During boiling $\mathrm{Mg}{\left({\mathrm{HCO}}_3\right)}_2$ is converted to insoluble $\mathrm{Mg}{\left(\mathrm{OH}\right)}_2$

$\mathrm{Mg}{\left({\mathrm{HCO}}_3\right)}_2\stackrel{\mathrm{Boiling}}{\to }\mathrm{Mg}{\left(\mathrm{OH}\right)}_2\downarrow +2{\mathrm{CO}}_2$

$\mathrm{Ca}{\left({\mathrm{HCO}}_3\right)}_2\stackrel{\mathrm{boiling}}{\to }{\mathrm{CaCO}}_3\downarrow +{\mathrm{H}}_2\mathrm{O}+{\mathrm{CO}}_2$

$\mathrm{Ca}{\left({\mathrm{HCO}}_3\right)}_2$ changes to insoluble ${\mathrm{CaCO}}_3$

Mg(OH)_2​ is less soluble than MgCO_3. Hence, the products formed are CaCO₃ and Mg(OH)₂.​