เราได้ว่า E [ X 0 ] = X 0 = S {\displaystyle E[X_{0}]=X_{0}=S\,}
พิจารณา E [ X n | X n − 1 = c ] {\displaystyle E[X_{n}\ |\ X_{n-1}=c]\,} เราได้ว่า E [ X n | X n − 1 = c ] = 1 2 ⋅ 2 c + 1 2 ⋅ c 4 = 9 8 c {\displaystyle E[X_{n}\ |\ X_{n-1}=c]={\frac {1}{2}}\cdot 2c+{\frac {1}{2}}\cdot {\frac {c}{4}}={\frac {9}{8}}c\,}
ฉะนั้น จาก law of total expectation, E [ X n ] = ∑ c Pr ( X n − 1 = c ) E [ X n | X n − 1 = c ] = ∑ c Pr ( X n − 1 = c ) ⋅ 9 8 c = 9 8 ∑ c c Pr ( X n − 1 = c ) = 9 8 E [ X n − 1 ] {\displaystyle E[X_{n}]=\sum _{c}\Pr(X_{n-1}=c)E[X_{n}\ |\ X_{n-1}=c]=\sum _{c}\Pr(X_{n-1}=c)\cdot {\frac {9}{8}}c={\frac {9}{8}}\sum _{c}c\Pr(X_{n-1}=c)={\frac {9}{8}}E[X_{n-1}]}
![{\displaystyle E[X_{0}]=X_{0}=S\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a4b5d825f33d33ad278a1ba55c367d12a6ac9875)
![{\displaystyle E[X_{n}\ |\ X_{n-1}=c]\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ad46e4c71c2141012cc06915bd9c052deacb7006)
เราได้ว่า ![{\displaystyle E[X_{n}\ |\ X_{n-1}=c]={\frac {1}{2}}\cdot 2c+{\frac {1}{2}}\cdot {\frac {c}{4}}={\frac {9}{8}}c\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4ad1157bcad1271f50101033fd2b0bdd28166a22)
![{\displaystyle E[X_{n}]=\sum _{c}\Pr(X_{n-1}=c)E[X_{n}\ |\ X_{n-1}=c]=\sum _{c}\Pr(X_{n-1}=c)\cdot {\frac {9}{8}}c={\frac {9}{8}}\sum _{c}c\Pr(X_{n-1}=c)={\frac {9}{8}}E[X_{n-1}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a55747530c132a69d01899669553552507acd566)
