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- STA256 Lecture 09

Expectation - Linearity

$k_{1}, k_{2} \in R$ , $g_{1}, g_{2}$ are functions
#tk theorem 36
- $g_{1} (X)$ and $g_{2} (X)$ are function of a random variable $X$
- Suppose expectations exists, then for any constants $k_{1}, k_{2}$ , the expectation $k_{1} g_{1} (X) + k_{2} g_{2} (X)$ exists and is given by
- $E (k_{1} g_{1} (X) + k_{2} g_{2} (X)) = k_{1} E (g_{1} (X)) + k_{2} E (g_{2} (X))$
- Proof:
  - Discrete
    - Let $k_{1}, k_{2} \in R, g_{1}, g_{2}$ are functions, on Discrete Random Variable $X$
    - $E (k_{1} g_{1} (x) + k_{2} g_{2} (x))$
    - We know the Expected Value is $E (x) = \sum x_{i} p (x_{i})$
    - $= \sum ((k_{1} g_{1} (x_{i}) + k_{2} g_{2} (x_{i})) p (x_{i}))$
    - $= \sum ((k_{1} g_{1} (x_{i})) p (x_{i}) + (k_{2} g_{2} (x_{i})) p (x_{i}))$
    - $= \sum (k_{1} g_{1} (x_{i})) p (x_{i}) + \sum (k_{2} g_{2} (x_{i})) p (x_{i})$
    - $= k_{1} \underset{E (g_{1} (X))}{\underset{⏟}{\sum (g_{1} (x_{i})) p (x_{i})}} + k_{2} \underset{E (g_{2} (X))}{\underset{⏟}{\sum (g_{2} (x_{i})) p (x_{i})}}$
    - $= k_{1} E (g_{1} (X)) + k_{2} E (g_{2} (X))$
    - proven
  - Continuous #tk
    - Replace summation with integrals