OK, this is what I would like to make sure it's true. Please check my math and let me know if I'm wrong.

Let $\displaystyle X$ be a Pareto r.v., with pdf $\displaystyle {{f}_{X}}(x)=a\theta^ax^{-a-1}$ defined for all $\displaystyle x\geq\theta$. Notice that the pdf of a Pareto is influenced by its domain. Now let's do the shift $\displaystyle Y=X+k$.

The shifted Pareto should have the pdf $\displaystyle {{f}_{Y}}(y)={{f}_{X}}(y-k)=a\theta^a(y-k)^{-a-1}$ defined for all $\displaystyle y\geq\theta+k$. Is $\displaystyle Y$ still Pareto?

If it is, shouldn't its pdf be $\displaystyle {{f}_{Y}}(y)=a(\theta+k)^a(y-k)^{-a-1}$?

Mr F says: Where has this come from? It's integral over $\displaystyle \, [\theta + k, \, + \infty)$ is not equal to 1. It's not a pdf.
If this is correct, what I'm quoting from you above should be incorrect, and here's the counterexample. If the math is wrong, what's the silly mistake? It's driving me nuts

Yep, that's what I mentioned in my original post. However, if the math is OK, then not even this might be true in general. The shifted Pareto should be distorted due to the $\displaystyle (\theta+k)^a$ rescaling, vs. the original $\displaystyle \theta^a$ factor.