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
If you want to look at the truncated Pareto-distribution on an interval $(a,b), b>a>0$ (which means you cut off the part before $a$ and after $b$), you forgot to divide through the probability that $X$ actually falls into the intervall. Let $F(x;k;\theta)$ be the cdf of the Pareto distribution, then the pdf of the truncated Pareto distribution wrt the interval $(a,b)$ is given through $$\frac{f(x,k,\theta)}{F(b,k,\theta)-F(a,k,\theta)}, \qquad x \in (a,b) $$ Moreover the corresponding expectation is $$\int_a^bx\frac{f(x,k,\theta)}{F(b,k,\theta)-F(a,k,\theta)}dx $$ For $b,k,\theta$ fixed the latter expression increases as $a$ increases.