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$$  Пусть: $E(e^{2}_{t})=a; \delta^{2}_{E}=b; \delta^{2}_{N}=c;$ Тогда:  $$  E(e^{2}_{t+1})=(1-\frac{E(e^{2}_{t})+\delta^{2}_{E}}{E(e^{2}_{t})+\delta^{2}_{E}+\delta^{2}_{N}})^{2}\cdot (E(e^{2}_{t})+\delta^{2}_{E})+(\frac{E(e^{2}_{t})+\delta^{2}_{E}}{E(e^{2}_{t})+\delta^{2}_{E}+\delta^{2}_{N}})^{2}\cdot \delta^{2}_{N}=\frac{(\delta^{2}_{N})^{2}\cdot (E(e^{2}_{t})+\delta^{2}_{E})}{E(e^{2}_{t})+\delta^{2}_{E}+\delta^{2}_{N})^{2}}+\frac{\delta^{2}_{N}\cdot (a+b)^{2}}{(a+b+c)^{2}}=\\=\frac{c\cdot (E(e^{2}_{t})+\delta^{2}_{E})^{2}}{(E(e^{2}_{t})+\delta^{2}_{E}+\delta^{2}_{N})^{2}}=\\=\frac{c\cdot  (a+b)\cdot (c+a+b)}{(a+b+c)^{2}}=\frac{c\cdot (a+b)}{a+b+c}=\frac{\delta^{2}_{N}\cdot (E(e^{2}_{t})+\delta^{2}_{E})}{E(e^{2}_{t})+\delta^{2}_{E}+\delta^{2}_{N}} $$  Таким образом, получаем:  $$