Electronegativity in krypton and xenon?
Why do Krypton and Xenon have high electronegativity?


As you point out, krypton ($\ce{Kr}$) and xenon ($\ce{Xe}$) are members of the Noble gas family.  They are generally unreactive (noble) because all of their occupied orbitals are filled with electrons - they really don't want to gain or lose an electron.  However, in the 1960's researchers found that $\ce{Kr}$ and $\ce{Xe}$ did react with extremely electronegative elements such as fluorine and oxygen to form new molecules.  In the reaction process, $\ce{Kr}$ and $\ce{Xe}$ basically gave up an electron to the very electronegative elements they reacted with.

$$\ce{Xe + 3F2 → XeF6}~~(1)$$
$$\ce{XeF6 + 3H2O → XeO3 + 6HF}~~(2)$$

The most common definition of electronegativity is based on Pauling's work and is given by the equation

$${\chi_{\rm A} - \chi_{\rm B} = ({\rm eV})^{-1/2} \sqrt{E_{\rm d}({\rm AB}) - [E_{\rm d}({\rm AA}) + E_{\rm d}({\rm BB})]/2}}$$

where ${\chi_{\mathrm{A}}}$ and ${\chi_{\mathrm{B}}}$ are the electronegativities of atoms $\ce{A}$ and $\ce{B}$ and ${E_{d}(AB)}$ represents the bond dissociation energy of molecule $\ce{A-B}$.  

Using $\ce{HBr}$ as an example, if we know the bond dissociation energies of $\ce{HBr,~ H2}$ and $\ce{Br2}$, then we can calculate the difference in electronegativity between $\ce{H}$ and $\ce{Br}$.  From this, if we know the electronegativity of hydrogen, then we can determine the electronegativity of bromine.

In essence, for any element that can react and form a molecule, we can calculate its electronegativity.  Since Kr and Xe do react to form molecules, we can use the above methodology to calculate its electronegativity.

Note: In the case of atoms like Kr and Xe that do not form a diatomic species like $\ce{Xe2}$, we need an extra step to find their electronegativity.  Since there is no $\ce{E_{d}(Xe-Xe)}$, we must use Pauling's equation twice, once with reaction (1) and once with reaction (2).  We now have 2 equations and two unknowns (${\chi_{\mathrm{Xe}}}$ and ${E_{d}(Xe-Xe)}$) so we can solve for ${\chi_{\mathrm{Xe}}}$, the electronegativity of $\ce{Xe}$.