จาก E [ X | X 1 i s e v e n ] = E [ X 1 | X 1 i s e v e n ] + E [ X 2 | X 1 i s e v e n ] {\displaystyle E[X|X_{1}iseven]=E[X_{1}|X_{1}iseven]+E[X_{2}|X_{1}iseven]} จาก linearity of expectations
P r ( X 1 i s e v e n ) = P r ( X 1 = 2 ) + P r ( X 1 = 4 ) + P r ( X 1 = 6 ) = 1 6 + 1 6 + 1 6 = 1 2 {\displaystyle Pr(X_{1}iseven)=Pr(X_{1}=2)+Pr(X_{1}=4)+Pr(X_{1}=6)={\frac {1}{6}}+{\frac {1}{6}}+{\frac {1}{6}}={\frac {1}{2}}}
จากโจทย์ X 1 , X 2 {\displaystyle X_{1},X_{2}} เป็นอิสระจากกัน
หาค่า
E [ X 1 | X 1 i s e v e n ] = 2 × 1 3 + 4 × 1 3 + 6 × 1 3 = 4 {\displaystyle E[X_{1}|X_{1}iseven]=2\times {\frac {1}{3}}+4\times {\frac {1}{3}}+6\times {\frac {1}{3}}=4}
E [ X 2 | X 1 i s e v e n ] = 3.5 {\displaystyle E[X_{2}|X_{1}iseven]=3.5}
ดังนั้น E [ X | X 1 i s e v e n ] = E [ X 1 | X 1 i s e v e n ] + E [ X 2 | X 1 i s e v e n ] = 4 + 3.5 = 7.5 {\displaystyle E[X|X_{1}iseven]=E[X_{1}|X_{1}iseven]+E[X_{2}|X_{1}iseven]=4+3.5=7.5}
![{\displaystyle E[X|X_{1}iseven]=E[X_{1}|X_{1}iseven]+E[X_{2}|X_{1}iseven]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f69373679c09e5784f6ce15c5f832642cfb0d7f4)
จาก linearity of expectations
เป็นอิสระจากกัน
![{\displaystyle E[X_{1}|X_{1}iseven]=2\times {\frac {1}{3}}+4\times {\frac {1}{3}}+6\times {\frac {1}{3}}=4}](https://wikimedia.org/api/rest_v1/media/math/render/svg/78f990300ff3f1cec9dc9e61d4da88fe792bfd04)
![{\displaystyle E[X_{2}|X_{1}iseven]=3.5}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b0f3c5aae352b46591b8c0c99d7720e89530c70c)
![{\displaystyle E[X|X_{1}iseven]=E[X_{1}|X_{1}iseven]+E[X_{2}|X_{1}iseven]=4+3.5=7.5}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b6cda5b7ac9a0824e05f26eece4c192d8e67bab5)
