Question

Question: Prove the law of total expectations.

Short answer

Answer

It is Proved that\(\sum\limits_{i = I}^n {E\left( {X|{S_i}} \right)} \cdot P\left( {{S_i}} \right)\).

Step-by-step solution

Given Information

The Sample space S is the disjoint union of the events \({S_1},{S_2},............{S_n}\)and X is a random variable.

Definition of a Conditional Probability

The law of total expectation, also known as the law of iterated expectations (or LIE) and the “tower rule”, states that for random variables and , E ( X ) = E { E ( X | Y ) } , provided that the expectations exist.

Proving the law of total expectations

\(\begin{array}{l}P\left( {\mathop \cap \limits_{i = I}^n {X_i} = {x_i}} \right) = \prod\nolimits_{i = I}^n {P\left( {\pi {X_i} = {x_i}} \right)} \\E\left( X \right) = \sum {x \cdot P\left( {X = x} \right)} \\ = \sum\limits_{x \in U_{i = I}^n{S_i}}^n {x \cdot P\left( {X = x} \right)} \\ = \sum\limits_{i = I}^n {\sum\limits_{x \in {S_i}}^{} {x \cdot P\left( {X = x} \right)} } \\ = \sum\limits_{i = I}^n {\sum\limits_{x \in {S_i}}^{} {x \cdot \left( {P\left( {X = x} \right) \cap \left( {x \in {S_i}} \right)} \right)} } \\ = \sum\limits_{i = I}^n {\sum\limits_{x \in {S_i}}^{} {x \cdot \left( {P\left( {X = x} \right)\left( {x \in {S_i}} \right) \cdot \left( {x \in {S_i}} \right)} \right)} } \\ = \sum\limits_{i = I}^n {\sum\limits_{x \in {S_i}}^{} {x \cdot \left( {P\left( {X = \left( {X|{S_i}} \right)} \right) \cdot P\left( {{S_i}} \right)} \right)} } \\ = \sum\limits_{i = I}^n {E\left( {X|{S_i}} \right)} \cdot P\left( {{S_i}} \right)\end{array}\)

Hence, Proved.