=3
k = 1.1

a= seq(3,10,0.1)

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

plot(a,mm)

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

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

Thank you

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.