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\selectlanguage{russian}Пусть: \selectlanguage{english}$E(e^{2}(t))=a; \delta^{2}_{E}=b; \delta^{2}_{N}=c;$  \selectlanguage{russian}Тогда:  \selectlanguage{english}$$  E(e^{2}(t+1))=(1-\frac{a+b}{a+b+c})^{2}*(a+b)+(\frac{a+b}{a+b+c})^{2}*c=\frac{c^{2}*(a+b)}{(a+b+c)^{2}}+\frac{c*(a+b)^{2}}{(a+b+c)^{2}}=\frac{c*(a+b)*(c+a+b)}{(a+b+c)^{2}}=\frac{c*(a+b)}{a+b+c}=\frac{\delta^{2}_{N}*(E(e^{2}(t))+\delta^{2}_{E})}{E(e^{2}(t))+\delta^{2}_{E}+\delta^{2}_{N}} E(e^{2}(t+1))=(1-\frac{a+b}{a+b+c})^{2}*(a+b)+(\frac{a+b}{a+b+c})^{2}*c=\frac{c^{2}*(a+b)}{(a+b+c)^{2}}+\frac{c*(a+b)^{2}}{(a+b+c)^{2}}=\\=\frac{c*(a+b)*(c+a+b)}{(a+b+c)^{2}}=\frac{c*(a+b)}{a+b+c}=\frac{\delta^{2}_{N}*(E(e^{2}(t))+\delta^{2}_{E})}{E(e^{2}(t))+\delta^{2}_{E}+\delta^{2}_{N}}  $$