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Maxwell edited untitled.tex
almost 8 years ago
Commit id: f16e8d0e3efa6503068951f04c77126806643f00
deletions | additions
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index 67916d7..17defa6 100644
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Пусть: $E(e^{2}_{t})=a; \delta^{2}_{E}=b; \delta^{2}_{N}=c;$ Тогда:
$$
E(e^{2}_{t+1})=(1-\frac{E(e^{2}_{t})+\delta^{2}_{E}}{E(e^{2}_{t})+\delta^{2}_{E}+\delta^{2}_{N}})^{2}\cdot (E(e^{2}_{t})+\delta^{2}_{E})+(\frac{E(e^{2}_{t})+\delta^{2}_{E}}{E(e^{2}_{t})+\delta^{2}_{E}+\delta^{2}_{N}})^{2}\cdot \delta^{2}_{N}=\frac{c^{2}\cdot (E(e^{2}_{t})+\delta^{2}_{E})}{E(e^{2}_{t})+\delta^{2}_{E}+\delta^{2}_{N})^{2}}+\frac{\delta^{2}_{N}\cdot
(E(e^{2}_{t})+\delta^{2}_{E})^{2}}{(a+b+c)^{2}}=\\=\frac{c\cdot (a+b)\cdot (E(e^{2}_{t})+\delta^{2}_{E})^{2}}{($E(e^{2}_{t})+\delta^{2}_{E}+\delta^{2}_{N}=c)^{2}}=\\=\frac{\delta^{2}_{N}=\delta^{2}_{N}\cdot (E(e^{2}_{t})+\delta^{2}_{E})\cdot (c+a+b)}{(a+b+c)^{2}}=\frac{c\cdot (a+b)}{a+b+c}=\frac{\delta^{2}_{N}\cdot (E(e^{2}_{t})+\delta^{2}_{E})}{E(e^{2}_{t})+\delta^{2}_{E}+\delta^{2}_{N}}
$$
Таким образом, получаем:
$$